The sun shines at a 60° angle to the ground. How long is the shadow cast
by a 20 ft tall flagpole?

Answer:

1.15

Step-by-step explanation:

The required constant of proportionality of equation P = 1.15 n is k = 1.15.

An equation is a combination of different variables, in which two mathematical expressions are equal to each other.

Given that,

The equation for the price of the kiwis is,

P = 1.15 n

Here, P, shows the price and n shows the number of kiwis.

The ratio that establishes a proportionate link between any two given values is referred to as the constant of proportionality.

Let a proportionality equation is,

y = k x,

k is the constant of proportionality here,

To find the constant of proportionality,

Compare the given equation P = 1.15 n with standard proportionality equation,

k = 1.15

The constant of proportionality is 1.15 unit.

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Answer:

a) 2.81 b)-2.33 c) -2.88 d)3.09

Step-by-step explanation:

The complete question is:

Determine the value of z so that the area under the standard normal curve

a. in the right tail is 0.025 Round your answer to two decimal places.

b. in the left tail is 0.01 Round your answer to two decimal places.

c. in the left tail is 0.002 Round your answer to two decimal places.

d. in the right tail is 0.01 Round your answer to two decimal places.

a)P( Z> ???)=  0.0025

P(Z> ???)= 1-P(Z<???)

P(Z<???)= 1-0.0025

P(Z<??)= 0.9975

From Z distribution table,

Z = 2.81

b) P(Z<???)= 0.01

From Z distribution table

Z= -2.33

c) P(Z< ??? ) = 0.002

From Z distribution table

Z= -2.88

d) P( Z> ???)=  0.001

P(Z> ???)= 1-P(Z<???)

P(Z<???)= 1-0.001

P(Z<??)= 0.999

From Z distribution table,

Z=3.09

Answer:

2nd, 4th and last one are correct answers

Step-by-step explanation:

[tex] {16}^{ \frac{5}{2} } \\ = ( {4}^{2} )^{ \frac{5}{2} } \\ = {4}^{{2} \times \frac{5}{2} } \\ = {4}^{5} \\ [/tex]

Options: 2nd, 4th and last one are correct answers

Answer:

It equals 2/5.

Step-by-step explanation:

Basically you ignore the 5 because it stays the same. All you have to add are the numerators. 1+1=2. So it would be 2/5.

A scenario:

2 pizzas cut into 5th's. Everyone ate 4 in each and kept one slice. When you add those 2 together you would be 2 slices.

Answer:

dS/dt ≈ 0.0343

Step-by-step explanation:

We are given;

C = 1.32 x 10^(5)

R = 1.3 x 10^(-2)

dR/dt = 1.0 x 10^(-5)

The function is: S = C(R² - r²)

We want to find dS/dt when r is constant.

Thus, let's differentiate since we have dR/dt;

dS/dR = 2CR

So, dS = 2CR.dR

Let's accommodate dt. Thus, divide both sides by dt to obtain;

dS/dt = 2CR•dR/dt

Plugging in the relevant values to get;

dS/dt = 2(1.32 x 10^(5))x 1.3 x 10^(-2) x 1.0 x 10^(-5)

dS/dt = 3.432 x 10^(-2)

dS/dt ≈ 0.0343

Answer:

We need to conduct a hypothesis in order to check if the mean the actual mean length of movies released during 2010-2015 is more than the data base value (a;ternative hypothesis) ,so then the system of hypothesis would be:  

Null hypothesis:[tex]\mu \leq 117[/tex]  

Alternative hypothesis:[tex]\mu > 117[/tex]  

Step-by-step explanation:

Data given and notation  

[tex]\bar X=118.44[/tex] represent the sample mean

[tex]s=8.82[/tex] represent the sample standard deviation

[tex]n=160[/tex] sample size  

[tex]\mu_o =117[/tex] represent the value that we want to test

[tex]\alpha[/tex] represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

[tex]p_v[/tex] represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean the actual mean length of movies released during 2010-2015 is more than the data base value (a;ternative hypothesis) ,so then the system of hypothesis would be:  

Null hypothesis:[tex]\mu \leq 117[/tex]  

Alternative hypothesis:[tex]\mu > 117[/tex]  

Step-by-step explanation:

[tex]1)\:v = 2 {e}^{3t} + 5 {e}^{ - 3t} \\ differentiating \: w.r.t.t \: on \: both \: sides \\ acceleration =\\ \frac{dv}{dt} = \frac{1}{dt} (2 {e}^{3t} + 5 {e}^{ - 3t} ) \\ = 2 \times \frac{1}{dt} {e}^{3t} + 5 \times \frac{1}{dt} {e}^{ - 3t} \\ \\ = 2 \times {e}^{3t} \times 3 + 5 \times {e}^{ - 3t} \times ( - 3) \\ = 6{e}^{3t} - 15{e}^{ - 3t} \\ \therefore \frac{dv}{dt} = 6{e}^{3t} - 15{e}^{ - 3t} \\ \therefore \bigg(\frac{dv}{dt} \bigg) _{t=1} = 6 {e}^{3 \times 1} - 15 {e}^ { - 3 \times 1} \\ \bigg(\frac{dv}{dt} \bigg) _{t=1} = 6 {e}^{3} - 15 {e}^ { - 3 } \\ acceleration = \\ \purple{ \boxed{ \bold{\bigg(\frac{dv}{dt} \bigg) _{t=1} = \bigg(\frac{6 {e}^{6} - 15}{ {e}^{3} } \bigg) \: m {s}^{ - 2} }}} \\ \\ 2) \: let \:s \: be \: the \: total \: distance \: travelled \\ \therefore \: s = v \times t \\ \therefore \: s= (2 {e}^{3t} + 5 {e}^{ - 3t}) \times t \\ \therefore \: (s)_{t=2} = (2 {e}^{3 \times 2} + 5 {e}^{ - 3 \times 2}) \times 2 \\ \therefore \: (s)_{t=2} = (2 {e}^{6} + 5 {e}^{ - 6}) \times 2 \\ \therefore \: (s)_{t=2} = 4 {e}^{6} + 10{e}^{ - 6} \\ \red{ \boxed{ \bold{\therefore \: (s)_{t=2} = \bigg(\frac{4 {e}^{12} + 10}{{e}^{ 6}} \: \bigg)m}}}\\ [/tex]

Answer:

[tex] \epsilon = Y -X\beta[/tex]

And the expected value for [tex] E(\epsilon) = 0[/tex] a vector of zeros and the covariance matrix is given by:

[tex] Cov (\epsilon) = \sigma^2 I[/tex]

So we can see that the error terms not have a variance of 0. We can't assume that the errors are assumed to have an increasing mean, and we other property is that the errors are assumed independent and following a normal distribution so then the best option for this case would be:

The regression model assumes the errors are normally distributed.

Step-by-step explanation:

Assuming that we have n observations from a dependent variable Y , given by [tex] Y_1, Y_2,....,Y_n[/tex]

And for each observation of Y we have an independent variable X, given by [tex] X_1, X_2,...,X_n[/tex]

We can write a linear model on this way:

[tex] Y = X \beta +\epsilon [/tex]

Where [tex]\epsilon_{nx1}[/tex] i a matrix for the error random variables, and for this case we can find the error ter like this:

[tex] \epsilon = Y -X\beta[/tex]

And the expected value for [tex] E(\epsilon) = 0[/tex] a vector of zeros and the covariance matrix is given by:

[tex] Cov (\epsilon) = \sigma^2 I[/tex]

So we can see that the error terms not have a variance of 0. We can't assume that the errors are assumed to have an increasing mean, and we other property is that the errors are assumed independent and following a normal distribution so then the best option for this case would be:

The regression model assumes the errors are normally distributed.

Final answer:

The best statement is that the regression model assumes the errors are normally distributed. In regression analysis, it is essential that the errors are independent, normally distributed, and have constant variance, which supports the validity of the model's predictions.

Explanation:

The correct statement among the provided options is that the regression model assumes the errors are normally distributed. This is a fundamental assumption of linear regression analysis, where it's assumed that the residuals or errors of the regression model are randomly distributed about an average of zero. These error terms must be independent, normal, and have constant variance (homoscedasticity) across all levels of the independent variables.

According to the theoretical foundation of regression, it is not assumed that errors have an increasing mean, nor that they have zero variance, as some diversity in errors is expected. Additionally, the assumption that errors are indeed dependent would violate the principles of ordinary least squares (OLS) regression, making the model invalid.

Normality, independence, and equal variance are key premises in regression analysis to ensure the validity of the model's inferences. Indeterminate errors that affect the dependent variable 'y' are assumed to be normally distributed and independent of the independent variable 'x'. This maintains the integrity of the regression model.

Answer:

1. Sample Mean = 2.50

2. Sample Variance = 1.35

3. Standard Deviation = 1.16

Step-by-step explanation:

Note: As this question is not complete, similar can be found on internet where following are asked to calculate and here I will calculate these as well.

1. Sample Mean.

2. Sample Variance.

3. Sample Standard Deviation.

So, for sample mean to calculate from the given data points. We need to apply to following formula to calculate sample mean.

Sample Mean = (X1 + X2 + .... + XN) divided by Total number of data points.

Sample Mean = (0.17 + 1.94 + 2.62 + 2.35 + 3.05 + 3.15 + 2.53 + 4.81 + 1.92) divided by (9).

Sample Mean = 2.50

Likewise, from the answer of sample mean we can calculate sample variance by following steps.

Solution:

1.      Subtract the obtained mean from each of the data point given.

       (0.17 - 2.50) = -2.33

       (1.94 - 2.50) = -0.56

       (2.62 - 2.50) = 0.12

       (2.35 - 2.50) = -0.15

       (3.05 - 2.50) = 0.55

       (3.15 - 2.50) = 0.65

       (2.53 - 2.50) = 0.03

       (4.81 - 2.50) =  2.31

       (1.92 - 2.50) = -0.58

2.      Square each of the differences obtained in step 1.

            [tex]-2.33^{2}[/tex] = 5.43

            [tex]-0.56^{2}[/tex] = 0.31

               [tex]0.12^{2}[/tex] = 0.01

            [tex]-0.15^{2}[/tex] = 0.02

              [tex]0.55^{2}[/tex]  = 0.30

              [tex]0.65^{2}[/tex]  = 0.42

              [tex]0.03^{2}[/tex] = 0.01

              [tex]2.31^{2}[/tex] = 5.34    

           [tex]-0.58^{2}[/tex] = 0.34    

3.  Sum all of these squares.

(5.43 + 0.31 + 0.01 + 0.02 + 0.30 + 0.42 + 0.01 + 5.34 + 0.34) = 12.18

4. Divide 12.18 by (n-1), where n = 9.

Sample Variance =  12.18/9 = 1.35

Now, by using sample variance, we can calculate standard deviation with easy simple steps.

So, in order to find standard deviation, we just need to take the square root of sample variance that we have already calculated above.

Standard Deviation = [tex]\sqrt{1.35}[/tex] = 1.16

Good Luck!

Answer:

Step-by-step explanation:

Sample mean = (14 + 10 + 13 + 10 + 11)/5 = 11.6

Sample standard deviation,s = √(summation(x - mean)/n

Summation(x - mean) = (14 - 11.6)^2 + (10 - 11.6)^2 + (13 - 11.6)^2 + (10 - 11.6)^2 + (11 - 11.6)^2 = 13.2

s = √(13.2/5) = 1.62

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

µ = 8.9

For the alternative hypothesis,

µ > 8.9

it is a right tailed test because of >.

Since the number of samples is 5 and no population standard deviation is given, the distribution is a student's t.

Since n = 5,

Degrees of freedom, df = n - 1 = 5 - 1 = 4

t = (x - µ)/(s/√n)

Where

x = sample mean = 11.6

µ = population mean = 8.9

s = samples standard deviation = 1.62

t = (11.6 - 8.9)/(1.62/√5) = 3.73

We would determine the p value using the t test calculator. It becomes

p = 0.01

Since alpha, 0.05 > than the p value, 0.01, then we would reject the null hypothesis. Therefore, At a 5% level of significance, the sample data showed significant evidence that mean anger score of the marines is greater than that of college men.

Therefore, the marines's negative moods increased and it is higher than that of college men.

Answer:

243

Step-by-step explanation:

3x3=9

9x3=27

27x3=81

81x3=243

Answer:

We can select a lower confidence level and increase the sample size.

Step-by-step explanation:

The precision of the confidence interval depends on the margin of error ME = Zcritical * Sqrt[(p(1-p)/n]

In this Zcritical value is in the numerator. Z critical decreases as Confidence level decreases. (Zc for 99% = 2.576, Zc for 95% is 1.96, Zc for 90% = 1.645). Therefore decreasing the Confidence level decreases ME.

Also we see that sample size n is in the denominator. So the ME decreases as sample size increases.

Therefore, We can select a lower confidence level and increase the sample size.

The best option that could be used to produce higher precision in our estimates of the population proportion is;

Option C; We can select a lower confidence level and increase the sample size.

Formula for confidence interval is given as;

CI = p^ ± z√(p^(1 - p^)/n)

Where;

p^ is the sample proportion

z is the critical value at given confidence level

n is the sample size

Now, the margin of error from the CI formula is:

MOE = z√(p^(1 - p^)/n)

Now, the lesser the margin of error, the narrower the confidence interval and thus the more precise is the estimate of the population proportion.

Now, looking at the formula for MOE, two things that could change aside the proportion is;

z and n.

Now, the possible values of z are;

At CL of 99%; z = 2.576

At CL of 95%; z = 1.96

At CL of 90%; z = 1.645

We can see that the higher the confidence level, the higher the critical value and Invariably the higher the MOE.

Thus, to have a narrow CI, we need to use a lower value of CL and increase the sample size.

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Answer:

The probability that the sample mean would differ from the population mean by greater than 0.8 kg is P=0.3843.

Step-by-step explanation:

We have a population with mean 60 kg and a variance of 100 kg.

We take a sample of n=118 individuals and we want to calculate the probability that the sample mean will differ more than 0.8 from the population mean.

This can be calculated using the properties of the sampling distribution, and calculating the z-score taking into account the sample size.

The sampling distribution mean is equal to the population mean.

[tex]\mu_s=\mu=60[/tex]

The standard deviation of the sampling distribution is equal to:

[tex]\sigma_s=\sigma/\sqrt{n}=\sqrt{100}/\sqrt{118}=10/10.86=0.92[/tex]

We have to calculate the probability P(|Xs|>0.8). The z-scores for this can be calculated as:

[tex]z=(X-\mu_s)/\sigma_s=\pm0.8/0.92=\pm0.87[/tex]

Then, we have:

[tex]P(|X_s|>0.8)=P(|z|>0.87)=2*(P(z>0.87)=2*0.19215=0.3843[/tex]

Given Information:

Population = n = 800

Probability = p = 68% = 0.68

Answer:

We can say with 68% confidence that 568 lies in the range of (518, 570) therefore, it would not be unusual to get 568 consumers who recognize the Dull Computer Company name.

Step-by-step explanation:

Let us first find out the mean and standard deviation.

mean = μ = np

μ = 800*0.68

μ = 544

standard deviation = σ = √np(1-p)

σ = √800*0.68(1-0.68)

σ = 13.2

we know that 68% of data fall within 2 standard deviations from the mean

μ ± 2σ = 544-2*13.2, 544+2*13.2

μ ± 2σ = 544-26.4 , 544+26.4

μ ± 2σ =  517.6, 570.4

μ ± 2σ = 518, 570

We can say with 68% confidence that 568 lies in the range of (518, 570) therefore, it would not be unusual to get 568 consumers who recognize the Dull Computer Company name.

Answer:

y = x+1

Step-by-step explanation:

You should use the point-slope form of y-y1 = m(x-x1) to solve this.

First you use the slope formula to find your slope. (y2-y1)/(x2-x1)

(-1 - -3)/(-2 - -4) = 2/2 = 1

With the slope of 1 found, just plug a point into the point-slope form.

y-(-1) = (1)(x-(-2)) ->

y+1 = x+2 ->

y = x+1

y=x+1 is the equation of a line that passes through (-2, -1) and (-4, -3)

The slope of the line is the ratio of the rise to the run, or rise divided by the run. It describes the steepness of line in the coordinate plane.

The slope intercept form of a line is y=mx+b, where m is slope and b is the y intercept.

The slope of line passing through two points (x₁, y₁) and (x₂, y₂) is

m=y₂-y₁/x₂-x₁

The slope of line passing through  (-2, -1) and (-4, -3)

m = -3+1/-4+2

=-2/-2= 1

Now let us find y intercept

-1=1(-2)+b

-1=-2+b

b=1

Hence, y=x+1 is the equation of a line that passes through (-2, -1) and (-4, -3)

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Answer:

The critical value of t at 0.01 level of significance is 2.66.

Step-by-step explanation:

The hypothesis for the two-tailed population mean can be defined as:

H₀: μ = μ₀ vs. H₀: μ ≠ μ₀

It is provided that the population standard deviation is not known.

Since there is no information about the population standard deviation, we will use a t-test for single mean.

The test statistic is defined as follows:

[tex]t_{cal.}=\frac{\bar x-\mu}{s/\sqrt{n}}\sim t_{\alpha/2, (n-1)}[/tex]

The information given is:

n = 55

α = 0.01

Compute the critical value of t as follows:

[tex]t_{\alpha/2, (n-1)}= t_{0.01/2, (55-1)}=t_{0.005, 54}=2.66[/tex]

*Use a t-table for the value.

If the desired degrees of freedom are not provided consider he next highest degree of freedom.

Thus, the critical value of t at 0.01 level of significance is 2.66.

The equation in slope-intercept from is [tex]y = \frac{9}{8}x - \frac{3}{8}[/tex].

Step-by-step explanation:

Given that,

x = 8t+3

y = 9t+ 3

Now, getting the value of t on both side to compare the equations.

[tex]x = 8t+3[/tex]

[tex]x -3 = 8t[/tex]

[tex]t = \frac{x-3}{8}[/tex]

also

[tex]y = 9t+ 3[/tex]

[tex]t= \frac{y-3}{9}[/tex]

Now, by comparing the values of t, we get

[tex]\frac{x-3}{8} = \frac{y-3}{9}[/tex]

[tex]9x - 27 = 8y -24[/tex]

[tex]9x - 3 = 8y[/tex]

[tex]y = \frac{9}{8}x - \frac{3}{8}[/tex]

Hence, the equation in slope-intercept from is [tex]y = \frac{9}{8}x - \frac{3}{8}[/tex].

Answer:

Step-by-step explanation:

REcall the following definition of induced operation.

Let * be a binary operation over a set S and H a subset of S. If for every a,b elements in H it happens that a*b is also in H, then the binary operation that is obtained by restricting * to H is called the induced operation.

So, according to this definition, we must show that given two matrices of the specific subset, the product is also in the subset.

For this problem, recall this property of the determinant. Given A,B matrices in Mn(R) then det(AB) = det(A)*det(B).

Case SL2(R):

Let A,B matrices in SL2(R). Then, det(A) and det(B) is different from zero. So

[tex]\text{det(AB)} = \text{det}(A)\text{det}(B)\neq 0[/tex].

So AB is also in SL2(R).

Case GL2(R):

Let A,B matrices in GL2(R). Then, det(A)= det(B)=1 is different from zero. So

[tex]\text{det(AB)} = \text{det}(A)\text{det}(B)=1\cdot 1 = 1[/tex].

So AB is also in GL2(R).

With these, we have proved that the matrix multiplication over SL2(R) and GL2(R) is an induced operation from the matrix multiplication over M2(R).

The anchor drops 63 meters in the first 5 seconds.

The given function E(t) = -2.4t + 75 represents the elevation of the ship's anchor relative to the water's surface at any given time t. To find how far the anchor drops every 5 seconds, substitute t = 5 into the function:

E(5) = -2.4(5) + 75

E(5) = -12 + 75

E(5) = 63

Therefore, after 5 seconds, the anchor has dropped 63 meters relative to the water's surface. This indicates the change in elevation during this time period. The negative coefficient of t in the function implies a downward motion, and the constant term (75) represents the initial height of the anchor above the water. So, the anchor drops 63 meters in the first 5 seconds.

Answer:

=2p(-2p-5)

Step-by-step explanation:

2p(–3p2 + 4p – 5)

=2p(4p-6p-5)

=2p(-2p-5)

Answer:

598

Step-by-step explanation:

Answer:

598

Step-by-step explanation:

[tex]23[14 + 4(36 \div 12)] \\ = 23[14 + 4(3)] \\ = 23[14 + 12] \\ = 23 \times 26 \\ = 598 \\ [/tex]

Answer:

86.96

Step-by-step explanation:

not sure what is height or base.

Answer:

86.96

Step-by-step explanation:

Answer:

x = -7

Step-by-step explanation:

2x + 3 = x - 4

→ Minus x from both sides to isolate -4

x + 3 = -4

→ Minus 3 from both sides to isolate x and henceforth find the value of x

x = -7

Answer:

  1

Step-by-step explanation:

You "complete the square" by adding the square of half the x-term coefficient. Here, that is ...

 ((-2)/2)² = 1 . . . . value added to complete the square

If you want to keep 0 on the right, you must also subtract this value:

  x² -2x -36 = 0

  x² -2x +1 -36 -1 = 0 . . . . . . add and subtract 1 on the left

  (x -1)² -37 = 0 . . . . . . . . . . . written as a square

Answer:

Yes, it is convincing evidence to conclude that the proportion of red cars that drive too fast on this

highway is greater than the proportion of non-red cars that drive too fast.

Step-by-step explanation:

From the, we wish to first test;

H0; P_r - P_o = 0

And; H0; P_r - P_o > 0

Where; P_r and P_o are the proportion of red cars and other cars, respectively, who are driving too fast.

We will use a significance level of a = 0.05.

Thus;

The procedure is a two-sample z-test for the difference of proportions.

For, Random Conditions: The policemen chose cars randomly by rolling a die.

10%: We can safely assume that the number of cars driving past the rest area is essentially infinite, so the 10% restriction does not apply.

Large counts: The number of successes and failures in the two groups are 18, 10, 75, and 130—all of which are at least 10.

So, P_r = 18/28 = 0.64

P_o = 75/205 = 0.37

P_c = (18 + 75)/(28 + 205) = 0.4

Thus:

z = [(0.64 - 0.37) - 0]/√[[(0.4 x 0.6)/28] + [(0.4 x 0.6)/205]]

z = 2.73

From the one tailed z-score calculator online, I got P value = 0.003167

Thus, the P-value of 0.0032 is less than a = 0.05, so we reject H0. We

have sufficient evidence to conclude that the proportion of red cars that drive too fast on this

highway is greater than the proportion of non-red cars that drive too fast.

Answer:

The sampling distribution of the sample mean is:

[tex]\bar X\sim N(\mu_{\bar x}=2.4,\ \sigma_{\bar x}=0.40)[/tex]

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Then, the mean of the distribution of sample means is given by,

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

And the standard deviation of the distribution of sample means is given by,

[tex]\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}[/tex]

Let X = number of cars running a red light in a day, at a given intersection.

The information provided is:

[tex]E(X)=\mu=2.4\\SD(X)=\sigma=4\\n=100[/tex]

The sample selected is quite large, i.e. n = 100 > 30.

The Central limit theorem can be used to approximate the sampling distribution of the sample mean number of cars running a red light in a day, by the Normal distribution.

The mean of the sampling distribution of the sample mean is:

[tex]\mu_{\bar x}=\mu=2.4[/tex]

The standard deviation of the sampling distribution of the sample mean is:

[tex]\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{4}{\sqrt{100}}=0.40[/tex]

The sampling distribution of the sample mean is:

[tex]\bar X\sim N(\mu_{\bar x}=2.4,\ \sigma_{\bar x}=0.40)[/tex]

Final answer:

The sampling distribution of the sample mean is approximately normally distributed, with a mean of 2.4 cars and a standard deviation of 0.4 cars.

Explanation:

The sampling distribution of the sample mean can be described as follows:

Answer:

Length of the chord: 24 units

Angles are equal

Step-by-step explanation:

Drop a Perpendicular from the centre onto the chord. It will divide the chord into two equal parts, d units each

d² + 16² = r²

d² = 20² - 16²

d² = 144

d = 12

Chord = 2d = 24 units

Measure of central angle and the corresponding chord related to the measure of the arc intercepted by the chord are the same, they're equal

Answer:

The entire chord length is 12*2 = 24

The degree measure of a minor arc is equal to the measure of the central angle that intercepts it.

Step-by-step explanation:

We can make a right triangle to solve for 1/2 of the chord length.  The hypotenuse is 20 and one of the legs is 16

a^2+b^2 = c^2

16^2 + b^2 = 20^2

256 +b^2 = 400

Subtract 256 from each side

b^2 = 400-256

b^2 =144

Take the square root of each side

b = 12

That means 1/2 of the chord length is 12

The entire chord length is 12*2 = 24

Answer:

The proportion of chips that do not fail in the first 1000 hours of their use is 52%.

Step-by-step explanation:

The claim made by the company is that 52% of the chips do not fail in the first 1000 hours of their use.

A quality control manager wants to test the claim.

A one-proportion z-test can be used to determine whether the proportion of chips do not fail in the first 1000 hours of their use is 52% or not.

The hypothesis can be defined as:

H₀: The proportion of chips do not fail in the first 1000 hours of their use is 52%, i.e. p = 0.52.

Hₐ: The proportion of chips do not fail in the first 1000 hours of their use is different from 52%, i.e. p ≠ 0.52.

The information provided is:

[tex]\hat p=0.49\\n=900\\\alpha =0.02[/tex]

The test statistic value is:

[tex]z=\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}=\frac{0.49-0.52}{\sqrt{\frac{0.52(1-0.52)}{900}}}=-1.80[/tex]

The test statistic value is -1.80.

Decision rule:

If the p-value of the test is less than the significance level then the null hypothesis will be rejected and vice-versa.

Compute the p-value as follows:

[tex]p-value=2\times P(Z<-1.80)\\=2\times 0.03593\\=0.07186[/tex]

*Use a z-table.

The p-value = 0.07186 > α = 0.02.

The null hypothesis was failed to be rejected at 2% level of significance.

Conclusion:

The proportion of chips that do not fail in the first 1000 hours of their use is 52%.

Answer:

Step-by-step explanation:

1/8

1/2 multiply 1/4