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Mademuasel [1]
3 years ago
9

(URGENT) need answer for today​

Mathematics
1 answer:
Harman [31]3 years ago
8 0

<u>Given</u>:

Given that a basketball court is in the shape of a rectangle that is 94 feet long and 50 feet wide.

We need to determine the length of the diagonal to the court.

<u>Length of the diagonal:</u>

The length of the diagonal can be determined using the Pythagorean theorem.

Thus, we have;

d^2=l^2+w^2

where d is the diagonal, l is the length and w is the width of the rectangle.

Substituting the values, we have;

d^2=94^2+50^2

d^2=8836+2500

d^2=11336

d=\sqrt{11336}

d=106.5

Thus, the length of the diagonal of the court is 106.5 feet.

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$85 ÷ 5 = 17

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Anastaziya [24]

Answer:

im not writing all that lol

Step-by-step explanation:

So 1 would be 100%, 10/10 and 0 would be, 0/10 and 0% so for 50% for example would be 5/10, 0.5 and so on..

then just think of things that have a 100% chance and all the other percentages... example The sun rising tomorrow. and no chance 0% example the school flying away

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A school club is comparing prices at two movie theaters. Theater 1 charges $7.50 for a movie ticket and $5.75 per item at the co
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Step-by-step explanation:

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3 years ago
At Western University the historical mean of scholarship examination scores for freshman applications is 900. A historical popul
vampirchik [111]

Answer:

a) Null Hypothesis: \mu =900

Alternative hypothesis: \mu \neq 900

b) The 95% confidence interval would be given by (910.05;959.95)    

c) Since we confidence interval not ocntains the value of 900 we fail to reject the null hypothesis that the true mean is 900.

d) z=\frac{935 -900}{\frac{180}{\sqrt{200}}}=2.750

Since is a bilateral test the p value is given by:

p_v =2*P(Z>2.750)=0.0059

Step-by-step explanation:

a. State the hypotheses.

On this case we want to check the following system of hypothesis:

Null Hypothesis: \mu =900

Alternative hypothesis: \mu \neq 900

b. What is the 95% confidence interval estimate of the population mean examination  score if a sample of 200 applications provided a sample mean x¯¯¯= 935?

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X=935 represent the sample mean for the sample  

\mu population mean (variable of interest)

\sigma=180 represent the population standard deviation

n=200 represent the sample size  

The confidence interval for the mean is given by the following formula:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}} (3)  

The mean calculated for this case is \bar X=3278.222

The sample deviation calculated s=97.054

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that z_{\alpha/2}=1.96

Now we have everything in order to replace into formula (1):

935-1.96\frac{180}{\sqrt{200}}=910.05    

935+1.96\frac{180}{\sqrt{200}}=959.95    

So on this case the 95% confidence interval would be given by (910.05;959.95)    

c. Use the confidence interval to conduct a hypothesis test. Using α= .05, what is your  conclusion?

Since we confidence interval not ocntains the value of 900 we fail to reject the null hypothesis that the true mean is 900.

d. What is the p-value?

The statistic is given by:

z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

If we replace we got:

z=\frac{935 -900}{\frac{180}{\sqrt{200}}}=2.750

Since is a bilateral test the p value is given by:

p_v =2*P(Z>2.750)=0.0059

So then since the p value is less than the significance we can reject the null hypothesis at 5% of significance.

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3 years ago
SEAGULLS A biologist studying seagulls estimates that 5% of the seagulls in a flock have been banded. If 22 of the seagulls have
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Answer:

440 seagulls

Step-by-step explanation:

Total number of seagulls in flock = Number of seagulls have banded / Percentage of seagulls have banded

Total number of seagulls in flock = 22 / 5%

Total number of seagulls in flock = 22 / 0.05

Total number of seagulls in flock = 440 seagulls

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2 years ago
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