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1 year ago

7

A proofreader earns $8.50 for each page that he proofreads. If he works 7.5 hours and proofreads 14 pages each day on average, w

hich proportion can be used to calculate his average hourly wage,w?

2 answers:

dimulka [17.4K]1 year ago

7 0

Answer:

$15.8666666667 an hour on average

Step-by-step explanation:

14/7.5=1.8666666666

this means that for every hour the proofreader is working he proof reads 1.8666667 pages since he makes 8.5 for every page we have to multiply the pages per hour times the salary of every page to get his average hourly wage

1.8666666666 * 8.5 = $15.8666666667 an hour on average

Greeley [361]1 year ago

6 0

Answer:

119/7.5 = w/1

Step-by-step explanation:

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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

Which phrase best describes the translation from the graph y = 2(x – 15)2 + 3 to the graph of y = 2(x – 11)2 + 3? 4 units to the

Mama L [17]

Left is plus right is negative -11-(-15)=4 so you know 4 to the left

8 0

1 year ago

Read 2 more answers

Suppose a shipment of 400 components contains 68 defective and 332 non-defective computer components. From the shipment you take

MrRissso [65]

Answer:

mean (μ) = 4.25

Step-by-step explanation:

Let p = probability of a defective computer components = $\frac{68}{400} = 0.17$

let q = probability of a non-defective computer components = $\frac{332}{400} = 0.83$

Given random sample n = 25

we will find mean value in binomial distribution

The mean of binomial distribution = np

here 'n' is sample size and 'p' is defective components

mean (μ) = 25 X 0.17 = 4.25

<u>Conclusion</u>:-

mean (μ) = 4.25

6 0

2 years ago

Fudgin [204]

Answer:

A. $301

B. $721

Step-by-step explanation:

Let $x be the amount of money they raised.

Rowena tried to put the $1 bills into two equal piles and found one left over at the end, then

$x=2q_1+1$

Polly tried to put the $1 bills into three equal piles and found one left over at the end, then

$x=3q_2+1$

Frustrated, they tried 4, 5, and 6 equal piles and each time had $1 left over, then

$x=4q_3+1\\ \\x=5q_4+1\\ \\x=6q_5+1$

Finally Rowena put all the bills evenly into 7 equal piles, and none were left over, then

$x=7q_6$

This means $x-1$ is divisible by 2, 3, 4, 5 and 6 without remainder, so

$x-1=2\cdot 3\cdot 2\cdot 5n=60n$

Hence,

$x=60n+1, \ n\in N$

The smallest amount of money they could have raised is $301, because

$x=60\cdot 5+1=301$ is divisible by 7.

Now, the number $x=60n+1$ should be divisible by 7 and must be greater than 500.

So,

$60n+1>500\\ \\60n>499\\ \\n>8$

When n = 9,

$x=60\cdot 9+1=541$ is not divisible by 7.

When n = 10,

$x=60\cdot 10+1=601$ is not divisible by 7.

When n = 11,

$x=60\cdot 11+1=661$ is not divisible by 7.

When n = 12,

$x=60\cdot 12+1=721$ is divisible by 7.

B. The least amount of money they could have raised is $721

7 0

2 years ago

Read 2 more answers

In right triangle ABC, mC - 90° and AC BC. Which trigonometric ratio is cquivalent to sin b?

vaieri [72.5K]

Answer:

All trigonometric Ratios are $SinB = \frac{AC}{AB}$ , $SinA= \frac{CB}{AB}$ , $CosA= \frac{AC}{AB}$

And $Cos B = \frac{CB}{AB}$ .

Step-by-step explanation:

Given that,

A right angle triangle ΔABC, ∠C =90°.

Diagram of the given scenario shown below,

In triangle ΔABC :-

$Hypotenuse = AB\\Base = CB\\Perpendicular = AC$

So, $Sin\theta = \frac{perpendicular}{hypotenuse}$

$SinB = \frac{AC}{AB}$

Now, for ∠A the dimensions of trigonometric ratios will be changed.

Here the base for ∠A is AC , perpendicular side is CB and hypotenuse will be same for all ratios.

$SinA= \frac{CB}{AB}$

Again, $Cos\theta= \frac{base}{hypotenuse}$

Then, $CosA= \frac{AC}{AB}$

And $Cos B = \frac{CB}{AB}$ .

Hence,

All trigonometric Ratios are $SinB = \frac{AC}{AB}$ , $SinA= \frac{CB}{AB}$ , $CosA= \frac{AC}{AB}$

And $Cos B = \frac{CB}{AB}$ .

8 0

2 years ago