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3 years ago

54 is 90% of what number

2 answers:

6 0

$\begin{array}{ccc}90\%&-&54\\\\100\%&-&x\end{array}\ \ \ |cross\ multiply\\\\\\90x=54\cdot100\ \ \ \ |divide\ both\ sides\ by\ 90\\\\x=\frac{54\cdot100}{90}\\\\x=\frac{54\cdot10}{9}\\\\x=\frac{6\cdot10}{1}\\\\\boxed{x=60}\leftarrow answer$

denpristay [2]3 years ago

5 0

54 is n of 90

54= nx90

n=54/90= .6 is 60%

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Write parametric equations of the line with the equation 4x-6y=-12

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Answer:

no the answer is 2

Step-by-step explanation:

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3 years ago

You are wrapping a gift box that is 16 inches long, 7 inches wide, and 9 inches tall. Find the amount of wrapping paper you need

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638 in.²

Step-by-step explanation:

Assuming you need to cover just the surface area of the box with no overlaps, and the shape is a rectangular prism, then you need to find the surface area of a rectangular prism.

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3 years ago

Many studies have found an association between red meat consumption and an increased risk of chronic diseases. What is the relat

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Answer:

1. This is an observational study.

2. The explanatory variable is A. red meat consumption

3. The response variable is B. Whether or not a subject dies.

Step-by-step explanation:

1. An observational study is a type of study where the researcher would observe the effects and he would also make no attempt to change the outcome

2. The explanatory variable is also called the independent variable and in this case it is red meat consumption because it is what would be used to predict the dependent variable

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4 years ago

TV advertising agencies face increasing challenges in reaching audience members because viewing TV programs via digital streamin

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Answer:

[0.5235, 0.5765]

To interpret this result, we could say there is a 99% of probability that the proportion of American adults who have watched digitally streamed TV programming on some type of device is between 52.35% and 57.65%

b) 1,843 American adults

Step-by-step explanation:

The 99% confidence interval is given by

$\bf p\pm z^*\sqrt{p(1-p)/n}$

where

p = the proportion of American adults surveyed who said they have watched digitally streamed TV programming on some type of device = 55% = 0.55

$\bf z^*$ the z-score for a 99% confidence level associated with the Normal distribution N(0,1). We can do this given that the sample size (2,341) is big enough

n = sample size = 2,341

We can find the $\bf z^*$ value either with a table or with a spreadsheet.

In Excel use NORM.INV(0.995,0,1)

In OpenOffice Calc use NORMINV(0.995;0;1)

We get a value of $\bf z*$ = 2.576

and our 99% confidence interval is

$\bf 0.55\pm 2.576\sqrt{0.55*0.45/2341}=0.55\pm 2.576*0.0103=0.55\pm 0.265 = [0.5235, 0.5765]$

To interpret this result, we could say there is a 99% of probability that the proportion of American adults who have watched digitally streamed TV programming on some type of device is between 52.35% and 57.65%

We are 99% confident that this interval contains the true population proportion.

(b) What sample size would be required for the width of a 99% CI to be at most 0.03 irrespective of the value of p?? (Round your answer up to the nearest integer.)

The sample size n in a simple random sampling is given by

$\bf n=\frac{(z^*)^2p(1-p)}{e^2}$