What is the length of segment SR? units

A new club sent out 172 coupons to boost sales for next year's memberships. They provided 3 times as many to potential members

cricket20 [7]

Answer: 172: 4= 43

43× 3= 129

172-129= 43

Odp.Wysłali 43 kupony do istniejących członków.

Step-by-step explanation:1.bo wysłali 3×wiecej i jeszcze są istniejący członkowie

2.ile wyslali potencialnym

3.i ile wyslali do istniejacych

4 0

2 years ago

What is the quotient of 2x^(3)+3x^(2)+5x divided by x^(2)+x+1

Anit [1.1K]

<h3><u>Answer</u><u>:</u></h3>

$\large{ \underline{ \boxed{ \green{ \tt{2x + 1 + \frac{2x - 5}{ {x}^{2} + x + 1} }}}}}$

7 0

2 years ago

What Is 1 - 1/3? <br> Fraction Skills

ANEK [815]

Answer:

2/3

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

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.

8 0

2 years ago

The Tennessee Tourism Institute (TTI) plans to sample information center visitors entering the state to learn the fraction of vi

baherus [9]

Answer: 2185

Step-by-step explanation:

Let p be the proportion of visitors are campers.

Given : The Tennessee Tourism Institute (TTI) plans to sample information center visitors entering the state to learn the fraction of visitors who plan to camp in the state.

The prior proportion of visitors are campers : p=0.35

Allowable error : E= 2%= 0.02

We know that the z-value for 95% confidence = $z_c=1.96$