Solve 0 = (x – 4)2 – 1 by graphing the related function.
What are the solutions to the equation?

3 and 5 is the answer
AND THAT'S JUST ON PERIOD POOH!

Answer:

[tex]n^{2} - 20n -96 = 0[/tex]

use product and sum method

product = -96

sum = -20

numbers needed = ( -24 , 4)

n - 24 = 0

n + 4 = 0

hence n = 24 and n = -4

[tex]x^{2} + 12 x = 48[/tex]

in the form [tex]ax^{2} +bx +c = 0[/tex]

= [tex]x^{2} +12x - 48[/tex]

make use of the formula :

[tex]\frac{-b+-\sqrt{b^{2} -4ac} }{2a}[/tex]

replace values to make 2 equations :

1.[tex]\frac{-12+\sqrt{12^{2} -4*1*-48} }{2*1}[/tex] = 3.17

2.[tex]\frac{-12-\sqrt{12^{2} -4*1*-48} }{2*1}[/tex] = -15.2

hence x = 3.17 and x = -15.2

[tex]x^{2} -14x+40=0[/tex]

use product and sum method

product = 40

sum = -14

numbers needed = (-10 , -4)

x - 10 = 0

x - 4 = 0

hence x = 10 and x = 4

[tex]5b^{2} -20b-18 = 7[/tex]

in the form [tex]ax^{2} +bx +c = 0[/tex]

this becomes [tex]5b^{2} -20b-18-7[/tex]

= [tex]5b^{2} -20b-25[/tex]

can simplify by 5

= [tex]b^{2} -4b-5 =0\\[/tex]

use product and sum method

product = -5

sum = -4

numbers needed (-5 , 1)

b-5 = 0

b + 1 = 0

hence b = 5 and b = -1

Answer:

Step-by-step explanation:

1) n² - 20n - 96 = 0

n² - 20n + (- 20/2)² = 96 + (- 20/2)²

(n - 10)² = 96 + 100

(n - 10)² = 196

Taking square root of both sides

n - 10 = √196 = 14

n = 14 + 10

n = 24

2) x² + 12x = 48

x² + 12x + (12/2)² = 48 + (12/2)²

(x + 6)² = 48 + 36 = 84

Taking square root of both sides,

x + 6 = 9.2

x = 9.2 - 6

x = 3.2

3) x² - 14x + 40 = 0

x² - 14x = - 40

x² - 14x + (- 14/2)² = - 40 + (- 14/2)²

(x - 7)² = - 40 + 49 = 9

Taking square root of both sides,

x - 7 = 3

x = 3 + 7

x = 10

4) 5b² - 20b - 18 = 7

5b² - 20b = 7 + 18

5b² - 20b = 25

Dividing both sides by 5, it becomes

b² - 4b = 5

b² - 4b + (-4/2)² = 5 + (-4/2)²

(b - 2)² = 5 + 4 = 9

Taking square root of both sides

b - 2 = 3

b = 3 + 2

b = 5

Answer:

0.8 is the probability that a single detection system will detect a missile attack.

Step-by-step explanation:

We are given the following information:

We treat detection a missile attack as a success.

P(detecting a missile attack) = 80% = 0.8

Then the number of missile attack follows a binomial distribution, where

[tex]P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}[/tex]

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 1

We have to evaluate:

[tex]P(x = 1)\\= \binom{1}{1}(0.8)^1(1-0.8)^0\\= 0.8[/tex]

0.8 is the probability that a single detection system will detect a missile attack.

Answer:

(a) Probability that a single detection system will detect an attack is 0.80

Step-by-step explanation:

We are given that a reliability question is whether a detection system will be able to identify an attack and issue a warning. Assuming that a particular detection system has a 0.80 probability of detecting a missile attack.

The above situation can be represented through Binomial distribution;

[tex]P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....[/tex]

where, n = number of trials(samples) taken = 1 detection system

            r = number of success

            p = probability of success which in our question is probability of

                   detecting a missile attack, i.e., 80%

LET X = a particular detection system

Also, it is given that a single detection system is taken,

So, it means X ~ [tex]Binom(n=1,p= 0.80)[/tex]

Now, Probability that a single detection system will detect an attack is given by = P(X = 1)  

  P(X = 1) = [tex]\binom{1}{1}0.8^{1} (1-0.8)^{1-1}[/tex]

               = [tex]1 \times 0.8 \times 1[/tex] = 0.80 .

Using the empirical rule for a bell-shaped distribution, which states that nearly all data for a normal distribution falls within three standard deviations of the mean, approximately 95% of undergraduate students at this university have grade point averages between 1.76 and 3.28.

The student is asking for the percentage of GPA between 1.76 and 3.28 using the Empirical Rule which applies to a Bell-shaped distribution or normal distribution. The empirical rule states that for a normal distribution, nearly all of the data will fall within three standard deviations of the mean.

The mean in this case is 2.52 and the standard deviation is 0.38. Therefore the values 1.76 and 3.28 fall within the mean minus two standard deviations and mean plus two standard deviations respectively. Therefore, by the empirical rule, these values represent approximately 95% of the data.

In other words, according to the empirical rule, approximately 95% of undergraduate students at this university have grade point averages between 1.76 and 3.28.

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Answer:

a) [tex]\mu[/tex]

b) [tex]\bar{x}[/tex]

c) (0.152, 0.214)

Step-by-step explanation:

We are given the following in the question:

Sample mean = 0.183 ppm

Standard error = 0.016

a) quantity being estimated

We have to estimate the population mean.

Notation for population mean:

[tex]\mu[/tex]

b) Best estimate for population mean is the sample mean

Notation for sample mean:

[tex]\bar{x}[/tex]

The point estimate for population mean is

[tex]\mu = \bar{x} = 0.183[/tex]

c) 95% confidence interval

[tex]\bar{x}\pm z_{critical}(\text{Standard Error})[/tex]

[tex]z_{critical}\text{ at}~\alpha_{0.05} = 1.96[/tex]

Putting values, we get,

[tex]0.183 \pm 1.96(0.016)\\=0.183\pm 0.03136\\=(0.15164, 0.21436)\\\approx (0.152, 0.214)[/tex]

Thus, the 95% confidence interval is:

(0.152, 0.214)

Interpretation:

After treatment with moose drool, we are 95% certain or 95% confident that the interval (0.152, 0.214) contains the true mean of the population that is the mean level of the toxin ergovaline on the grass.

The quantity being estimated is the mean level of the toxin ergovaline on the grass after treatment with moose drool. The best estimate for this quantity is 0.183 ppm. The 95% confidence interval for this estimate is (0.15164, 0.21436) ppm.

a. The quantity being estimated is the mean level of the toxin ergovaline on the grass after treatment with moose drool. The parameter used is the standard error, which measures the variability of the estimate.

b. The notation for the quantity that gives the best estimate is the mean level of the toxin ergovaline after treatment with moose drool, denoted as μ.

c. To calculate a 95% confidence interval, we need to determine the margin of error. Since the standard error is given as 0.016, the margin of error is 1.96 times the standard error, which is approximately 0.03136. The 95% confidence interval is then calculated by subtracting and adding the margin of error from the mean level of the toxin ergovaline after treatment with moose drool, resulting in an interval of (0.15164, 0.21436). This means that we are 95% confident that the true mean level of the toxin ergovaline after treatment with moose drool is between 0.15164 and 0.21436 ppm.

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Answer:

See step by step explanations to get answer.

Step-by-step explanation:

Given that:

Our linear approximation will be of the form y = c + mx. Use the values of f at the point a and the nearby point a h to find a good approximation. (You will need to set up a linear system involving c, m,a, h, f(a), f(a+ h), where c and m are the variables. You should also be able to solve this linear system exactly for the vector.

Answer:

No, the manager is not correct based on the 95% confidence interval.

Step-by-step explanation:

We are given that the average daily revenue over the past five-week period has been $1,080 with a standard deviation of $260, i.e.; X bar = $1080 and s = $260 and sample size, n = 35 .

The Pivotal quantity for 95% confidence interval is given by;

                [tex]\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]

where, X bar = sample mean = $1080

                s  = sample standard deviation = $260

                 n = sample size = 35 {five-week}

So, 95% confidence interval for average daily revenue, [tex]\mu[/tex] is given by;

P(-2.032 < [tex]t_3_4[/tex] < 2.032) = 0.95

P(-2.032 < [tex]\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.032) = 0.95

P(-2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] < [tex]{Xbar - \mu}[/tex] < 2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] ) = 0.95

P(X bar - 2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] < [tex]\mu[/tex] < X bar + 2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] ) = 0.95

95% confidence interval for [tex]\mu[/tex] = [ X bar - 2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] , X bar + 2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] ]

                                            = [ 1080 - 2.032 * [tex]{\frac{260}{\sqrt{35} }[/tex] , 1080 + 2.032 * [tex]{\frac{260}{\sqrt{35} }[/tex] ]

                                             = [ 990.70 , 1169.30 ]

No, the manager is not correct based on the fact that the coffee and pastry strategy would lead to an average daily revenue of $1,200 because the calculate 95% confidence interval does not include value of $1200.

Therefore, the store manager believe is not correct.

The 95% confidence interval for the store's average daily revenue is calculated to be approximately ($993.97, $1166.03). Since $1200 is outside this interval, the manager's belief that the coffee and pastry strategy will lead to an average daily revenue of $1200 is not backed by this confidence level.

In the field of statistics, a confidence interval (CI) is a type of interval estimate that is used to indicate the reliability of an estimate. The method for calculating a 95% confidence interval for the average daily revenue involves the sample mean, the standard deviation, and the z-score associated with a 95% confidence level, which is approximately 1.96. Let's use the provided data to calculate:

The range of this 95% confidence interval is from $993.97 to $1166.03. This means we are 95% confident that the true average daily revenue lies within this interval. Since $1200 lies outside this interval, the manager's belief is not supported by this confidence interval.

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Answer:

80 feet

Step-by-step explanation:

Given:

Initial speed of the car ([tex]v_0[/tex]) = 40 ft/sec

Deceleration of the car ([tex]\frac{dv}{dt}[/tex]) = -10 ft/sec²

Final speed of the car ([tex]v_x[/tex]) = 0 ft/sec

Let the distance traveled by the car be 'x' at any time 't'. Let 'v' be the velocity at any time 't'.

Now, deceleration means rate of decrease of velocity.

So, [tex]\frac{dv}{dt}=-10\ ft/sec^2[/tex]

Negative sign means the velocity is decreasing with time.

Now, [tex]\frac{dv}{dt}=\frac{dv}{dx}(\frac{dx}{dt})[/tex] using chain rule of differentiation. Therefore,

[tex]\frac{dv}{dx}\cdot\frac{dx}{dt}= -10\\\\But\ \frac{dx}{dt}=v.\ So,\\\\v\frac{dv}{dx}=-10\\\\vdv=-10dx[/tex]

Integrating both sides under the limit 40 to 0 for 'v' and 0 to 'x' for 'x'. This gives,

[tex]\int\limits^0_{40} {v} \, dv=\int\limits^x_0 {-10} \, dx\\\\\left [ \frac{v^2}{2} \right ]_{40}^{0}=-10x\\\\-10x=\frac{0}{2}-\frac{1600}{2}\\\\10x=800\\\\x=\frac{800}{10}=80\ ft[/tex]

Therefore, the car travels a distance of 80 feet before stopping.

Answer:

The total possible number of complete councils that could be selected is 24,174,150.

Step-by-step explanation:

The order of the men and the women is not important. So we use the combinations formula to solve this question.

Combinations formula:

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

What is the total possible number of complete councils that could be selected

3 women from a set of 15

5 men from a set of 25

[tex]T = C_{15,3}*C_{25,5} = \frac{15!}{3!12!}*\frac{25!}{5!20!} = 455*53130 = 24174150[/tex]

The total possible number of complete councils that could be selected is 24,174,150.

Answer:

E (X) = 6.4

Step-by-step explanation:

SOLUTION:

A random variable x = number of cups of coffee consumed on an average day.

∴Let x = 4 represent four or more cups. Round your answers to four decimal places.

X          Probability (X)

0             0.1  

1              0.15

2             0.3

3             0.75

4             0. 25

5             0.21

∴ E (X) = Ux(Mean)

0x.0.1 + 1 x.15 + 2 x 0.3 + 3 x 0.75 + 4 x 0.25 + 5 x 0.21 =  6.4

Answer:

[tex] \bar X = \frac{\sum_{i=1}^n x_i f_i}{N}= \frac{1385.5}{81}= 17.1[/tex]

Now  in order to calculate the variance we can use the following formula:

[tex] s^2 = \frac{\sum fx^2 -[\frac{(\sum fx)^2}{N}]}{N-1}[/tex]

and replacing we got:

[tex] s^2 = \frac{32790.25 -\frac{(1385.5)^2}{80}}{79}= 111.3[/tex]

And the deviation would be the square root of the variance:

[tex] s = \sqrt{111.33} = 10.6[/tex]

Step-by-step explanation:

We assume the following dataset

Age range (years)               1-10        11-20       21-30       >31

Number of individuals        30            18            23          10

Solution to the problem

We can solve the problem creating the following table:

Class      Midpoint(xi)   fi        xi*fi        xi^2 *fi

1-10             5.5            30       165        907.5

11-20           15.5           18       279       4324.5

21-30          25.5          23       586.5   14955.75

>31              35.5          10        355      12602.5

___________________________________

Total                            81      1385.5    32790.25

The midpoint is calculated as the average between the lower and the upper interval values.

We can calculate the mean with the following formula:

[tex] \bar X = \frac{\sum_{i=1}^n x_i f_i}{N}= \frac{1385.5}{81}= 17.1[/tex]

Now  in order to calculate the variance we can use the following formula:

[tex] s^2 = \frac{\sum fx^2 -[\frac{(\sum fx)^2}{N}]}{N-1}[/tex]

and replacing we got:

[tex] s^2 = \frac{32790.25 -\frac{(1385.5)^2}{80}}{79}= 111.3[/tex]

And the deviation would be the square root of the variance:

[tex] s = \sqrt{111.33} = 10.6[/tex]

Answer:

Question is not clear please post question clearly lots of question marks.

Your differential equation is not displayed well. It though looks like this:

2d²y/dx² - 20dy/dx + 136y = 0

If this is not the differential equation, the method of solving this would still be used in solving the correct one.

We first write an auxiliary equation to the differential equation.

The auxiliary equation is:

2m² - 20m + 136 = 0

Dividing by 2, we have

m² - 10m + 68 = 0

Next, we solve the auxiliary equation to obtain the values of m.

Solving using the quadratic formula

m = [-b ± √(b² - 4ac)]/2a

Where a = 1, b = -10, and c = 68

m = [10 ± √(100 - 272)]/2

= 5 ± (1/2)√(-172)

= 5 ± (1/2)i√172

= 5 ± 6.6i

For solutions of the form a ± ib, the complimentary solution is

y = e^(ax)[C1cosbx + C2sinbx]

Therefore, the complimentary solution is

y = e^(5x)[C1cos(6.6x) + C2sin(6.6x)]

Answer:

[tex](60)^{\frac{1}{2}} = \sqrt{60}[/tex]

Step-by-step explanation:

We are given the following expression:

[tex](60)^{\frac{1}{2}}[/tex]

We are given the following options:

[tex]A.~\dfrac{60}{2}\\\\B.~\sqrt{60}\\\\C.~(\dfrac{1}{60})^{2}\\\\D.~\dfrac{1}{\sqrt{60}}[/tex]

Exponent Properties:

  [tex]x^{-a} = (\dfrac{1}{x})^{a}\\\\(x^m)^n = x^{mn}\\\\\dfrac{x^m}{x^n} = x^{m-n}[/tex]

Thus, the correct answer is

[tex](60)^{\frac{1}{2}} = \sqrt{60}[/tex]

Answer:

B or [tex]\sqrt{60}[/tex]

Step-by-step explanation:

Answer:

V₂=1061.85 ft³

Step-by-step explanation:

To determine where more wood is found, just find the volume of each log and see which one has the largest volume

knowing that the volume of a straight cylinder is

V=πR²h, where R=radius and h=height

[tex]V_{1}=\pi (2.2)^{2}60= 912.31ft^{3}\\V_{2}=\pi (2.6)^{2}50= 1061.85ft^{3}[/tex]

As we can observe

V₂>V₁

Answer:

Let x is the number of the cannon and y is the number of the row boat.Thus the equation will become

8x+5y=110

Answer: 8C+5R≤110

Step-by-step explanation: Let hours of fabrication of canoes = C

Hours of fabrication of rowboat = R

If a canoe requires 8 hours of fabrication. And a row boat requires 5 hours of fabrication. Then

8C + 5R

The fabrication department has at most 110 hours to labor each week. 

Therefore

8C+5R≤110

At most means less than or equal to.

There are [tex]\binom{52}3=\frac{52!}{3!(52-3)!}[/tex] (or "52 choose 3") ways of drawing any 3 cards from the deck.

There are 13 hearts in the deck, and 26 cards with a black suit. So there are [tex]\binom{13}3[/tex] and [tex]\binom{26}3[/tex] ways of drawing 3 hearts or 3 black cards, respectively. Then the probability of drawing 3 hearts is

[tex]P(\text{3 hearts})=\dfrac{\binom{13}3}{\binom{52}3}=\dfrac{11}{850}[/tex]

and the probability of drawing 3 black cards is

[tex]P(\text{3 black})=\dfrac{\binom{26}3}{\binom{52}3}=\dfrac2{17}[/tex]

All other combinations can be drawn with probability [tex]1-\frac{11}{850}-\frac2{17}=\frac{739}{850}[/tex].

Let [tex]W[/tex] be the random variable for one's potential winnings from playing the game. Then

[tex]P(W=w)=\begin{cases}\frac{11}{850}&\text{for }w=\$50\\\frac2{17}&\text{for }w=\$25\\\frac{739}{850}&\text{otherwise}\end{cases}[/tex]

a. For a single game, one can expect to win

[tex]E[W]=\displaystyle\sum_ww\,P(W=w)=\frac{\$50\cdot11}{850}+\frac{\$20\cdot2}{17}+\frac{\$0\cdot739}{850}=\$3[/tex]

b. For a single game, one's winnings have a variance of

[tex]V[W]=E[(W-E[W])^2]=E[W^2]-E[W]^2[/tex]

where

[tex]E[W^2]=\displaystyle\sum_ww^2\,P(W=w)=\frac{\$50^2\cdot11}{850}+\frac{\$20^2\cdot2}{17}+\frac{\$0^2\cdot739}{850}=\$^2\frac{1350}{17}\approx\$^279.41[/tex]

so that [tex]V[W]=\$^2\frac{1197}{17}\approx\$^270.41[/tex]. (No, that's not a typo, variance is measured in squared units.) Standard deviation is equal to the square root of the variance, so it is approximately $8.39.

c. With a $5 buy-in, the expected value of the game would be

[tex]E[W-\$5]=E[W]-\$5=-\$2[/tex]

i.e. a player can expect to lose $2 by playing the game (on average).

d. With the $5 cost, the variance of the winnings is the same, since the variance of a constant is 0:

[tex]V[W-\$5]=V[W][/tex]

so the standard deviation is the same, roughly $8.39.

e. You shouldn't play this game because of the negative expected winnings. The odds are not in your favor.

The expected winning for a single game defined is : $3.59

The standard deviation of winning is : $10.11

Expected winning if game costs $5 to play is :

- $0.79

The standard deviation of winning if game costs $5 to play is : $11.33

The game should not be played with a game play fee of -$5 as the expected winning value is negative.

Recall : selection is done without replacement :

Number of hearts in a deck = 13

Probability of drawing 3 hearts :

P(drawing 3 Hearts)

First draw × second draw × third draw

13/52 × 12/51 × 11/50 = 1716/132600 = 858/66300

Probability of selecting 3 black cards :

Number of black cards in a deck = 26

P(drawing 3 black cards) :

First draw × second draw × third draw

26/52 × 25/51 × 24/50 = 15600 / 132600 = 7800/66300

Probability of making any other draw :

P(3 hearts) + P(3 blacks) + P(any other draw) = 1

858/66300 + 7800/66300 + P(any other draw) = 1

P(any other draw) = 57642/66300

For a single game :

X _______ $50 ________ $25 _______ $0

P(X)_ 858/66300__ 7800/66300_57642/66300

E(X) = Σ[ X × p(X)]

E(X) =Σ[(50 × 858/66300)+(25 × 7800/66300)+0]

E(X) = $3.588

The standard deviation = √Var(X)

Var(X) = Σ[ X² × p(X)] - E(X)

Σ[ X²×p(X)] = Σ[(50² × 858/66300)+(25² × 7800/66300)+0] = 105.88235

Var(X) = 105.88235 - 3.588 = 102.29435

Standard deviation of winning = √102.29435 = $10.114

If the game cost $5 to play :

Net amount won if :

3 hearts are drawn = $50 - $5 = $45

3 blacks are drawn = $25 - $5 = $20

Any other combination are drawn= $0 - $5 = -$5

The distribution becomes :

X _______ $45 ________ $20 _______ -$5

P(X)_ 858/66300__ 7800/66300_57642/66300

E(X) = Σ[ X × p(X)]

E(X) =Σ[(50 × 858/66300)+(25 × 7800/66300)+ (-5 × 57642/66300)]

E(X) = - $0.7588

Standard deviation of winning :

The standard deviation = √Var(X)

Var(X) = Σ[ X² × p(X)] - E(X)

Σ[ X²×p(X)] = Σ[(50² × 858/66300)+(25² × 7800/66300)+ (-5² × 57642/66300)] = 127.61764

Var(X) = 127.61764 - (-0.7588) = 128.37644

Standard deviation of winning :

Std(X) = √Var(X) = √128.37644 = $11.330

With a game cost of - $5 ; the expected winning for a single game gives a negative value, therefore you should not play the game.

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Answer:

(a) the dependent variable here are the Participants.

(b) the dependent variable is discrete.

(c) The scale measurement of measurement used is interval/ratio to measure the dependent variable.

(d) the independent variable is Vitamin C

(e) The research method being used is experimental.

Step-by-step explanation:

The dependent variable (sometimes known as the responding variable) is what is being studied and measured in the experiment.

Examples of continuous dependent variable may include costs, profits and sales.But some dependent variables are discrete – that is, they take on a relatively small number of integer values.

Interval scale and ratio scale are the two variable measurement scales where they define the attributes of the variables quantitatively.A ratio scale is a measurement scale which has more or less all the properties of an interval scale. Ratio data on this scale has measurable intervals.

Experimental research is a study that strictly adheres to a scientific research design. It includes a hypothesis, a variable that can be manipulated by the researcher, and variables that can be measured, calculated and compared.

Answer: 1. The dependent variable is the common cold.

2. It is a discrete variable.

3. The interval/ratio scale.

4. The independent value is the large doses of Vitamin C administered.

5. The research method used is Experimental.

Step-by-step explanation:

1. The dependent variable is the factor that we are trying to understand. In this case, it is the common cold and how it is affected by large doses of vitamins.

2. A discrete value is computed by counting. So the dependent variable - the common cold was computed by counting the number of occurrences.

3. The Interval/ratio scale is used because both the order of measurement and the differences between them are observed.

4. The Independent variable is the control in the experiment which can be compared to changes in the dependent variable. The large doses of vitamin C affects the common cold.

5. Correlational research observes patterns in variables that occur naturally while Experimental research introduces a change and monitors its effect. The research is Experimental because a change in the form of the placebo is introduced and observed.

Answer:

The scores that are less than or equal to 10.8 are considered significantly low.

The scores that are greater than or equal to 32.4 are considered significantly high.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 21.6

Standard Deviation, σ = 5.4

We are given that the distribution of score is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

Significantly low score:

[tex]z \leq -2\\z = \displaystyle\frac{x-21.6}{5.4} \leq -2\\\\\displaystyle\frac{x-21.6}{5.4} \leq -2\\\\x\leq -2(5.4) + 21.6\\\Rightarrow x \leq 10.8[/tex]

Thus, scores that are less than or equal to 10.8 are considered significantly low.

Significantly high score:

[tex]z \geq 2\\z = \displaystyle\frac{x-21.6}{5.4} \geq 2\\\\\displaystyle\frac{x-21.6}{5.4} \geq 2\\\\x\geq 2(5.4) + 21.6\\\Rightarrow x \geq 32.4[/tex]

Thus, scores that are greater than or equal to 32.4 are considered significantly high.

Answer:

The volume of a pyramid is 1/3 Sh, where S is the area of the base and h is the height.  Since the area of base (Square) is , S = [tex]s^{2}[/tex], where s is the side of the base.  So the volume is

V = 1/3 [tex]s^{2}[/tex]h.

Further solution is on paper (Pictures attached)

Answer:

(a) 0.1326

(b) 0.2732

(c) 0.0410

Step-by-step explanation:

Let X = number of defective components.

The probability of X is, P (X) = p = 0.02.

The random variable X follows a Binomial distribution with parameters n and p. The probability mass function of a Binomial distribution is:

[tex]P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3,...[/tex]

(a)

Compute the probability that the 100 orders can be filled without reordering components as follows:

n = 100

[tex]P(X=0)={100\choose 0}0.02^{0}(1-0.02)^{100-0}=1\times1\times0.13262=0.1326[/tex]

Thus, the probability that the 100 orders can be filled without reordering components is 0.1326.

(b)

Compute the probability that out of 102 orders 2 orders needs reordering as follows:

n = 102

[tex]P(X=2)={102\choose 2}0.02^{2}(1-0.02)^{102-2}=5151\times0.0004\times0.13262=0.2732[/tex]

Thus, the probability that out of 102 orders 2 orders needs reordering is 0.2732.

(c)

Compute the probability that out of 105 orders 2 orders needs reordering as follows:

n = 105

[tex]P(X=5)={105\choose 5}0.02^{5}(1-0.02)^{105-5}=96560646\times0.0000000032\times0.13262=0.0410[/tex]

Thus, the probability that out of 105 orders 5 orders needs reordering is 0.0410.

Answer:

Step-by-step explanation:

Hello!

The parameter of interest in this exercise is the population proportion of Asians that would welcome a person of other races in their family. Using the race of the welcomed one as categorizer we can define 3 variables:

X₁: Number of Asians that would welcome a white person into their families.

X₂: Number of Asians that would welcome a Latino person into their families.

X₃: Number of Asians that would welcome a black person into their families.

Now since we are working with the population that identifies as "Asians" the sample size will be: n= 251

Since the sample size is large enough (n≥30) you can apply the Central Limit Theorem and approximate the variable distribution to normal.

[tex]Z_{1-\alpha /2}= Z_{0.975}= 1.965[/tex]

1. 95% CI for Asians that would welcome a white person.

If 79% would welcome a white person, then the expected value is:

E(X)= n*p= 251*0.79= 198.29

And the Standard deviation is:

V(X)= n*p*(1-p)= 251*0.79*0.21=41.6409

√V(X)= 6.45

You can construct the interval as:

E(X)±Z₁₋α/₂*√V(X)

198.29±1.965*6.45

[185.62;210.96]

With a 95% confidence level, you'd expect that the interval [185.62; 210.96] contains the number of Asian people that would welcome a White person in their family.

2. 95% CI for Asians that would welcome a Latino person.

If 71% would welcome a Latino person, then the expected value is:

E(X)= n*p= 251*0.71= 178.21

And the Standard deviation is:

V(X)= n*p*(1-p)= 251*0.71*0.29= 51.6809

√V(X)= 7.19

You can construct the interval as:

E(X)±Z₁₋α/₂*√V(X)

178.21±1.965*7.19

[164.08; 192.34]

With a 95% confidence level, you'd expect that the interval [164.08; 192.34] contains the number of Asian people that would welcome a Latino person in their family.

3. 95% CI for Asians that would welcome a Black person.

If 66% would welcome a Black person, then the expected value is:

E(X)= n*p= 251*0.66= 165.66

And the Standard deviation is:

V(X)= n*p*(1-p)= 251*0.66*0.34= 56.3244

√V(X)= 7.50

You can construct the interval as:

E(X)±Z₁₋α/₂*√V(X)

165.66±1.965*7.50

[150.92; 180.40]

With a 95% confidence level, you'd expect that the interval [150.92; 180.40] contains the number of Asian people that would welcome a Black person in their family.

I hope it helps!

Answer:

E. Change in the central tendency of the process output.

Step-by-step explanation:

The x-bar chart is a control chart for the central tendency of the process output. Therefore it's purpose is to determine whether there has been a change in the central tendency of the process output.

Answer:

The change in the central tendency of the process output.

Step-by-step explanation:

Lets examine each of the listed options

The change in the percent defective in a sample is associated with the change in average outgoing quality (AOQ).

The change in the dispersion of the process output is associated with R-chart since the function of R-chart is to detect changes in the dispersion.

The change in the number of defects in a sample is associated with C-chart since the function of a C-chart is to show the number of flaws per unit in a sample.

The change in the central tendency of the process output is associated with X-bar chart since the function of X-bar chart is to check whether the values are within the appropriate limits (central tendency) of the process output or not.

Therefore, the purpose of an X-bar chart is to determine whether there has been a change in the central tendency of the process output.

Answer:

Part a: The probability is 0.3436

Part b: The probability is 0.5381

Part c: The probability is 0.7600

Step-by-step explanation:

As per the given data

[tex]p_{T | SP} = e^{0.1}\\p_{T | Good} = e^{0.3}\\[/tex]

(a)

Probability that train is delayed by more than 1 minute = P(T > 1) = P(SP) * P(T > 1 | SP) + P(Good) * P(T > 1 | Good)

[tex]= 0.3 e^{-.1 \times 1 }+ 0.7 * e^{-.3 \times 1}\\ = 0.3 e^{-.1} + 0.7 e^{-.3}[/tex]

Probability that after 1 minute of waiting, probability that train has signal problems = P(SP | T > 1)

= P(T > 1 | SP) * P(SP) / P(T > 1) (By Bayes theorem)

[tex]= \dfrac{0.3 e^{-.1 }}{ 0.3 e^{-.1 }+ 0.7 e^{-.3 }}\\\\= \dfrac{0.2714512}{0.790024}\\\\= 0.3435987[/tex]

(b)

Probability that train is delayed by more than 5 minutes = P(T > 5) = P(SP) * P(T > 5 | SP) + P(Good) * P(T > 5 | Good)

[tex]= 0.3 e^{-.1 \times 5 }+ 0.7 * e^{-.3 \times 5}\\ = 0.3 e^{-.5} + 0.7 e^{-1.5}[/tex]

Probability that after 5 minute of waiting, probability that train has signal problems = P(SP | T > 5)

= P(T > 5 | SP) * P(SP) / P(T > 5) (By Bayes theorem)

[tex]= \dfrac{0.3 e^{-.5 }}{ 0.3 e^{-.5 }+ 0.7 e^{-1.5 }}\\\\= \dfrac{0.1819592}{0.3381503}\\\\= 0.5381015[/tex]

(c)

Probability that train is delayed by more than 10 minutes = P(T > 10) = P(SP) * P(T > 10 | SP) + P(Good) * P(T > 10 | Good)

[tex]= 0.3 e^{-.1 \times 10 }+ 0.7 * e^{-.3 \times 10}\\ = 0.3 e^{-1.0} + 0.7 e^{-3.0}[/tex]

Probability that after 5 minute of waiting, probability that train has signal problems = P(SP | T > 10)

= P(T > 10 | SP) * P(SP) / P(T > 10) (By Bayes theorem)

[tex]= \dfrac{0.3 e^{-1.0 }}{ 0.3 e^{-1.0 }+ 0.7 e^{-3.0 }}\\\\= \dfrac{0.1103638}{0.1452148}\\\\= 0.7600038[/tex]

(a) The probabilities recalculated based on the waiting time are: 21.4% after 1 minute,

(b) 54.0% after 5 minutes, and

(c) 75.9% after 10 minutes.

These calculations use Bayes' theorem and the exponential distribution functions provided.

The question pertains to conditional probability and involves exponential distributions.

Let’s solve this step-by-step:

Part (a)

Given:
pT|SP(t) = [tex]0.1e^(-0.1t)[/tex]
pT|Good(t) = [tex]0.3e^(-0.3t)[/tex]
P(Signal Problems) = [tex]0.30[/tex]
P(Good Service) = [tex]0.70[/tex]

We need to find P(Signal Problems | T ≥ 1). Using Bayes’ theorem:

Using the law of total probability:

P(T ≥ 1) = P(T ≥ 1 | Signal Problems)P(Signal Problems) + P(T ≥ 1 | Good Service)P(Good Service)
= [tex]e^(-0.1)[/tex] [tex]* 0.30 +[/tex] [tex]e^(-0.3)[/tex] [tex]* 0.70[/tex]

Now applying Bayes’ theorem:
P(Signal Problems | T ≥ 1) = [tex][[/tex][tex]e^(-0.1)[/tex] [tex]* 0.30] / [[/tex][tex]e^(-0.1)[/tex] [tex]* 0.30 +[/tex] [tex]e^(-0.3)[/tex] [tex]* 0.70][/tex]
[tex]\approx [0.9048 * 0.30] / [0.9048 * 0.30 + 0.7408 * 0.70][/tex]
[tex]\approx 0.214[/tex]

Part (b)

Similarly for T ≥ 5, the probabilities become:

Using the law of total probability:
P(T ≥ 5) = [tex]e^(-0.5)[/tex] [tex]* 0.30 +[/tex] [tex]e^(-1.5)[/tex][tex]* 0.70[/tex]

Applying Bayes’ theorem:
P(Signal Problems | T ≥ 5) = [tex][[/tex][tex]e^(-0.5)[/tex] [tex]* 0.30] / [[/tex][tex]e^(-0.5)[/tex] [tex]* 0.30 +[/tex] [tex]e^(-1.5)[/tex] [tex]* 0.70][/tex]
≈ [tex][0.6065 * 0.30] / [0.6065 * 0.30 + 0.2231 * 0.70][/tex]
≈ [tex]0.540[/tex]

Part (c)

For T ≥ 10:

Using the law of total probability:
P(T ≥ 10) = [tex]e^{-1} \cdot 0.30 + e^{-3} \cdot 0.70[/tex]

Applying Bayes’ theorem:
P(Signal Problems | T ≥ 10) = [tex]\frac{e^{-1} \cdot 0.30}{e^{-1} \cdot 0.30 + e^{-3} \cdot 0.70}[/tex]
≈ [tex][0.3679 * 0.30] / [0.3679 * 0.30 + 0.0498 * 0.70][/tex]
≈ [tex]0.759[/tex]

Answer:

[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]  

For this case the value of r = -0.66

Now we can calculate the determination coeffcient:

[tex] r^2 = (-0.66)^2 = 0.4356[/tex]

And then we can conclude that 43.56% of the variation in y can be explained by the explanatory variable

And then 100-43.56 = 56.44 % of the variation in y that cannot be explained by the explanatory variable

Step-by-step explanation:

For this case we need to calculate the slope with the following formula:

[tex]m=\frac{S_{xy}}{S_{xx}}[/tex]

Where:

[tex]S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}[/tex]

[tex]S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}[/tex]

[tex]\bar x= \frac{\sum x_i}{n}[/tex]

[tex]\bar y= \frac{\sum y_i}{n}[/tex]

And we can find the intercept using this:

[tex]b=\bar y -m \bar x[/tex]

And the model obtained for this case is:

[tex] y = -3.4 +5.2 x[/tex]

The correlation coefficient is a "statistical measure that calculates the strength of the relationship between the relative movements of two variables". It's denoted by r and its always between -1 and 1.

And in order to calculate the correlation coefficient we can use this formula:  

[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]  

For this case the value of r = -0.66

Now we can calculate the determination coeffcient:

[tex] r^2 = (-0.66)^2 = 0.4356[/tex]

And then we can conclude that 43.56% of the variation in y can be explained by the explanatory variable

And then 100-43.56 = 56.44 % of the variation in y that cannot be explained by the explanatory variable

The Coefficient of determination gives the proportion of variation which can be explained by the regression line.

The proportion of variation that cannot be explained by the given regression line is 56.44%

From the question, the correlation Coefficient, R = - 0.66

This means that :

Therefore, proportion of the variation in y that cannot be explained is 56.44%

Learn more :https://brainly.com/question/18405415

Answer:

The cutoff for an A grade is 82.8.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

[tex]\mu = 70, \sigma = 10[/tex]

a. What is the cutoff for an A grade?

The top 10% of the class get an A grade. So the cutoff is the value of X when Z has a pvalue of 1-0.1 = 0.9. So it is X when Z = 1.28

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]1.28 = \frac{X - 70}{10}[/tex]

[tex]X - 70 = 1.28*10[/tex]

[tex]X = 82.8[/tex]

The cutoff for an A grade is 82.8.

Answer:  

Step-by-step explanation:  

so from the question i have here, i will be giving a step by step analysis of the question.  

(a). Here we are showing a relationship, i.e.  

P(Granite ║ R₁ ∠ R₂ ∠ R₃) ˃ P (Basalt ║  R₁ ∠ R₂ ∠ R₃)  

from the LHS;  

P(Granite ║  R₁ ∠ R₂ ∠ R₃) = P(Granite ║  R₁ ∠ R₂ ∠ R₃) / P(R₁ ∠ R₂ ∠ R₃)  

= P(Granite)P(R₁ ∠ R₂ ∠ R₃ ║  Granite) / [ P(Granite  R₁ ∠ R₂ ∠ R₃) + P(Basaltic  R₁ ∠ R₂ ∠ R₃) ]  

= 0.25 × 0.6 / [(0.25×0.6)+(0.75×0.1)] = 0.667  

from the RHS;  

P (Basalt ║  R₁ ∠ R₂ ∠ R₃) = P(Basalt  R₁ ∠ R₂ ∠ R₃) / P(R₁ ∠ R₂ ∠ R₃)  

= P(Basalt)P(R₁ ∠ R₂ ∠ R₃ ║  Basalt) / [ P(Basalt  R₁ ∠ R₂ ∠ R₃) + P(Granite  R₁ ∠ R₂ ∠ R₃) ]  

= 0.75 × 0.1 / [(0.25 × 0.6)+(0.75 × 0.1)] = 0.333  

Therefore from this we can infer that;  

P(Granite ║  R₁ ∠ R₂ ∠ R₃) ˃ P (Basalt ║  R₁ ∠ R₂ ∠ R₃)  

(b). here we are asked to classify the rocks considering the measurement yield.  

Measurement yielded R1 < R3 < R2  

P(Granite ║  R₁ ∠ R₃ ∠ R₂) = P(Granite  R₁ ∠ R₃ ∠ R₂) / P(R₁ ∠ R₃ ∠ R₂)  

= P(Granite)P(R₁ ∠ R₃ ∠ R₂ ║  Granite) / [ P(Granite  R₁ ∠ R₃ ∠ R₂) + P(Basaltic  R₁ ∠ R₃ ∠ R₂) ]  

= 0.25 × 0.25 / [(0.25 × 0.25)+(0.75 × 0.2)] = 0.294  

also for RHS;  

P (Basalt ║  R₁ ∠ R₃ ∠ R₂) = P(Basalt ║  ∠ R₃ ∠ R₂) / P(R₁ ∠ R₃ ∠ R₂)  

= P(Basalt)P(R₁ ∠ R₃ ∠ R₂ ║  Basalt) / [ P(Basalt  R₁ ∠ R₃∠ R₂) + P(Granite  R₁ ∠ R₃ ∠ R₂) ]  

= 0.75 × 0.2 / [(0.25 × 0.25)+(0.75 × 0.2)] = 0.706  

from this we can infer that;

P(Granite ║  R₁ ∠ R₃ ∠ R₂) ∠ P (Basalt ║  R₁ ∠ R₃ ∠ R₂)  

Also considering measurements yielded R₃ ∠ R₁ ∠ R₂  

P(Granite ║  R₃ ∠ R₁ ∠ R₂) = P(Granite  R₃ ∠ R₁ ∠ R₂) / P(R₃ ∠ R₁ ∠ R₂)  

= P(Granite)P(R₃ ∠ R₁ ∠ R₂ ║  Granite) / [ P(Granite  R₃ ∠ R₁ ∠ R₂) + P(Basaltic  R₃ ∠ R₁ ∠ R₂) ]  

= 0.25 × 0.15 / [(0.25×0.15)+(0.75×0.7)] = 0.067

from the RHS;

P (Basalt ║  R₃ ∠ R₁ ∠ R₂) = P(Basalt  R₃ ∠ R₁ ∠ R₂) / P(R₃ ∠ R₁ ∠ R₂)  

= P(Basalt)P(R₃ ∠ R₁ ∠ R₂ ║  Basalt) / [ P(Basalt  R₃ ∠ R₁ ∠ R₂) + P(Granite  R₃ ∠ R₁ ∠ R₂) ]  

1 - 0.067 = 0.933  

from this we can infer that;

P(Granite ║  R₃ ∠ R₁ ∠ R₂) ∠ P (Basalt ║  R₃ ∠ R₁ ∠ R₂)  

cheers i hope this helps  

Final answer:

Measurements yielding R1 < R2 < R3 suggest a granitic composition, while the pattern R1 < R3 < R2 leans granitic, and R3 < R1 < R2 lean towards basaltic. The rock type can be further supported by the appearance of the rock, with granitic rocks being coarse and light-colored due to feldspar and quartz, while basaltic rocks are fine-grained and dark with ferromagnesian minerals. If granite contains basaltic inclusions, the granite is younger.

Explanation:

In analyzing rock compositions from infrared spectrum measurements, if measurements yield R1 < R2 < R3, based on typical patterns, the rock would more likely be classified as granitic because, for granite, the intensity measurements usually increase in that order. Conversely, when measurements produce R1 < R3 < R2, it does not fit neatly into the categories provided for basalt or granite, but it leans closer to a granitic composition given that R3 is not the least among the three, which is a characteristic of basaltic rocks. Similarly, if R3 presents a lower intensity compared to R1 and R2, and you have to classify without further information, it might suggest a basaltic nature following the typical pattern for basalt (R3 < R1 < R2), although it is important to note that real-world applications may require additional contextual information to accurately determine the rock type.

When studying rock samples, you should also consider differences in mineral size (coarse-textured like granite versus fine-textured like basalt), mineral content (dark, mafic minerals vs light, felsic minerals), and elemental content of the igneous rock types. For instance, granitic rock contains feldspar and quartz, which is reflected in its coarse and light-colored appearance. In contrast, basaltic rock, which is fine-grained and dark-colored, includes ferromagnesian minerals and feldspars. Additionally, if a granitic rock has inclusions of basalt (xenoliths), this implies that the granitic rock would typically be younger than the basalt since the inclusions must have been present already for the granite to incorporate them during its formation process.

Answer: Jeremy drove 84 miles.

Step-by-step explanation:

Let x represent the number of miles that Brenda drove.

If Jeremy drove twice

as far as Brenda, it means that the distance covered by Jeremy would be 2x miles

When they stopped after some time, they were already

126 miles apart. This means that the total distance covered by both of them is 126 miles. Therefore,

x + 2x = 126

3x = 126

x = 126/3

x = 42 miles

The number of miles that Jeremy drove is

42 × 2 = 84 miles

Answer:

Range = The difference between the highest and lowest which is

The mean is the sum of all values divided by the number of values

= 59+19+30+75+42+49+57+81+11+87+91 divided by n (which is 11)

601÷11

= 54.6

Variance = sum of squared deviations from the mean divided by n-1

=679.65

Standard Deviation = This is the square root of variance which gives us;

26.07

Answer:

Step-by-step explanation:

Mathematically, if two triangles are similar, then all the three of their angles are congruent to each other and their corresponding sides are in the same proportion. This means that the ratio of their corresponding sides are equal to each other.

Answer: A trigonometric table will provide all the angles in respect to the ratios of the sides.

Step-by-step explanation:

A trigonometric table would provide all of the ratios of the sides in respect to the angles. Sin, cos, and tan are parts of trigonometric table.

Trigonometry table, tabulated values for some or all of the six trigonometric functions for various angular values.

Since the ratios for sides of triangles are the same, when the angles are the same, trigonometric tables have been developed which show these ratios for every angle.

To solve the initial-value problem for the differential equation y''' + y' = 0 with the general solution y = c1 + c2 cos(x) + c3 sin(x), we differentiate y to find y' and y'', apply the initial conditions at x = π, and solve for c1, c2, and c3. The specific solution is y = 1 + cos(x) - 9 sin(x).

We're given the general solution of the third-order differential equation y''' + y' = 0 as y = c1 + c2 cos(x) + c3 sin(x), and we need to find specific constants c1, c2, and c3 that satisfy the initial conditions y(π) = 0, y'(π) = 9, and y''(π) = -1.

First, differentiate y = c1 + c2 cos(x) + c3 sin(x) to find y' and y''.

Substitute x = π into the resulting equations to apply the initial conditions.

Solve the system of equations to find the values of c1, c2, and c3.

Let's carry out these steps:

Differentiation gives us:

y' = -c2 sin(x) + c3 cos(x)

y'' = -c2 cos(x) - c3 sin(x)

Applying initial conditions at x = π:

y(π) = c1 - c2 = 0

y'(π) = -c3 = 9

y''(π) = -c2 = -1

Solve the system:

c1 = c2

c2 = 1

c3 = -9

Therefore, the specific solution to the given initial-value problem is y = 1 + cos(x) - 9 sin(x).

Answer:

Step-by-step explanation:

Given that a college senior who took the Graduate Record Examination exam scored 530 on the Verbal Reasoning section and 600 on the Quantitative Reasoning section.

Quantitative reasons scores are N (429, 148)

Hence Z score for the college senior = [tex]\frac{530-429}{148} \\=0.6824[/tex]

The mean score for Verbal Reasoning section was 460 with a standard deviation of 105

Verbal reasoning score was N(460, 105)

The score of college senior = 530

Z score for verbal reasoning =[tex]\frac{530-460}{105} \\=0.6667[/tex]

comparing we can say he scored more on quantiative reasoning.

Thus comparison was possible only by converting to Z scores.