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ElenaW [278]
3 years ago
8

At a summer camp there is one counselor for every 5 campers.

Mathematics
1 answer:
Flura [38]3 years ago
4 0

Answer:

Then if there were 10 campers, there would be 5 counselors

Step-by-step explanation:

For every five campers, there is one counselor. Divide the number of campers by 2 two find how many counselors there are.

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A sandwich decrease from $8 to $6 what is the percentage decrease
kati45 [8]

Answer:

25%

Step-by-step explanation:

First find the difference in price= $8-$6 = $2

Now, find percentage decrease in the price of sandwich.

Percentage decrease= (difference in price/ original price) *100

                                   = \frac{2}{8} *100

                                   =\frac{200}{8}

Percentage decrease=25%

                                                   

4 0
3 years ago
Will give brainliest lots of points A is the midpoint of LB,Lb =6x-17 and AB=2x+3
NARA [144]
A is the mid point ==> AL = AB = 2AB
6x-17=2(2x+3) => 6x -17 = 4x +6 => 2x = 23 =>x=23/2
then just plug x into LB and AB
5 0
3 years ago
Please help I’ll give brainliest
bekas [8.4K]

Answer:

Total area:75

Step-by-step explanation:

Rectangle area=6×5=30

triangle area=(18×5)/2=45

total area: 30+45=75

5 0
3 years ago
Read 2 more answers
The question is above in the picture​
anzhelika [568]
Yea his answer is right ^^
6 0
3 years ago
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
3 years ago
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