Consider the vector b⃗ b→b_vec with length 4.00 mm at an angle 23.5∘∘ north of east. What is the y component bybyb_y of this vector?

Answer: 0.332 < p < 0.490

Step-by-step explanation:

We know that the confidence interval for population proportion is given by :-

[tex]\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]

, where n= sample size

[tex]\hat{p}[/tex] = sample proportion

z* = critical z-value.

As per given , we have

n= 258

Sample proportion of college students who own a car = [tex]\hat{p}=\dfrac{106}{258}\approx0.411[/tex]

Critical z-value for 99% confidence interval is 2.576. (By z-table)

Therefore , the  99% confidence interval for the true proportion(p) of all college students who own a car will be :[tex]0.411\pm (2.576)\sqrt{\dfrac{0.411(1-0.411)}{258}}\\\\=0.411\pm (2.576)\sqrt{0.00093829}\\\\= 0.411\pm (2.576)(0.0306315197142)\\\\=0.411\pm 0.0789=(0.411-0.0789,\ 0.411+0.0789)\\\\=(0.3321,\ 0.4899)\approx(0.332,\ 0.490)[/tex]

Hence, a 99% confidence interval for the true proportion of all college students who own a car : 0.332 < p < 0.490

Answer:

A

Step-by-step explanation:

The relative frequency is calculated by dividing the frequency of that class to the sum of frequencies. It can be represented as

[tex]Relative frequency=\frac{f}{sum(f)}[/tex]

Hence, the relative frequency of class is the percentage or proportion of data lies in that class.

The cumulative frequency of a class is computed by adding the frequency of the respective class to the frequencies of all previous classes. The cumulative frequency of first class will always be equal to the frequency of first class.

Hence, the cumulative frequency  for a class is the sum of frequency for that class and the frequencies of all previous classes.

Relative frequency is the proportion of total data that falls into a class, while cumulative frequency is the sum of frequencies of that class and all previous classes. Therefore, relative frequency illustrates the percentage of data in a certain class, while cumulative frequency shows the accumulation of data up to that point.

The terms relative frequency and cumulative frequency both pertain to statistics; however, they represent different concepts. Relative frequency of a class is the percentage or proportion of the whole set of data that falls in that class. It's calculated by dividing the frequency of that class by the total number of data points.

On the other hand, cumulative frequency of a class is the sum of the frequencies of that class and all previous classes in a dataset. Essentially, it accumulates counts as you proceed through the dataset.

For instance, if you have five classes with frequencies of 1, 2, 3, 4 and 5, the relative frequencies would be 0.0667, 0.1333, 0.2, 0.2667, and 0.3333 respectively (assuming we divide each frequency by the total data points, which is 15 in this case). In contrast, the cumulative frequencies would be 1, 3, 6, 10 and 15, indicating the total count up to each class.

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Answer:

Step-by-step explanation:

1. sinФ = cos 25

25 ° is in the between 0 and 90°

therefore it can simply represent

cosФ= sin (90-Ф) = sin (90 - 25) = sin 65

2. sin(Ф/3 + 10) = cos Ф

cos Ф = sin (90 -Ф)

sin(Ф/3 + 10) = cos Ф

sin(Ф/3 + 10) = cos Ф = sin(90-Ф)

Ф/3+10=90-Ф

10Ф+3/30 = 90-Ф

10Ф+3 = 30(90-Ф)

10Ф+3 = 2700-30Ф

10Ф+30Ф=2700-3

40Ф = 2697

Ф = 2697 / 40 = 67.425 ≅ 67.4°

Answer:

f'(π) = (-5/3)

Step-by-step explanation:

for the equation

∫₀ˣ (3f(t)+5t)dt=sin(x)

3*∫₀ˣ f(t) dt+5*∫₀ˣ t*dt=sin(x)

then

∫₀ˣ t*dt = (t²/2) |₀ˣ = x²/2 - 0²/2 = x²/2

thus

3*∫₀ˣ f(t) dt + 5* x²/2 =  sin(x)

∫₀ˣ f(t) dt =  1/3*sin(x) - 5/6*x²

then applying differentiation

d/dx ( ∫₀ˣ f(t) dt) = f(x) - f(0)  ( from the fundamental theorem of calculus)

d/dx (1/3*sin(x) - 5/6*x²) = 1/3*cos(x) - 5/3*x

therefore

f(x) - f(0) =  1/3*cos(x) - 5/3*x

f(x) =  1/3*cos(x) - 5/3*x + f(0)

applying differentiation again (f'(x) =df(x)/dx)

f'(x) =  -1/3*sin(x) - 5/3

then

f'(π) =  -1/3*sin(π) - 5/3 = 0 - 5/3 = -5/3

f'(π) = (-5/3)

Given expression is:

→ [tex]\int x_0 (3f(t)+5t) dt = sin(x)[/tex]

or,

→ [tex]3\times \int x_0 f(t) dt+5\times \intx_0 t\times dt = sin(x)[/tex]

then,

→ [tex]\int x_0 t\times dt = (\frac{t^2}{2} ) x_0[/tex]

      [tex]\frac{x^2}{2} -\frac{0^2}{2} = \frac{x^2}{2}[/tex]

thus,

→ [tex]3\times \int x_0 f(t) +5\times \frac{x^2}{2} = sin(x)[/tex]

  [tex]\int x_0 f(t) dt = \frac{1}{3}\times sin(x) - \frac{5}{6}\times x^2[/tex]

By applying differentiation, we get

→ [tex]\frac{d}{dx} (\int x_0 f(t) dt) = f(x) - f(0)[/tex]

  [tex]\frac{d}{dx}(\frac{1}{3} sin(x)) - \frac{5}{6}x^2= \frac{1}{3} cos(x) - \frac{5}{3} x[/tex]

Therefore,

→ [tex]f(x) -f(0) = \frac{1}{3}cos (x) - \frac{5}{3} x[/tex]

  [tex]f(x) = \frac{1}{3}cos (x) -\frac{5}{3} x+f(0)[/tex]

By differentiating again, we get

→ [tex]f'(x) = -\frac{1}{3}sin(\pi) -\frac{5}{3}[/tex]

           [tex]= 0-\frac{5}{3}[/tex]

           [tex]= -\frac{5}{3}[/tex]

Thus the response above is correct.

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Explanation on determining which type of cameras sold more in 2001 and when digital camera sales surpassed film camera sales.

(a) To show that more film cameras than digital cameras were sold in 2001, we need to substitute t = 0 into both functions:

For digital cameras: f(0) = 3.06(0) + 6.84 = 6.84 million units
For film cameras: g(0) = -1.85(0) + 16.48 = 16.48 million units
Therefore, more film cameras (16.48 million units) were sold in 2001 than digital cameras (6.84 million units).

(b) To find when digital camera sales first exceeded film camera sales, we need to set the two functions equal to each other and solve for t:
3.06t + 6.84 = -1.85t + 16.48
4.91t = 9.64
t ≈ 1.96 years, which is approximately at the end of 2003.

SR =sin (angle) = opposite leg/ hypotenuse

sin(52) = SR/ 7.6

SR = 7.6 x sin(52)

SR = 5.988 ( round answer as needed.)

QR = cos(angle) = adjacent leg / hypotenuse

Cos(52) = QR/ 7.6

QR = 7.6 x cos(52)

QR = 4.679. ( round answer as needed).

Angle S = 180 - 90 -52 = 38 degrees

Answer: lowering

Step-by-step explanation:

Statins are drugs that are administered in order to help reduce cholesterol level in the body. Statins achieves this by making it impossible for the body to have access to substances that helps to produce cholesterol. This drug reduces cholesterol up to 50% and are taken once every day. Examples includes rosuvastatin, pitavastatin, simvastatin.

Final answer:

To create a histogram showing the probability distribution of the number of illiterate people out of eleven people in a sample, you can use the binomial probability formula. Calculate the probability for each possible number of illiterate people and represent it in a histogram.

Explanation:

This question is asking for a histogram showing the probability distribution of the number of illiterate people out of eleven people in a sample, based on the information that about 20% of all people in the United States are illiterate.

To create the histogram, we can use the binomial probability formula. The formula is P(X = k) = C(n, k) * p^k * (1 - p)^(n - k), where C(n, k) represents combinations, n is the number of trials, k is the number of successes, and p is the probability of success.

We can calculate the probability for each possible number of illiterate people out of the eleven, ranging from 0 to 11, and represent it in a histogram.

The dimensions of the parcel are 80 ft (width) and 88 ft (length).

To find the dimensions of the parcel, we can set up a system of equations using the given information. Let's denote the width of the parcel as x ft. Since the parcel is 8 ft longer than it is wide, the length can be represented as x + 8 ft. We can use the Pythagorean theorem to find the length of the diagonal: x^2 + (x + 8)^2 = 232^2. Solving this equation will give us the width and length of the parcel.

Expanding the equation, we get: x^2 + (x^2 + 16x + 64) = 53824. Combining like terms, we have: 2x^2 + 16x + 64 = 53824. Rearranging the equation, we have a quadratic equation: 2x^2 + 16x + (-53760) = 0. To solve this quadratic equation, we can use factoring, completing the square, or the quadratic formula.

By factoring, we find that the dimensions of the parcel are 80 ft (width) and 88 ft (length).

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Answer:

$12,400

Step-by-step explanation:

Given:

Week earning + 9% commission = $1,916

Weekly Earning = $800

Commission for the Week = $1,916 - $800 = $1,116

Let total sales for the week = x

Therefore, x = ($1,116 * 100)/9

x = $12,400

Answer:

$12400

Step-by-step explanation:

Wages = Basic pay + Commission

1916 = 800 + 9%A

Let A be the total sales

9%A = 1916 - 800

9%A = 1116

A = $12400

Answer:

The answer to your question is 7345 l

Step-by-step explanation:

Data

Convert 7kl to 345 L

Process

- To solve this question we must know the equivalence of kl to L.

                    1 kl -----------------  1000 l

- Now, use proportions and cross multiplication to solve it

                    1 kl ----------------  1000 l

                    7 kl----------------   x l

                    x = (7 x 1000) / 1

                    x = 7000 / 1

                    x = 7000 l

- Calculate the result

                   7000 + 345 = 7345 l

Answer:

$2 640

Step-by-step explanation:

Principal = $16 500; Time = 2 years Rate = 8%

Simple Interest = P × R × T/100

= $16 500 × 8 × 2/100

=$264 000/100

Simple Interest = $2 640

Answer:

[tex] \bar X= \frac{1930055}{110}=17545.95[/tex]

Step-by-step explanation:

For this case we can construct the following table in order to find the mean for the grouped data:

  Interval           Frequency (fi)    Midpoint (Xi)      Xi *fi

5001-10000              27                   7500.5          202513.5

10001-15000             23                  12500.5         287511.5

15001-20000             12                  17500.5         210006

20001-25000            18                  22500.5        405009  

25001-30000            30                 27500.5         825015

Total                          110                                         1930055

And the mean is calculated with the following formula:

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

Where [tex] n = \sum_{i=1}^n f_i = 110[/tex]

So then if we replace into the formula we got:

[tex] \bar X= \frac{1930055}{110}=17545.95[/tex]

Answer:

a) [tex] P(X \geq 3) = 1- P(X<3) = 1-P(X \leq 2) = 1-[P(X=0)+P(X=1) +P(X=2)][/tex]

The individual pprobabilities are:

[tex]P(X=0)=(300000000C0)(\frac{1}{500000000})^3 (1-\frac{1}{500000000})^{300000000-0}=0.548812[/tex]

[tex]P(X=1)=(300000000C0)(\frac{1}{500000000})^3 (1-\frac{1}{500000000})^{300000000-0}=0.329287[/tex]

[tex]P(X=2)=(300000000C0)(\frac{1}{500000000})^3 (1-\frac{1}{500000000})^{300000000-0}=0.098786[/tex]

And if we replace we got:

[tex] P(X \geq 3) = 1- [0.548812+0.329287+0.098786]=0.023115[/tex]

b) [tex] P(X \geq 3) = 1- P(X<3) = 1-P(X \leq 2) = 1-[P(X=0)+P(X=1) +P(X=2)][/tex]

And we can calculate this with the following excel formula:

[tex] =1-BINOM.DIST(2;300000000;(1/500000000);TRUE)[/tex]

And we got:

[tex] P(X\geq 3) = 0.023115[/tex]

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

[tex]X \sim Binom(n=300000000, p=\frac{1}{500000000})[/tex]

The probability mass function for the Binomial distribution is given as:

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

Where (nCx) means combinatory and it's given by this formula:

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

(a) Write down the exact expression for the probability P(3 or more people will be killed by lightning in the U.S. next year). Evaluate this expression to 6 decimal places.

For this case we want this probability, we are going to use the complement rule:

[tex] P(X \geq 3) = 1- P(X<3) = 1-P(X \leq 2) = 1-[P(X=0)+P(X=1) +P(X=2)][/tex]

The individual probabilities are:

[tex]P(X=0)=(300000000C0)(\frac{1}{500000000})^3 (1-\frac{1}{500000000})^{300000000-0}=0.548812[/tex]

[tex]P(X=1)=(300000000C0)(\frac{1}{500000000})^3 (1-\frac{1}{500000000})^{300000000-0}=0.329287[/tex]

[tex]P(X=2)=(300000000C0)(\frac{1}{500000000})^3 (1-\frac{1}{500000000})^{300000000-0}=0.098786[/tex]

And if we replace we got:

[tex] P(X \geq 3) = 1- [0.548812+0.329287+0.098786]=0.023115[/tex]

(b) Write down a relevant approximate expression for the probability from (a). Justify briefly the approximation. Evaluate this expression to 6 decimal places.

For this case we want this probability, we are going to use the complement rule:

[tex] P(X \geq 3) = 1- P(X<3) = 1-P(X \leq 2) = 1-[P(X=0)+P(X=1) +P(X=2)][/tex]

And we can calculate this with the following excel formula:

[tex] =1-BINOM.DIST(2;300000000;(1/500000000);TRUE)[/tex]

And we got:

[tex] P(X\geq 3) = 0.023115[/tex]

Answer:

Step-by-step explanation:

3x²-12x+7=0

        x = (-b ± √ b²-4ac) / 2a

  from these equation: a = 3 , b = 12 , c = 7

Accordingly, b²  -  4ac =   12²- 4 (3 x 7) = 144 - 4(21) = 144 - 84 =   60

                        2a= 2 x 3 = 6

Applying the quadratic formula :

             x =  (-12 ± √ 60 )/6

  √ 60 rounded to 2 decimal digit using calculator 7.75

x =  (-12 ± 7.75 )/6

(-12 ± 7.75 ) = (-12 + 7.75 ) = 5.75

                      (-12 - 7.75 )  = -19.75

x =  5.75/6 = 0.96

       -19.75/6 = 3.29

sum of the square of roots = 0.96² + 3.29²

                                                    0.9261 +   10.8241 = 11.75

The sum of the squares of its roots of the quadratic equation

3x² - 12x + 7 = 0 is p² + q² = 34/3.

A quadratic equation is an algebraic expression in the form of variables and constants.

A quadratic equation has two roots as its degree is two.

We have a quadratic equation 3x² - 12x + 7 = 0.

We know a quadratic equation ax² - bx + c = 0 can be written as,

x² - (b/a)x + (c/a) = 0 ⇒ x² - (p + q)x + pq = 0 where p and q are the roots.

∴ 3x² - 12x + 7 = 0.

x² - (12/3)x + 7/3 = 0.

Hence (p + q) = 4 and pq = 7/3.

Now, (p + q)² = p² + q² + 2pq.

16 = p² + q² + 14/3.

p² + q² = 16 - 14/3.

p² + q² = (48 - 14)/3.

p² + q² = 34/3.

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Answer:

a. [tex]S\times T=\{(2,1),(2,3),(2,5),(4,1),(4,3),(4,5),(6,1),(6,3),(6,5)\}[/tex] . Set S × T has 9 elements.

b. [tex]T\times S=\{(1,2),(1,4),(1,6),(3,2),(3,4),(3,6),(5,2),(5,4),(5,6)\}[/tex]. Set T × S has 9 elements.  

c. [tex]S\times S=\{(2,2),(2,4),(2,6),(4,2),(4,4),(4,6),(6,2),(6,4),(6,6)\}[/tex] . Set S × S has 9 elements.

d. [tex]T\times T=\{(1,1),(1,3),(1,5),(3,1),(3,3),(3,5),(5,1),(5,3),(5,5)\}[/tex] . Set T × T has 9 elements.

Step-by-step explanation:

The given sets are S={2,4,6} and T={1,3,5}.

We need to find the set-roster notation to write each of the following sets, and the number of elements that are in each set.

a.

[tex]S\times T=\{(s,t)|s\in S and t\in T\}[/tex]

[tex]S\times T=\{(2,1),(2,3),(2,5),(4,1),(4,3),(4,5),(6,1),(6,3),(6,5)\}[/tex]

Set S × T has 9 elements.

b.

[tex]T\times S=\{(t,s)|s\in S and t\in T\}[/tex]

[tex]T\times S=\{(1,2),(1,4),(1,6),(3,2),(3,4),(3,6),(5,2),(5,4),(5,6)\}[/tex]

Set T × S has 9 elements.  

c.

[tex]S\times S=\{(s,s)|s\in S \}[/tex]

[tex]S\times S=\{(2,2),(2,4),(2,6),(4,2),(4,4),(4,6),(6,2),(6,4),(6,6)\}[/tex]

Set S × S has 9 elements.

d.

[tex]T\times T=\{(t,t)|t\in T\}[/tex]

[tex]T\times T=\{(1,1),(1,3),(1,5),(3,1),(3,3),(3,5),(5,1),(5,3),(5,5)\}[/tex]

Set T × T has 9 elements.

Using the set roster notation, the number of elements in each of the sets would be the product of the number of elements in each individual set, which is 9.

A.) S × T :

{2, 4, 6} × {1, 3, 5}

{(2, 1), (2, 3), (2, 5), (4, 1), (4, 3), (4, 5), (6, 1), (6, 3), (6, 5)}

Number is of elements = 9

B.) T × S :

{1, 3, 5} × {2, 4, 6}

{(1,2), (1,4), (1,6), (3,2), (3,4), (3, 6), (5, 2), (5,4), (5,6)}

Number is of elements = 9

C.) S × S :

{1, 3, 5} × {1, 3, 5}

{(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 3), (5, 1), (5, 3), (5, 5)}

Number is of elements = 9

D.) T × T :

{2, 4,6} × {2, 4, 6}

{(2,2), (2, 4), (2, 6), (4,2), (4, 4), (4, 6), (6,2), (6, 4), (6,6)}

Number is of elements = 9

Hence, the distribution of values of each set.

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Answer:

a) Range = 25.6

b) Variance = 48.1446

c) s = 6.9386

d) See below

Step-by-step explanation:

(a)  the sample range

To obtain the sample range, we must sort the data increasingly :

23.7, 26.3, 28.3, 28.7, 29.6, 29.8, 30.1, 31, 33.5, 49.3

Then, find the distance between the largest and the lowest

Range = 49.3 - 23.7 = 25.6

(b) the sample variance [tex]s^2[/tex] from the definition

In order to find the variance from the definition, we need first the mean. The mean is defined as the average

[tex]\bar x=\displaystyle\frac{\displaystyle\sum_{i=1}^{n}x_i}{n}[/tex]

where the [tex]x_i[/tex] are the values of the data collected and n=10 the size of the sample.

So, the mean is

[tex]\bar x=31.03[/tex]

Now, the variance of the sample is defined as  

[tex]s^2=\displaystyle\frac{\displaystyle\sum_{i=1}^n(x_i-\bar x)^2}{n-1}[/tex]

and we have that the variance is

[tex]s^2=48.1446[/tex]

(c) the sample standard deviation

The sample standard deviation is nothing but the square root of the variance

[tex]s=\sqrt{48.1446}=6.9386[/tex]

d) [tex]s^2[/tex]  using the shortcut method

The shortcut method figures out the variance without having to compute the mean. The formula is

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

where n=10 is the sample size .

So, using the shortcut method,

[tex]s^2=\frac{10*10,061.91-96,286.09}{10*9}=48.1446[/tex]

Answer:

4%

Step-by-step explanation:

A 20% solution with a total volume of 400 mL has 20% * 400 mL of solute.

20% * 400 mL = 80 mL

When you dilute the solution to 2 L, you introduce additional water, but no additional solute, so now you have the same 80 mL of solute in 2 L of total solution.

The concentration is:

(80 mL)/(2 L) = (80 mL)/(2000 mL) = 0.04

As a percent it is:

0.04 * 100% = 4%

"4" will be the final percentage strength.

According to the question,

20% solution with 400 mL has 20% × 400 mL of solute

then,

→ [tex]20 \ percent\times 400=80[/tex]

hence,

The concentration will be:

→ [tex]\frac{80 \ mL}{2 \ L} = \frac{80 \ mL}{2000 \ mL}[/tex]

            [tex]= 0.04[/tex]

or,

            [tex]= 0.04\times 100[/tex]

            [tex]= 4[/tex] (%)

Thus the answer above is right.

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Answer:

$18

Step-by-step explanation:

The cost, c, of a ham sandwich at a deli varies directly with the number of sandwiches, n. c = $54 when n is 9

This means that to buy 9 sandwiches, it costs $54. So one sandwich costs 54/9 = $6.

What is the cost of the sandwiches when n is 3?

To buy three sandwiches, it costs $6*3 = $18.

So the correct answer is:

$18

Answer:A

Step-by-step explanation:

$18

The probability that a student selected at random majors in engineering is 30% which is 0.3.

The probability  that the student both majors in engineering and play club sports is 10% which is 0.1.

For a student who is selected at random to be one who majors in engineering, there are two possible ways.

The student majors in engineering OR the Student both majors in engineering and plays club

The Probabilty that the student majors in Enginnering =

The probability that the student majors in engineering plus the  probability that Student both majors in engineering and plays club sport

= 0.3+0.1= 0.4

Answer: Experimental study

Step-by-step explanation:

This is an experimental study because it involves determination of the effect of each treatment on separate individuals. It involves selecting the 81 individuals randomly to be treated with a particular treatment (either of the two treatment). In experimental study the researcher apply separate treatments on a different groups and measure its effect on each of the groups.

This is an experimental study where participants were randomly assigned to either naturopathic care or standardized psychotherapy to assess their effects on anxiety scores.

This is an example of an experimental study. In an experimental study, researchers assign participants to different groups and manipulate variables to determine cause-and-effect relationships. In this case, Cooley et al. randomly assigned individuals to either naturopathic care or standardized psychotherapy, which are the two treatments being compared. They then measured the effects of these treatments on anxiety scores.

The fact that participants were randomly assigned to the two treatments suggests that this study was experimental rather than observational.

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Answer:

The accrued value after 5 years is $1,605.95.

The accrued value after 10 years is $1,672.9.

The accrued value after 15 years is $1,739.85.

Step-by-step explanation:

This is a simple interest problem.

The simple interest formula is given by:

[tex]E = P*I*t[/tex]

In which E are the earnings, P is the principal(the initial amount of money), I is the interest rate(yearly, as a decimal) and t is the time.

After t years, the total amount of money is:

[tex]T = E + P[/tex].

In this problem, we have that:

[tex]P = 1539, I = 0.01[/tex]

Accrued value after 5 years

This T when t = 5. So

[tex]E = P*I*t[/tex]

[tex]E = 1339*0.01*5 = 66.95[/tex]

The total is

[tex]T = E + P = 66.95 + 1539 = 1605.95[/tex]

The accrued value after 5 years is $1,605.95.

Accrued value after 10 years

This T when t = 10. So

[tex]E = P*I*t[/tex]

[tex]E = 1339*0.01*10 = 133.9[/tex]

The total is

[tex]T = E + P = 133.9 + 1539 = 1672.9[/tex]

The accrued value after 10 years is $1,672.9.

Accrued value after 15 years

This T when t = 15. So

[tex]E = P*I*t[/tex]

[tex]E = 1339*0.01*15 = 200.85[/tex]

The total is

[tex]T = E + P = 200.85 + 1539 = 1739.85[/tex]

The accrued value after 15 years is $1,739.85.

Answer with Step-by-step explanation:

We are given that

P(A)=0.4 and P(B)=0.7

We know that

[tex]P(A)+P(B)+P(A\cap B)=P(A\cup B)[/tex]

We know that

Maximum value of [tex]P(A\cup B)[/tex]=1 and minimum value of [tex]P(A\cup B)[/tex]=0

[tex]0\leq P(A\cup B )\leq 1[/tex]

[tex]0\leq P(A)+P(B)-P(A\cap B)\leq 1[/tex]

[tex]0\leq 0.4+0.7-P(A\cap B)\leq 1[/tex]

[tex]0\leq 1.1-P(A\cap B)\leq 1[/tex]

[tex]0\leq 1.1-P(A\cap B)[/tex]

[tex]P(A\cap B)\leq 1.1[/tex]

It is not possible that [tex]P(A\cap B)[/tex] is equal to 1.1

[tex]1.1-P(A\cap B)\leq 1[/tex]

[tex]-P(A\cap B)\leq 1-1.1=-0.1[/tex]

Multiply by (-1) on both sides

[tex]P(A\cap B)\geq 0.1[/tex]

Again, [tex]P(A\cup B)\geq P(B)[/tex]

[tex]0.4+0.7-P(A\cap B)\geq 0.7[/tex]

[tex]1.1-P(A\cap B)\geq 0.7[/tex]

[tex]-P(A\cap B)\geq -1.1+0.7=-0.4[/tex]

Multiply by (-1) on both sides

[tex]P(A\cap B)\leq 0.4[/tex]

Hence, [tex]0.1\leq P(A\cap B)\leq 0.4[/tex]

Answer:

a. is below in the explanation

b. 1/3e^(-y/3)

c.0.05882

d. 0.2231

e. 7     f.25

Step-by-step explanation:

Let and be independent random variables representing the lifetime (in 100 hours) of Type A and Type B light bulbs, respectively. Both variables have exponential distributions, and the mean of X is 2 and the mean of Y is 3.

a can be solved as follows      [tex]\frac{1}{\alpha\beta } e^{\frac{-x-y}{\alpha\beta } }[/tex]

[tex]\frac{1}{6} e^{\frac{-\alpha }{2} -\beta/3 }[/tex]

1/6[tex]e^{\frac{-(3x+2y}{6 }[/tex]

b.

f(y/x)=f(x).f(y)/{f(y)}=f(y)

1/3e^(-y/3)

c. Find the probability that a Type A bulb lasts at least 300 hours and a Type B bulb lasts at least 400 hours.

[tex]\int\limits^\alpha _4 {1/6e^{-(3x+2y)/6} } \, dy[/tex]

1/2e^-(3x+8)/6

[tex]\int\limits^\alpha _3 {e^{-(3x+800)} } \, dx \\[/tex]

0.05882

Given that a type B fails at 300 hours . we find the probability that type A bulb lasts longer than 300hr

f(x>y|y=3)=f(x>3)

[tex]\int\limits^a_3 {1/2e^{-x/2} } \, dx[/tex]

0.2231

e.e) What is the expected total lifetime of two Type A bulbs and one Type B bulb?  

E(2A+B)

(2E(A)+E(y)

2*2+3

=7

f. variance of the total lifetime of two types A bulbs and one type B bulb

V(2x+y)=

4*4+9

=25

The joint pdf of X and Y is (1/6)e^(-x/2-y/3). The conditional pdf of Y given X = x is (1/2)e^(-y/3). The probability that a Type A bulb lasts at least 300 hours and a Type B bulb lasts at least 400 hours is 9/250. The probability that a Type A bulb lasts longer than 300 hours, given that a Type B bulb fails at 300 hours, is approximately 0.9999. The expected total lifetime of two Type A bulbs and one Type B bulb is 7 100 hours. The variance of the total lifetime of two Type A bulbs and one Type B bulb is 10 100 hours.

To find the joint pdf f(x|y) of X and Y, we need to find the product of the individual pdfs of X and Y since they are independent. Since X and Y follow exponential distributions with means of 2 and 3 respectively, their pdfs can be written as f(x) = (1/2)e^(-x/2) and f(y) = (1/3)e^(-y/3). Therefore, the joint pdf f(x|y) is given by:

f(x|y) = f(x)*f(y) = (1/2)e^(-x/2) * (1/3)e^(-y/3) = (1/6)e^(-x/2-y/3).

To find the conditional pdf f_2(y|x) of Y given X = x, we use Bayes' theorem:

f_2(y|x) = (f(x|y)*f(y))/f(x) = [(1/6)e^(-x/2-y/3) * (1/3)e^(-y/3)] / [(1/2)e^(-x/2)] = (1/2)e^(-y/3).

To find the probability that a Type A bulb lasts at least 300 hours and a Type B bulb lasts at least 400 hours, we integrate the joint pdf over the given range:

P(X >= 300, Y >= 400) = ∫∫(1/6)e^(-x/2-y/3)dxdy = ∫(3/2)e^(-x/2)dx * ∫e^(-y/3)dy. Solving the integrals, we get P(X >= 300, Y >= 400) = 9/250.

Given that a Type B bulb fails at 300 hours, we need to find the probability that a Type A bulb lasts longer than 300 hours. This is equivalent to finding P(X > 300 | Y = 300). Using the conditional pdf f_2(y|x) = (1/2)e^(-y/3), we integrate from 300 to infinity:

P(X > 300 | Y = 300) = ∫(1/2)e^(-y/3)dy = 1 - ∫(1/2)e^(-y/3)dy = 1 - (1/2)e^(-y/3) = 1 - (1/2)e^(-300/3) = 1 - e^(-100) ≈ 0.9999.

The expected total lifetime of two Type A bulbs and one Type B bulb is obtained by summing the means of X and Y twice and the mean of Y once:

E[Total Lifetime] = 2*E[X] + E[Y] = 2*2 + 3 = 7.

The variance of the total lifetime of two Type A bulbs and one Type B bulb is obtained by summing the variances of X and Y twice and the variance of Y once:

Var[Total Lifetime] = 2*Var[X] + Var[Y] = 2*(2^2) + 3^2 = 10.

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Final answer:

The equation that describes the situation is g + 300 - 1600 = 1500. Simplifying this equation gives g = 2800.

Explanation:

To write the equation that describes the situation, we start with the initial weight of the mound as g pounds. Throughout the day, 300 pounds are added to the mound, so the new weight is g + 300 pounds. Two orders of 800 pounds each are sold, so 1600 pounds are removed from the mound. At the end of the day, the weight of the mound is 1500 pounds. The equation that describes the situation is:

g + 300 - 1600 = 1500

Simplifying this equation, we have:

g - 1300 = 1500

Adding 1300 to both sides, the equation becomes:

g = 2800

The rate of change is A. constant, B. linear

Step-by-step explanation:

Rate of change is the ratio of change in value of y with corresponding value of x

Rate of change=Δy/Δx

Rate of change=5-2/4-1 =3/3=1

The rate of change is constant in the given interval and linear with a positive slope.

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Answer:

ljgiuipuiopu

Step-by-step explanation:

The new net worth after a gain in stock value is $74.2 billion, or in the scientific notation it is $7.42×10¹0.

In this problem, we have a person with a starting net worth of $71.0 billion, or $7.10×1010 in scientific notation. This person's stock has increased by $3.20 billion, or 3.20×109 in scientific notation. To find the new net worth, these two amounts should be added together.

Step 1: Write the starting net worth and stock increase in standard form: $71.0×10⁹ + $3.2×10⁹. Step 2: Add these two amounts together to find the new net worth. $74.2×10⁹ This is the new net worth in standard form.

To express this amount in scientific notation, it becomes $7.42×10¹0, which is your final answer.

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Answer: True

Step-by-step explanation:

A subset is a term used in the sets topic In mathematics to describe a Part of A larger values being considered.

Sample In experimental procedures is known as Part of what is to be used to determine the reaction of a larger mass of specimen.

Taking these two definitions by side, it is quite evident that both of them represent taking part of a larger picture to form a smaller picture which does not totally deviate from the larger picture being considered.