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lilavasa [31]
4 years ago
14

Find the values of the variables

Mathematics
1 answer:
Umnica [9.8K]4 years ago
7 0

Looks like you're given an equality between two matrices,

\begin{bmatrix}-12&-w^2\\2f&3\end{bmatrix}=\begin{bmatrix}2k&-81\\-14&3\end{bmatrix}

Two matrices are equal if the entries in the same positions (row and column) are equal. This means

\begin{cases}-12=2k\\-w^2=-81\\2f=-14\\3=3\end{cases}

We can ignore the last one, since it's clearly true. For the remaining, we get

-12=2k\implies k=-\dfrac{12}2=-6

-w^2=-81\implies w^2=81\implies w=\pm9

2f=-14\implies f=-\dfrac{14}2=-7

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6/25

Step-by-step explanation:

6 numbers (from 1 - 25) are less than 7.  There are 25 numbers in total.  This makes the probability 6/25.

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If the lateral area of a cube is 36 cm2. Find its total area.​
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3 years ago
12-4(5x+3y)= will mark as brainliest
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I got -20x-12y+12.......

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3 years ago
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A physical therapist wants to determine the difference in the proportion of men and women who participate in regular sustained p
Alekssandra [29.7K]

Using the z-distribution and the formula for the margin of error, it is found that:

a) A sample size of 54 is needed.

b) A sample size of 752 is needed.

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of \alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which z is the z-score that has a p-value of \frac{1+\alpha}{2}.

The margin of error is of:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

90% confidence level, hence\alpha = 0.9, z is the value of Z that has a p-value of \frac{1+0.9}{2} = 0.95, so z = 1.645.

Item a:

The estimate is \pi = 0.213 - 0.195 = 0.018.

The sample size is <u>n for which M = 0.03</u>, hence:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.03 = 1.645\sqrt{\frac{0.018(0.982)}{n}}

0.03\sqrt{n} = 1.645\sqrt{0.018(0.982)}

\sqrt{n} = \frac{1.645\sqrt{0.018(0.982)}}{0.03}

(\sqrt{n})^2 = \left(\frac{1.645\sqrt{0.018(0.982)}}{0.03}\right)^2

n = 53.1

Rounding up, a sample size of 54 is needed.

Item b:

No prior estimate, hence \pi = 0.05

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.03 = 1.645\sqrt{\frac{0.5(0.5)}{n}}

0.03\sqrt{n} = 1.645\sqrt{0.5(0.5)}

\sqrt{n} = \frac{1.645\sqrt{0.5(0.5)}}{0.03}

(\sqrt{n})^2 = \left(\frac{1.645\sqrt{0.5(0.5)}}{0.03}\right)^2

n = 751.7

Rounding up, a sample of 752 should be taken.

A similar problem is given at brainly.com/question/25694087

5 0
3 years ago
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