Compare each set of data. Determine which set shows greater variability.

Set A

5, 8, 9, 10, 12, 15, 16, 20

Set B

24, 34, 37, 42, 56, 57, 65

Answer:

quarter past 9

Step-by-step explanation:

Answer:09:10

Step-by-step explanation:

10 minutes after 9 can be written as

09:10

Answer:

solution is : x = -6 and x = 2

Step-by-step explanation:

Answer:

the answer is A

Step-by-step explanation:

Answer:

1 and 1/30 gallons of paint

Step-by-step explanation:

Answer:

4mm edge 2021

Step-by-step explanation:

a) The area of a kite is given by [tex]A=\frac{d_{1}d_{2}}{2}[/tex], where [tex]d_{1}, d_{2}[/tex] are the lengths of the diagonals. So the answer is [tex]\frac{1}{2} (40 \times 60)[/tex].

b) Evaluating the expression from part (a), we get the area to be [tex]\boxed{1200 \text{ cm}^{2}}[/tex]

Answer:

The team scored 50 points

Step-by-step explanation:

Let the points scored by team be x

We are given that sally was the top scorer woth 30% of the teams points

So, Sally scored points = [tex]30\% \times x[/tex]

Sally scored points = [tex]\frac{30}{100}x[/tex]

We are given that Sally scored 15 points

So, [tex]\frac{30}{100}x=15[/tex]

[tex]30x=1500[/tex]

[tex]x=\frac{1500}{30}[/tex]

x=50

Hence The team scored 50 points

Answer:

The solutions for the equation [tex]x^2\:-\:27=\:67[/tex] are [tex]x=\sqrt{94},\:x=-\sqrt{94}[/tex].

Step-by-step explanation:

To find the solutions for the equation [tex]x^2\:-\:27=\:67[/tex] you must:

[tex]\mathrm{Add\:}27\mathrm{\:to\:both\:sides}\\\\x^2-27+27=67+27[/tex]

[tex]\mathrm{Simplify}\\\\x^2=94[/tex]

[tex]\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{94},\:x=-\sqrt{94}[/tex]

Answer:

Step-by-step explanation:

Hello!

a)

One of the advantages of assigning the units randomly to the different treatments is to avoid selection bias. An example of selection bias would be selecting the best subjects at the barbell curl to the "encouraged" group and the worst ones to the "not encouraged" group.

Randomization also guarantees independence between the experimental units and treatment groups.

b) (see boxplots in attachment)

Encouraged group:

Box:

Q₁≅ 10

Q₂≅ 30

Q₃= 50

IQD= 50 - 10 = 40

The box seems symmetric, the distance between Q₁-Q₂ and Q₂-Q₃ is the same and the median is exactly in the middle.

Whiskers:

The left whisker is shorter than the right whisker, which is longer.

If you were to look only at the box, the distribution seems symmetrical but adding the whiskers there is a pronounced right asymmetry.

Min value ≅ 5

Max value ≅ 75

Range= 75 - 5 = 70

Not encouraged group:

Box:

Q₁≅ 10

Q₂≅ 20

Q₃= 35

IQD= 35 - 10 = 25

The box shows a little skewness to the right (the 2nd Quartile is closer to the first one than the 3rd) and is overall shorter compared to the first group.

Whiskers:

The left whisker is shorter than the right whisker, so in general, this distribution is also right-skewed.

Min value ≅ 0

Max value ≅ 60

Range= 60 - 0 = 60

Compared with the "encouraged" group, the IQD and the Range are shorter, which means that the "not encouraged" group shows less dispersion.

c) and d)

Daniel claims that people perform better in the barbell curl, on average, when being encouraged.

He divided the 29 subjects into two groups and measured their performance, then the study variables are:

X₁: Performance in the barber curl of a subject that was encouraged by his coach.

X₂: Performance in the barber curl of a subject that did not receive encouragement.

Logically, since Daniel wants to study the performance of both groups on average, you'd think that the parameters of interest will be the population means. But to study the average you need the population to have a normal distribution and looking at those boxplots, none of them is near the normal distribution and the size of both samples isn't big enough for an approximation.

The best option is to conduct a non-parametric analysis, for example, the Mann Whitney U-test and instead of comparing the population means, you'll compare the population medians: θ₁ vs. θ₂ ⇒ In the hypothesis of this test, you'll state that both samples come from the same population and if both have the same median, then each observation of the first sample x₁i will have an equal probability of being smaller or greater than each observation of the second sample x₂i (probability 0.5)

Depending of the statistics course, the hypotheses may change, I'm used at working directly with the medians, if they are equal, it will mean that the top 50% and bottom 50% of each population will be the same, which is the same as saying that P(x₁i > x₂i)= 0.5 vs. P(x₁i > x₂i)≠ 0.5

The conditions for the Mann-Whitney U-test are:

1) All observations on both groups should be independent. Check

2) The variables should be continuous or at least ordinal. Check

3) Both variables have the same distribution under the null hypothesis. Check (Looking at the boxplots, both are right-skewed)

4) Under the alternative hypothesis, the values of one of the populations exceed the other. In this case: H₁: θ₁ > θ₂

e)

H₀: θ₁ ≤ θ₂  vs. H₁: θ₁ > θ₂

Using the p-value approach, the decision rule is always the same:

If p-value ≤ α ⇒ Reject the null hypothesis.

If p-value > α ⇒ Do not reject the null hypothesis.

p-value: 0.107 > α: 0.05 ⇒ The decision is to not reject the null hypothesis.

At a 5% significance level, there is no significant evidence to reject the null hypothesis. Both samples seem to be from the same population. He can conclude that encouraging the subjects doesn't change significantly their performance in the barbell curl.

I hope you have a SUPER day!

Answer:

a) P(4)=0.03

b) P(x≥3)=0.15

c) P(x<3)=0.85

d) µX=1.52

e) σX=1.69

Step-by-step explanation:

The question is incomplete:

The variable X has the following probability distribution:

x  1  2  3  4   5

P(x) 0.7 0.15 0.10 0.03  0.02

a) The probability P(x=4) can be read from the table

[tex]P(4)=0.03[/tex]

b) The probability that there are at least 3 occupants in the car is P(x≥3).

[tex]P(x\geq3)=P(3)+P(4)+P(5)\\\\P(x\geq3)=0.10+0.03+0.02\\\\P(x\geq3)=0.15[/tex]

c) The probability that a car has fewer than 3 occupants (P(x<3)) is:

[tex]P(x<3)=1-P(x\geq3)=1-0.15=0.85[/tex]

d) The mean can be calculated as:

[tex]\mu_x=\sum_{i=1}^5p_i\cdot x_i\\\\\mu_x=0.7*1+0.15*2+0.10*3+0.03*4+0.02*5\\\\\mu_x=0.7+0.30+0.30+0.12+0.10\\\\ \mu_x=1.52[/tex]

e) The standard deviation can be calculated as:

[tex]\sigma_x=\sqrt{\sum_{i=1}^5p_i(x_i-\mu_x)^2}}\\\\ \sigma_x=\sqrt{0.7(1-1.52)^2+ 0.15(2-1.52)^2+ 0.1(3-1.52)^2+ 0.03(4-1.52)^2 +0.02(5-1.52)^2}\\\\ \sigma_x=\sqrt{0.7*0.2704+0.15*0.2304+0.1*2.1904+0.03*6.1504+0.03*12.1104}\\\\\sigma_x=\sqrt{0.18928+0.06912+0.65712+0.738048+1.21104}\\\\ \sigma_x=\sqrt{2.8646}\\\\ \sigma_x=1.69[/tex]

The graph of the probability distribution of X is a bar graph. The mean of X is 1.65 and the standard deviation can be calculated using the variance formula. The probability that a car has at least 3 occupants is 0.12 and the probability that a car has fewer than 3 occupants is 0.85.

Answer:

D

Step-by-step explanation:

SEX

Answer:

Yes

Step-by-step explanation:

Since we want to test if it is a right triangle, we can use the Pythagorean Theorem

a^2+b^2=c^2

where a and b are the legs (shortest sides) and c is the hypotenuse (longest side)

We have the sides: 24,18 and 30

24 and 18 are the shortest, therefore they are the legs. 30 is the longest, so it is the hypotenuse.

So, a is 24, b is 18 and c is 30

24^2+18^2=30^2

Evaluate the exponents

576+324=900

Add 576 and 324

900=900

This statement is true, so this is a right triangle

Answer:

The probability that a randomly selected monthly cell phone bill is less than $95 is 0.9772

Step-by-step explanation:

The average monthly household cellular phone bill is $73.

[tex]\mu = 73[/tex]

Local monthly household cell phone bills are normally distributed with a standard deviation of $11.

[tex]\sigma = 11[/tex]

We are supposed to find the probability that a randomly selected monthly cell phone bill is less than $95 i.e.P(x<95)

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

[tex]Z=\frac{95-73}{11}[/tex]

Z=2

Refer the z table for p value

So,p value = 0.9772

Hence the probability that a randomly selected monthly cell phone bill is less than $95 is 0.9772

Answer:

B. 280 units cubed

Step-by-step explanation:

Multiply all of the numbers together

Answer:

(-2,-6)

Step-by-step explanation:

The rule for reflecting across the x axis is multiply the y coordinate by -1.  Your x coordinate stays the same.

(x,y)---->(x,-y)

(-2,6)---> (-2,-6)

Answer:

0.545 = 54.5% probability that it came from machine A

Step-by-step explanation:

Bayes Theorem:

Two events, A and B.

[tex]P(B|A) = \frac{P(B)*P(A|B)}{P(A)}[/tex]

In which P(B|A) is the probability of B happening when A has happened and P(A|B) is the probability of A happening when B has happened.

In this question:

Event A: Defective.

Event B: Coming from machine A.

Machine A is responsible for 30%

This means that [tex]P(B) = 0.3[/tex]

10% of the output from machine A is defective

This means that [tex]P(B|A) = 0.1[/tex]

Probability of being defective:

Machine A is responsible for 30%. Of those, 10% are defective.

Machine B is responsible for 20%. Of those, 5% are defective.

Machine C is responsible for 100 - (30+20) = 50%. Of those, 3% are defective. Then

[tex]P(A) = 0.3*0.1 + 0.2*0.05 + 0.5*0.03 = 0.055[/tex]

Finally:

[tex]P(B|A) = \frac{P(B)*P(A|B)}{P(A)} = \frac{0.3*0.1}{0.055} = 0.545[/tex]

0.545 = 54.5% probability that it came from machine A

Answer:

I think it's 2,772

Step-by-step explanation:

I did

36-14=22

Then I did 22× $126

hi

if  14 = 126

   36 =  ?  

:  36*126 /14 = 324

To calculate the volume of the region W, set up and evaluate a triple integral over the given region. Identify the bounds, write the triple integral, find the limits of integration, and evaluate the integral.

To calculate the volume of the region W, we need to set up and evaluate a triple integral. We will integrate over the region defined by the given inequalities. Let's break it down step by step:

Identify the bounds of the region:

Write the triple integral:

∫∫∫ f(x,y,z) dz dy dx

Find the limits of integration:

Evaluate the integral:

Integrate the function f(x,y,z) over the given limits. The function f(x,y,z) will depend on the specific problem you are trying to solve.

https://brainly.com/question/32510822

#SPJ3

The volume of the region bounded by z = 100 - y², y = 10x², and the plane z = 0, is [tex]\(\frac{40000 \sqrt{10}}{3} \)[/tex] cubic units.

To find the volume of the region W bounded by z = 100 - y², y = 10x², and the plane z = 0, we need to evaluate a triple integral in the order dz dy dx. We can start by setting up the integral:

Thus, the triple integral is:

[tex]\[ \int_{-\sqrt{10}}^{\sqrt{10}} \int_{0}^{100} \int_{0}^{100-y^2} dz \, dy \, dx \][/tex]

First, we integrate with respect to z:

⇒ [tex]\[ \int_{0}^{100-y^2} dz = (100-y^2) - 0 = 100 - y^2 \][/tex]

This reduces our integral to:

[tex]\[ \int_{-\sqrt{10}}^{\sqrt{10}} \int_{0}^{100} (100 - y^2) \, dy \, dx \][/tex]

Next, we integrate with respect to y:

⇒ [tex]\[ \int_{0}^{100} (100 - y^2) \, dy = 100y - \frac{y^3}{3} \bigg|_{0}^{100}[/tex]

⇒ [tex]100(100) - \frac{(100)^3}{3} = 10000 - \frac{1000000}{3}[/tex]

⇒ 10000 - 333333.33 = [tex]\frac{20000}{3}[/tex]

Finally, we integrate with respect to x:

⇒ [tex]\[ \int_{-\sqrt{10}}^{\sqrt{10}} \left( \frac{20000}{3} \right)[/tex] [tex]dx = \frac{20000}{3} \left[ x \right]_{-\sqrt{10}}^{\sqrt{10}}[/tex]

⇒ [tex]\frac{20000}{3} ( 2\sqrt{10})[/tex] = [tex]\frac{40000 \sqrt{10}}{3}[/tex]

Therefore, the volume of W is [tex]\(\frac{40000 \sqrt{10}}{3} \)[/tex] cubic units.

Answer:

[tex]f(x)=2x^2-1\\f(5)=2(5)^2-1\\f(5)=2(25)-1\\f(5)=50-1\\f(5)=49[/tex]

Step-by-step explanation:

Answer:35 meters

Step-by-step explanation:

Answer:

35 meters

Formula:

a^2+b^2=c^2

12^2+b^2=37^2

Answer:

A not continueous function

Step-by-step explanation:

has a range

Answer:

There is no difference.  Because the mean of Ari's tests equals his median score of his six tests.  mean = 84% median = 84%

Step-by-step explanation:

88%, 73%, 97%, 76%, 90%, 80%

organize by low to high

73%,  76%,  80%, 88%, 90%, 97%

The median here is  (80 + 88)/2 = 84%

let us find the mean

mean = [73% +  76% +  80% + 88% + 90% + 97%  ]/6

mean =  504/6 =  84%

There is no difference.  Because the mean of Ari's tests equals his median score of his six tests.  mean = 84% median = 84%

Answer:

7, -7

Step-by-step explanation:

Answer:

∠1 = 31°

∠2 = 43°

Step-by-step explanation:

All triangles have a sum of 180°

Since we know this, then we know that in the end, the sum of the two triangles should both have a sum of 180° each.

∠2 equals 43° because it corresponds to ∠C and corresponding angles have the same measurement.

Now we can find the measurement of ∠1 by using the big triangle.

180 - (106 + 43)

180 - (149)

31°

So, the measurement of ∠1 is 31°.

Now, let's check our work.

Big triangle:

106 + 43 + 31 = 180

Small triangle:

43 + 31 + 106 = 180

Answer:

18

Step-by-step explanation:

6 ÷ 1/3

Change the whole number to a fraction

6/1  ÷ 1/3

Copy dot flip

6/1 * 3/1

18/1

18

Answer:

18

Step-by-step explanation:

6 ÷ 1/3

Change the whole number to a fraction

6/1  ÷ 1/3

Copy dot flip

6/1 * 3/1

18/1

18

The stockholders' equity section of Waterway Corporation consists of common stock ($10 par) $2,050,000 and retained earnings $520,000. A 10% stock dividend (20,500 shares) is declared when the market price per share is $14. The before-and-after effects of the dividend are as follows...

(a) Before the stock dividend, the components of stockholders' equity in Waterway Corporation are common stock ($10 par) $2,050,000 and retained earnings $520,000. After the stock dividend, the common stock would increase by 20,500 shares ($10 x 20,500 = $205,000) and retained earnings would decrease by the same amount since it is used to finance the dividend. So, the new components of stockholders' equity would be common stock ($10 par) $2,255,000 and retained earnings $315,000.

(b) Before the stock dividend, the shares outstanding in Waterway Corporation are not provided in the question. After the stock dividend, the number of shares outstanding would increase by 20,500 shares.

(c) The par value per share of the common stock remains the same before and after the stock dividend, which is $10 per share.

Final answer:

A 10% stock dividend for Waterway Corporation, with a par value of $10 and a market price of $14 per share, results in an increase in common stock by $205,000, a decrease in retained earnings by the same amount, an increase of 20,500 shares, while the par value per share remains at $10.

Explanation:

When a stock dividend is declared, it affects the stockholders' equity section of a corporation. In the case of Waterway Corporation, a 10% stock dividend is declared when each share has a market price of $14. Since the par value is $10 and there are 20,500 new shares being issued, the total value at par is $205,000 (20,500 shares x $10 par value). This amount is transferred from retained earnings to common stock. No additional capital is generated since it's a stock dividend, not a cash dividend.

Here's the before-and-after effects of the dividend:

(a) Components of stockholders' equity:
Before the dividend:
Common stock: $2,050,000
Retained earnings: $520,000
After the dividend:
Common stock: $2,255,000 (original $2,050,000 + $205,000 from the stock dividend)
Retained earnings: $315,000 (original $520,000 - $205,000 transferred to common stock)

(b) Shares outstanding:
Before the dividend: Assuming each common stock's par value aligns with one share, the company has 205,000 shares ($2,050,000 / $10 par value).
After the dividend: 225,500 shares (205,000 original shares + 20,500 new shares issued)

(c) Par value per share:
The par value per share remains unchanged at $10, as the stock dividend does not affect the par value.

Answer:

All the statement are true about these functions

Step-by-step explanation:

Given any monomial function where a=1 and n is even.

Attached is the graph of [tex]y=x^4 \:\&\: y=x^6[/tex] as a case study.

From the graphs, the following are observed.

To graph monomial functions where a = 1 and n is even, enter the equations into a graphing calculator and observe the graph. The graph opens up, passes through the origin, and is symmetric about the y-axis.

To graph the monomial functions where $a=1$ and $n$ is even, we can use a graphing calculator. For example, to graph $y=x^4, y=x^6$, and $y=x^8$, we can enter these equations into the graphing calculator and select an appropriate viewing window. By tracing along these functions, we can make the following observations:

Based on these observations, we can conclude that all of these statements are true about the graph of the monomial functions with $a=1$ and even $n$.

https://brainly.com/question/34351927

#SPJ3