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algol13
3 years ago
7

Silva set up a dog-walking business. He earned $60 in June. He is projected to earn 110% more in July. How much is Silva project

ed to make in July?
$126
$170
$176
$660
Mathematics
1 answer:
snow_lady [41]3 years ago
7 0

9514 1404 393

Answer:

  $126

Step-by-step explanation:

110% more than 100% is 210% of his June earnings:

  july = 2.10×$60 = $126

Silva is projected to make $126 in July.

_____

As always, you can think of % and /100 as being interchangeable.

  210% = 210/100 = 2.10

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Helen [10]

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6 0
3 years ago
According to the norms established for a reading comprehension test, eighth-graders should average 83.2 with a standard deviatio
joja [24]

Answer:

t=\frac{86.7-83.2}{\frac{8.6}{\sqrt{45}}}=2.73    

p_v =P(t_{44}>2.73)=0.0045    

If we compare the p value and a significance level given \alpha=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the population mean is signficantly higher than 83.2 at 1% of significance.

Step-by-step explanation:

Data given and notation    

\bar X=86.7 represent the average life for the sample    

s=8.6 represent the population standard deviation    

n=45 sample size    

\mu_o =83.2 represent the value that we want to test    

\alpha=0.01 represent the significance level for the hypothesis test.    

t would represent the statistic (variable of interest)    

p_v represent the p value for the test (variable of interest)    

State the null and alternative hypotheses.    

We need to apply a two tailed  test.  

What are H0 and Ha for this study?    

Null hypothesis:  \mu \leq 83.2  

Alternative hypothesis :\mu > 83.2  

Compute the test statistic  

The statistic for this case is given by:  

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

Calculate the statistic    

We can replace in formula (1) the info given like this:    

t=\frac{86.7-83.2}{\frac{8.6}{\sqrt{45}}}=2.73    

Give the appropriate conclusion for the test  

First we need to calculate the degrees of freedom given by:

df=n-1=45-1=44

Since is a one side right tailed test the p value would be:    

p_v =P(t_{44}>2.73)=0.0045    

Conclusion    

If we compare the p value and a significance level given \alpha=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the population mean is signficantly higher than 83.2 at 1% of significance.

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Answer:

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Step-by-step explanation:

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Answer:

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And we can assume a normal distribution and then we can solve the problem with the z score formula given by:

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And replacing we got:

z=\frac{38.6- 45}{16}= -0.4

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We can find the probability of interest using the normal standard table and with the following difference:

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Step-by-step explanation:

Let X the random variable who represent the expense and we assume the following parameters:

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And for this case we want to find the percent of his expense between 38.6 and 57.8 so we want this probability:

P(38.6

And we can assume a normal distribution and then we can solve the problem with the z score formula given by:

z=\frac{X -\mu}{\sigma}

And replacing we got:

z=\frac{38.6- 45}{16}= -0.4

z=\frac{57.8- 45}{16}= 0.8

We can find the probability of interest using the normal standard table and with the following difference:

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hope it helps:))

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3 years ago
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