Ask question

Ask question

All categories

3 years ago

6

A running track in the shape of an oval is shown. The ends of the track form semicircles.What is the perimeter of the inside of

the track? (π = 3.14) 260.00 m 347.92 m 372.00 m 435.84 m

2 answers:

morpeh [17]3 years ago

6 0

Answer:

435.84 is the answer.

Step-by-step explanation:

omeli [17]3 years ago

5 0

Two semi circles would equal one circle.

The perimeter of a circle is PI x diameter.

The diameter of the circle is shown as 56 m

The perimeter of the 2 ends = 56 x 3.14 = 175.84

The length of the inside of the track is given as 130 m, and there are 2 sides so 130 + 130 = 260 m.

Total perimeter = 260 + 175.84 = 435.84 m.

You might be interested in

How many zeros are at the end of the number 2^12 x 5^7 ?

Katena32 [7]

Answer:

7 zeroes

Step-by-step explanation:

I HOPE it will help you

4 0

3 years ago

Read 2 more answers

Suppose small aircraft arrive at a certain airport according to a Poisson process with rate α = 8 per hour, so that the number o

Ksenya-84 [330]

Answer:

Step-by-step explanation:

Step1:

We have Suppose small aircraft arrive at a certain airport according to a Poisson process with rate α =8 per hour, so that the number of arrivals during a time period of t hours is a Poisson rv with parameter μ = 8t

Step2:

Let “X” the number of small aircraft that arrive during time t and it follows poisson distribution parameter “”

The probability mass function of poisson distribution is given by

P(X) = , x = 0,1,2,3,...,n.

Where, μ(mean of the poisson distribution)

a).

Given that time period t = 1hr.

Then,μ = 8t

= 8(1)

= 8

Now,

The probability that exactly 6 small aircraft arrive during a 1-hour period is given by

P(exactly 6 small aircraft arrive during a 1-hour period) = P(X = 6)

Consider,

P(X = 6) =

=

=

= 0.1219.

Therefore,The probability that exactly 6 small aircraft arrive during a 1-hour period is 0.1219.

1).P(At least 6) = P(X 6)

Consider,

P(X 6) = 1 - P(X5)

= 1 - {+++++}

= 1 - (){+++++}

= 1 - (0.000335){+++++}

= 1 - (0.000335){1+8+32+85.34+170.67+273.07}

= 1 - (0.000335){570.08}

= 1 - 0.1909

= 0.8090.

Therefore, the probability that at least 6 small aircraft arrive during a 1-hour period is 0.8090.

2).P(At least 10) = P(X 10)

Consider,

P(X 10) = 1 - P(X9)

= 1 - {+++++

5 0

3 years ago

Suppose that a random sample of 10 newborns had an average weight of 7.25 pounds and sample standard deviation of 2 pounds. a. T

frosja888 [35]

Answer:

$z=\frac{7.25-7.5}{\frac{1.4}{\sqrt{10}}}=-0.565$

$p_v =P(Z$

a) If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true average is not significantly less than 7.5.

b) $\chi^2 =\frac{10-1}{1.96} 4 =18.367$

$p_v =P(\chi^2 >18.367)=0.0311$

If we compare the p value and the significance level provided we see that $p_v >\alpha$ so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population variance is not significantly higher than 1.96.

Step-by-step explanation:

Assuming this info: "Suppose birth weights follow a normal distribution with mean 7.5 pounds and standard deviation 1.4 pounds"

1) Data given and notation

$\bar X=7.25$ represent the sample mean

$s=1.2$ represent the sample standard deviation

$\sigma=1.4$ represent the population standard deviation

$n=10$ sample size

$\mu_o =7.5$ represent the value that we want to test

$\alpha=0.05,0.01$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is less than 7.5, the system of hypothesis would be:

Null hypothesis: $\mu \geq 7.5$

Alternative hypothesis: $\mu < 7.5$

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

3) Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{7.25-7.5}{\frac{1.4}{\sqrt{10}}}=-0.565$

4)P-value

Since is a left tailed test the p value would be:

$p_v =P(Z$

5) Conclusion

Part a

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true average is not significantly less than 7.5.

Part b

A chi-square test is "used to test if the variance of a population is equal to a specified value. This test can be either a two-sided test or a one-sided test. The two-sided version tests against the alternative that the true variance is either less than or greater than the specified value"

$n=10$ represent the sample size

$\alpha=0.01$ represent the confidence level

$s^2 =4$ represent the sample variance obtained

$\sigma^2_0 =1.96$ represent the value that we want to test

Null and alternative hypothesis

On this case we want to check if the population variance increase, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 \leq 1.96$

Alternative hypothesis: $\sigma^2 >1.96$

Calculate the statistic

For this test we can use the following statistic:

$\chi^2 =\frac{n-1}{\sigma^2_0} s^2$

And this statistic is distributed chi square with n-1 degrees of freedom. We have eveything to replace.

$\chi^2 =\frac{10-1}{1.96} 4 =18.367$

Calculate the p value

In order to calculate the p value we need to have in count the degrees of freedom , on this case 9. And since is a right tailed test the p value would be given by:

$p_v =P(\chi^2 >18.367)=0.0311$

In order to find the p value we can use the following code in excel:

"=1-CHISQ.DIST(18.367,9,TRUE)"

Conclusion

If we compare the p value and the significance level provided we see that $p_v >\alpha$ so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population variance is not significantly higher than 1.96.

4 0

3 years ago

One 8-ounce can of soda has 228 calories. If Shaun drinks the big 28-ounce size from the local mini-mart, how many calories wil

Rama09 [41]

Answer:

324.7 calories

6 0

2 years ago

Read 2 more answers

Who will win, maddy or cassie?

Nostrana [21]

Answer:

ME

Step-by-step explanation:

ME BC IMA GOD!

7 0

2 years ago