?/15•3/5=1/25 what does ? equal.

Last year, 20,820 people attended a kite festival. This year, 19779 people attended it. If attendance continues to decrease by t

ICE Princess25 [194]

Well you start by dividing 19779 by 20820 to see the percent of people compared to last year (it being 95%). Then you take 19,779 and multiply it by the same percentage, or 0.95. That means that there should be 18790 (it's technically 18790.1 but since that's people you round to the nearest whole number).

4 0

2 years ago

Use the exponential regression equation that best fits the data (10,4). (12,20), (13,35), and (16,300) to estimate y when x = 15

Aleonysh [2.5K]

Answer:

tbh idc

Step-by-step explanation:

5 0

2 years ago

If M is the midpoint of SR, and SR=24, what is the length of SM?

Montano1993 [528]

Do you have a picture you can upload? It’s most likely 24/2 = 12

7 0

2 years ago

Solve the equation <br>-5x+x over 2 = 9

Alekssandra [29.7K]

Ok soo

-5x+x/2=9

- 4x/2=9

-(2^2-1 *x)=9

2x/2=- 9/2

x=- 9/2

x= -4 1/2

5 0

2 years ago

Read 2 more answers

A survey collected data on annual credit card charges in seven different categories of expenditures: transportation, groceries,

Andru [333]

Answer:

A. Null and alternative hypothesis:

$H_0: \mu_d=0\\\\H_a:\mu_d\neq 0$

B. Yes. At a significance level of 0.05, there is enough evidence to support the claim that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

P-value = 0.00002

C. As the difference is calculated as (population 1 − population 2), being population 1: groceries and population 2: dinning out, and knowing there is evidence that the true mean difference is positive, we can say that the groceries annual credit card charge is higher than dinning out annual credit card charge.

The point estimate is the sample mean difference d=$840.

The 95% confidence interval for the mean difference between the population means is (490, 1190).

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

Then, the null and alternative hypothesis are:

$H_0: \mu_d=0\\\\H_a:\mu_d\neq 0$

The significance level is 0.05.

The sample has a size n=42.

The sample mean is M=840.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=1123.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{1123}{\sqrt{42}}=173.2827$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{840-0}{173.2827}=\dfrac{840}{173.2827}=4.848$

The degrees of freedom for this sample size are:

$df=n-1=42-1=41$

This test is a two-tailed test, with 41 degrees of freedom and t=4.848, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=2\cdot P(t>4.848)=0.00002$

As the P-value (0.00002) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

We have to calculate a 95% confidence interval for the mean difference.

The t-value for a 95% confidence interval and 41 degrees of freedom is t=2.02.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=2.02 \cdot 173.283=350$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 840-350=490\\\\UL=M+t \cdot s_M = 840+350=1190$

The 95% confidence interval for the mean difference is (490, 1190).

7 0

2 years ago