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KonstantinChe [14]
2 years ago
8

Innea's company's revenue in 201720172017 is \dfrac{36}{25} 25 36 ​ start fraction, 36, divided by, 25, end fraction of its reve

nue in 201620162016. What is Linnea's company's revenue in 201720172017 as a percent of its revenue in 201620162016 ?
Mathematics
1 answer:
cricket20 [7]2 years ago
6 0

Answer:

144%

Step-by-step explanation:

Complete question below:

Linnea's company's revenue in 2017 is 36/25 of its revenue in 2016. What is Linnea's company's revenue in 2017 as a percent of its revenue in 2016?

Solution

Let

36x= Linnea's company revenue in 2017

25x=Linnea's company revenue in 2016

Percent of Linnea's company's revenue in 2017 as a percent of its revenue in 2016= 36x / 25x × 100

36x / 25x * 100

=1.44 × 100

=144%

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Which arrangement shows −275 , −4.1 , −535 , and −3 in order from least to greatest?
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Answer:

-535, -275, -4.1, -3

Step-by-step explanation:

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Explain how you can use a model to find 6x17
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Maybe by useing cubes like 6 groups of 17 or u can also use a number line like the elementary skills
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Tomorrow, Mrs. Wendel's class will be using toothpicks for a science project. Each student must use at least 9 toothpicks for th
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Number of toothpicks required per student = At least 9

Total number of students = 30

Number of toothpicks required for the entire class = At least 9 × 30 = At least 270

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To plan the budget for next year a college must update its estimate of the proportion of next year's freshmen class that will ne
Naily [24]

Answer:

Null hypothesis:p\leq 0.35  

Alternative hypothesis:p > 0.35  

z=\frac{0.447 -0.35}{\sqrt{\frac{0.35(1-0.35)}{150}}}=2.491  

p_v =P(z>2.491)=0.0064  

So the p value obtained was a very low value and using the significance level assumed \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of applicants who request financial aid is significantly higher than 0.35

Step-by-step explanation:

Data given and notation

n=150 represent the random sample taken

X=67 represent the applicants who request financial aid

\hat p=\frac{67}{150}=0.447 estimated proportion of applicants who request financial aid

p_o=0.35 is the value that we want to test

\alpha represent the significance level

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion of applicants who request financial aid is higher than 0.35.:  

Null hypothesis:p\leq 0.35  

Alternative hypothesis:p > 0.35  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.447 -0.35}{\sqrt{\frac{0.35(1-0.35)}{150}}}=2.491  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The next step would be calculate the p value for this test.  

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

p_v =P(z>2.491)=0.0064  

So the p value obtained was a very low value and using the significance level assumed \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of applicants who request financial aid is significantly higher than 0.35

5 0
2 years ago
(b) we often read that iq scores for large populations are centered at 100. what percent of these 78 students have scores above
ss7ja [257]
As IQ scores for extensive populaces are focused at 100, the mean = 100. 
There ought to be around half scores above or underneath the mean score since mean and middle is about a similar when the populace is substantial. 
P(x > 100) = P( z> (100 - 100)/sd ) = P(z > 0)= 0.5 
The number of the student who has scored over 100 = 0.5 x 78 = 39 Therefore the answer is 39 students. 
4 0
2 years ago
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