You randomly draw a lane number for a 100-meter race. Then your friend randomly draws a

Lane number for the same race. Are these events independent or dependent?*

Answer:

Type 1 Error: Stating 'American families owing house< 67.4%', when it = 67.4%

And, implementing the tax break for first time home buyers, due to the error

Step-by-step explanation:

Null Hypothesis [H0] : American families owing house = 67.4%

Alternate Hypothesis [H1] : American families owing house < 67.4%

Type 1 error is he rejection of an actually true null hypothesis.

In this case, it means : Results reject H0 in favour of H1 & state that 'american families owing house < 67.4% ; when actually null hypothesis, i.e 'american families owing house = 67.4%'  is true.

This would imply that city council might extend the tax breaks for first time home buyers because of the type 1 error in the case. When, it is actually not needed as per the true data status.

Answer:

7.4

Step-by-step explanation:

To calculate the time taken to travel a certain distance at a certain speed, use the formula: Time = Distance/Speed. In this case, to travel 520 miles at 70 miles per hour, it would take approximately 7.43 hours or 7 hours and 26 minutes.

The subject of this question is about calculating time based on speed and distance in a real-world context. To find out how many hours it will take you to reach your destination travelling at 70 miles per hour for 520 miles, we use the formula: Time = Distance / Speed. In this context, the distance is 520 miles and the speed is 70 miles per hour. Insert these values into the formula, you get: Time = 520 / 70 which gives you approximately 7.43 hours. This means it would take you about 7 hours and 26 minutes to reach your destination, if you were traveling at a consistent speed of 70 miles per hour.

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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

Answer:

largest area that can be enclosed = 8,000,000 m²

Step-by-step explanation:

Since it is a rectangle plot, the area is expressed as; A = xy, where x is length and y is width.

Because it is next to the river, he only needs to fence three sides, so amount of fencing; F = x + 2y.

Since we know the amount of fencing available is 8000m, we get:

8000 = x + 2y

solving for x, we have;

x = 8000 - 2y

substitute 8000 - 2y for x into the area equation to give;

A = (8000 - 2y)y distribute

A = -2y² + 8000y

Now, due to the negative sign next to 2, this will be a parabola which opens down, meaning that the point of maximum area will be at the vertex,

Thus; y = -b/2a = -8000/[2(-2)] = 2000

x = 8000 - 2(2000) = 4000

A = 4000(2000) = 8,000,000 m²

Answer:

Step-by-step explanation:

let width=x

length=x+4

height=x-2

A=W×L

252=(x)(x+4)

252=x^2 +4x

0=x^2 +4x -252

use quadratic formula

x=14

W=14

L=18

H=12

V=W×L×H

V=(14)(18)(12)

Final answer:

To calculate the volume of the crate, solve for the width using the base area and the relation between length and width, then find the height. Multiply length, width, and height to obtain the crate's volume.

Explanation:

To find the volume of the crate, we must first determine the dimensions of the base. We know that the area of the base is 252 square inches, and that the length (L) is 4 inches greater than the width (W).

Therefore, we can express the length as L = W + 4. Since the area of a rectangle is given by length times width (A = L × W), we can write the equation W × (W + 4) = 252.

Step 1: Find the width (w)

Substitute l=w+4 into the first equation:

w² + 4w - 252= 0

Solve the quadratic equation using the quadratic formula:

w = −b ±√ b²−4ac​​/2a

where a=1, b=4, and c=−252.

w = −4 ± √4²−4 × 1 × −252​​/2 × 1

w = −4 ± √1024​​/2

w = −4 ± 32​/2

Since the width cannot be negative, we discard the negative solution.

Therefore, w=14 inches.

Step 2: Find the length (l) and height (h)

l=w+4=14+4=18 inches

h=w−2=14−2=12 inches

Step 3: Find the volume (V)

V = l × w × h = 18 × 14 × 12 = 3024 cubic inches.

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:

A not continueous function

Step-by-step explanation:

has a range

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:

D

Step-by-step explanation:

SEX

Complete Question

Which equation could you use to solve for x in the proportion StartFraction 4 over 5 EndFraction = StartFraction 9 over x EndFraction?

Answer:

(B) 4x = 45

Step-by-step explanation:

The given proportion is:

[Tex]\dfrac{4}{5}=\dfrac{9}{x}[/TeX]

To find an equivalent expression, let us cross multiply:

4*x=5*9

This gives:

4x=45

The correct option is B.

Answer:

7, -7

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:

1 and 1/30 gallons of paint

Step-by-step explanation:

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:

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 $n$.

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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:

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

Answer:

B. 280 units cubed

Step-by-step explanation:

Multiply all of the numbers together

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.

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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:

4mm edge 2021

Step-by-step explanation:

Answer:

100000

Step-by-step explanation:

[tex] {10}^{5} = 100000[/tex]

[tex]10 \times 10 \times 10 \times 10 \times 10 = 100000[/tex]

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:

(-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:

0.0228, Hope that useful for you.

Step-by-step explanation:

Because of " the scores...normally distributed of a mean 310 and the standard deviation 12", we can use the Central limit theorem.

That means:

([tex]P(X>334)= P(\frac{X-E(X)}{SD(X)}>\frac{334-310}{12})= P(Z>2)= 1-P(Z\leq 2)= \\1- 0.9772= 0.0228[/tex]

Answer:

solution is : x = -6 and x = 2

Step-by-step explanation:

Answer:

the answer is A

Step-by-step explanation:

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:

12.55

Step-by-step explanation:

Answer:

[tex]Area\,\,of\,\,the\,\, circle=12.56\,\, units ^2[/tex]

Step-by-step explanation:

Circumference of the circle= 12.56 units

[tex]Circumference=2\times\pi \times r[/tex]

As,

[tex]\pi =\dfrac{22}{7}=3.14[/tex]

     [tex]2\pi r=12.56\\\\2\times 3.14 \times r=12.56\\\\6.28\times r=12.56\\\\r=\dfrac{12.56}{6.28} \\\\r=2\,\,units[/tex]

Area of a circle= [tex]=\pi \times r^2[/tex]

                        [tex]=3.14\times 2^2\\\\\=3.14\times 4\\\\=12.56\,\, units ^2[/tex]

For a positive angle coterminal to 127°, select Graph B (with angle measurement of 492°). For the angle 79°, the nearest two positive coterminal angles are 439° and 799°, and the nearest two negative coterminal angles are -281° and 79°.

A positive angle coterminal to 127° can be found by adding 360° to it, since coterminal angles are separated by full rotations:

1.First Positive Coterminal Angle:

 [tex]\( 127° + 360° = 487° \)[/tex] (This is not one of the choices provided.)

2.Second Positive Coterminal Angle:

  [tex]\( 487° + 360° = 847° \)[/tex] (This is beyond the range of the options provided.)

However, if we look for a coterminal angle between 360° and 720°, we find:

 [tex]\( 127° + 360° = 487° \)[/tex], which is closer to 492° given in Graph B.

For the angle 79°:

1.Positive Coterminal Angles:

  - Add 360° for the first positive coterminal angle:

    [tex]\( 79° + 360° = 439° \)[/tex]

  - Add another 360° for the second positive coterminal angle:

 [tex]\( 79° + 720° = 799° \)[/tex]

2. Negative Coterminal Angles:

  - Subtract 360° for the first negative coterminal angle:

   [tex]\( 79° - 360° = -281° \)[/tex]

  - Subtract 720° for the second negative coterminal angle:

    [tex]\( 79° - 720° = -641° \)[/tex] (However, this is not the nearest negative coterminal angle.)

The first negative coterminal angle is already the nearest, so the second negative coterminal angle is actually the original angle itself, which is 79° (since subtracting 0° gives us the same angle).

So, the nearest two positive coterminal angles are 439° and 799°, and the nearest two negative coterminal angles are -281° and 79°.

complete question given below:

Answer:

Check the explanation

Step-by-step explanation:

Let X denotes steel ball and Y denotes diamond

[tex]\bar{x_1}[/tex] = 1/9( 50+57+......+51+53)

=530/9

=58.89

[tex]\bar{x_2}[/tex]= 1/9( 52+ 56+....+ 51+ 56)

=543/9

=60.33

difference = d =(60.33- 58.89)

=1.44

[tex]s^2=1/n\sum xi^2 - n/(n-1)\bar{x}^2[/tex]

s12 = 1/9( 502+572+......+512+532) -9/8 (58.89)2

=31686/8 - 9/8( 3468.03)

=3960.75 - 3901.53

=59.22

s1 = 7.69

s22 = 1/9( 522+ 562+....+ 512+ 562) -9/8 (60.33)2

=33295/8 - 9/8 (3640.11)

=4161.875 - 4095.12

=66.75

s2 =8.17

sample standard deviation for difference is

s=[tex]\sqrt{[(n1-1)s_1^2+ (n2-1)s_2^2]/(n1+n2-2)}[/tex]

 = [tex]\sqrt{[(9-1)*59.22+ (9-1)*66.75]/(9+9-2)}[/tex]

= [tex]\sqrt{1007.76/16}[/tex]

=7.93

sd = [tex]s*\sqrt{(1/n1)+(1/n2)}[/tex]

=[tex]7.93*\sqrt{(1/9)+(1/9)}[/tex]

=7.93* 0.47

=3.74

For 95% confidence level [tex]Z (\alpha /2)[/tex] =1.96

confidence interval is

[tex]d\pm Z(\alpha /2)*s_d[/tex]

=(1.44 - 1.96* 3.75 , 1.44+1.96* 3.75)

=(1.44 - 7.35 , 1.44 + 7.35)

=(-2.31, 8.79)

There is sufficient evidence to conclude that the two indenters produce different hardness readings.