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3 years ago

14

Hank spent $635.71 on groceries. Peggy spent $293.96 on groceries. They decide to split the entire grocery bill, which is $929.6

7. How much does Peggy still owe Hank if they split the grocery bills equally?

1 answer:

Serga [27]3 years ago

4 0

Answer:

170.87 or 10.875 is u round it then its 170.88

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Step-by-step explanation:

Using the Altitude- on- hypotenuse theorem

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144 = 6 DE ( divide both sides by 6 )

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Step-by-step explanation:

if I understand this correctly than you are looking for the inverse function of f(x) = y = 3x³.

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3 years ago

A color printer prints 21 pages in 9 minutes. How many pages does it print per minute? Round your answer to the nearest hundredt

hichkok12 [17]

Answer:

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3 years ago

A group of students estimated the length of one minute without reference to a watch or clock, and the times (seconds) are list

alexandr402 [8]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Yes the students are reasonably good at estimating one minute

a

Option A is correct

b

The test statistics is $t = 0.354$

Step-by-step explanation:

From the question we are told that

The set of data is

68, 82 , 38 , 62 , 41, 25 , 57 , 64, 67, 47, 61, 71, 91, 87, 64

The population mean is $\mu = 60$

The level of significance is given as $\alpha = 0.01$

The critical value for this level of significance obtained from the normal distribution table is

$Z_{\alpha } = 2.33$

The null hypothesis is

$H_o : \mu = 60 \ seconds$

The alternative hypothesis is

$Ha : \mu \ne 60 \ seconds$

Generally the sample mean is mathematically represented as

$\= x = \frac{\sum x_i }{n }$

where n = 15

So

$\= x = \frac{ 68+ 82 + 38 + 62 + 41+ 25 + 57 + 64+67+ 47+ 61+ 71+ 91+ 87+ 64}{15}$

$\= x = 61.67$

The standard deviation is mathematically represented as

$\sigma =\sqrt{ \frac{ \sum (x_i - \= x )^2}{n} }$

substituting values

$\sigma =\sqrt{ \frac{ \sum (68-61.67)^2 + ( 82-61.67 )^2 + (38 -61.67)^2 + 62-61.67)^2 + (41-61.67)^2 +(25-61.67)^2 +( 57+61.67)^2 }{15} }$ $\sqrt{ \frac{ ( \cdot \cdot + ( 64-61.67)^2 + (67-61.67)^2 +(47-61.67)^2 + (61-61.67)^2+ (71-61.67)^2 + (91 -61.67)^2+( 87-61.67)^2 + (64-61.67)^2}{15} }$ => $\sigma = 18.23$

The test statistics is evaluated as

$t = \frac{\= x - \mu }{ \frac{\sigma }{\sqrt{n} } }$

substituting values

$t = \frac{61.67 -60 }{ \frac{18.23 }{\sqrt{ 15} } }$

$t = 0.354$

Now comparing the statistics and the critical value of the level of significance we see that the the test statistics is less than the critical value

Hence the we fail to reject the null hypothesis which mean that these times are from a population with a mean equal to 60 seconds

So we can state that yes the students are reasonably good at estimating one minute given that the sample mean is not far from the population mean

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3 years ago