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givi [52]
2 years ago
9

Polly bought a new collar and leash for her dog. The total was $7.50. She paid with a ten-dollar bill.

Mathematics
1 answer:
Gelneren [198K]2 years ago
5 0
She got back $2.50 hope this helped
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What expression is not equivalent to 1/343
GREYUIT [131]
1 / 343
This is just equal to 
1 / 7^3 = 7^-3

A. 7^-1 * 7^-2 = 7^(-1 + -2) = 7^-3
B. 7^7 * 7^-10 = 7^( 7+ -10) = 7^-3
C. 7^-2 * 7^-5 = 7^(-2 + -5) = 7^-7
D. 7^-5 * 7^2 = 7^(-5 + 2) = 7^-3

So the correct answer is letter C. 7^-2 * 7^-5
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8\sqrt{2}

Step-by-step explanation:

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4 years ago
An educator claims that the average salary of substitute teachers in school districts is less than $60 per day. A random sample
valentina_108 [34]

Answer:

t=\frac{58.875-60}{\frac{5.083}{\sqrt{8}}}=-0.626    

The degrees of freedom are given by:

df=n-1=8-1=7  

The p value would be given by:

p_v =P(t_{(7)}  

Since the p value is higher than 0.1 we have enough evidence to FAIl to reject the null hypothesis and we can't conclude that the true mean is less than 60

Step-by-step explanation:

Information given

60, 56, 60, 55, 70, 55, 60, and 55.

We can calculate the mean and deviation with these formulas:

\bar X= \frac{\sum_{i=1}^n X_i}{n}

s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

Replacing we got:

\bar X=58.875 represent the mean

s=5.083 represent the sample standard deviation for the sample  

n=8 sample size  

\mu_o =60 represent the value that we want to test

\alpha=0.1 represent the significance level

t would represent the statistic

p_v represent the p value

Hypothesis to test

We want to test if the true mean is less than 60, the system of hypothesis would be:  

Null hypothesis:\mu \geq 60  

Alternative hypothesis:\mu < 60  

The statistic would be given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

Replacing the info we got:

t=\frac{58.875-60}{\frac{5.083}{\sqrt{8}}}=-0.626    

The degrees of freedom are given by:

df=n-1=8-1=7  

The p value would be given by:

p_v =P(t_{(7)}  

Since the p value is higher than 0.1 we have enough evidence to FAIl to reject the null hypothesis and we can't conclude that the true mean is less than 60

3 0
3 years ago
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