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Ghella [55]
4 years ago
13

Which is the correct answer

Mathematics
1 answer:
VashaNatasha [74]4 years ago
7 0

Let s denote the number of pages in each issue of the sports magazine.

Let t denote the number of pages in each issue of the technology magazine.

Given that a sports magazine prints 12 issues per year and a technology magazine prints 10 issues per year.The total number of pages in all the issues of the sports magazine for one year is 32 more than the total number of pages in all the issues of the technology magazine for one year.

This can be written in equation as,

12s=10t+32

Since, it is given that, "Each issue of the sports magazine has 18 fewer pages than each issue of the technology magazine".

This can be written in equation as,

s=t-18

Thus, the set of equations that represent the given data is

12s=10t+32 and s=t-18

Hence, Option A is the correct answer.

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Combine like terms to get 2x + 6x which is 8x and 3y + 9y which is 12y so the simplified answer is 8x + 12y

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Mr. Bond’s class wanted to estimate the mean mass of Snickers Fun Size bars. They randomly selected 74 bars and recorded the mas
AleksAgata [21]

Step 1: Find the standard error (SE)

The standard error is given by

SE=\frac{s}{\sqrt[]{n}}\begin{gathered} \text{ Where } \\ SE=\text{ the standard error} \\ s=\text{ the sample standard deviation} \\ n=\text{ the sample size} \end{gathered}

In this case,

n=74,s=0.76

Therefore,

SE=\frac{0.76}{\sqrt[]{74}}\approx0.0883

Step 2: Find the alpha level (α)

\alpha=1-\frac{(\text{Confidence level})}{100}\alpha=1-0.99=0.01

Step 3: Find the critical probability (P*)

P^{\prime}=1-\frac{\alpha}{2}

Therefore,

P^{\cdot}=1-\frac{0.01}{2}=0.995

Step 4: Find the critical value (CV)

The critical value the z-score having a cumulative probability equal to the critical probability (P*).

Using the cumulative z-score table we will find the z-score with value of 0.995

Hence,

CV=2.576

Step 5: Find the margin of error (ME)

ME=SE\times CV

Therefore,

ME=0.0883\times2.576=0.2275

Step 6: Find the confidence interval (CI)

\begin{gathered} CI\text{ is given by} \\ CI=(\bar{x}-ME,\bar{x}+ME) \\ \text{ In this case} \\ \bar{x}=17.1 \end{gathered}

Therefore,

CI=(17.1-0.2275,17.1+0.2275)=(16.8725,17.3275)

Hence there is a 99% probability that the true mean will lie in the confidence interval

(16.8725, 17.3275)

4 0
1 year ago
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