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3 years ago

Simplify 8 over negative 4 divided by negative 3 over 9. 6 −6 12 −12

Mathematics

1 answer:

skelet666 [1.2K]3 years ago

4 0

Answer:

Step-by-step explanation:

Given

$\frac{8}{-4} / \frac{-3}{9}$

Required

Solve

$\frac{8}{-4} / \frac{-3}{9}$

We start by converting the division sign to multiplication

$\frac{8}{-4} / \frac{-3}{9} =\frac{8}{-4} * \frac{-9}{3}$

Then, we divide -9 by 3

$\frac{8}{-4} / \frac{-3}{9}=\frac{8}{-4} * -3$

Then, we divide 8 by -4

$\frac{8}{-4} / \frac{-3}{9}=-2 * -3$

Multiply -2 and -3 to give final result

$\frac{8}{-4} / \frac{-3}{9} = 6$

Hence, $\frac{8}{-4} / \frac{-3}{9}$ is equivalent to 6

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3 years ago

(Based on Q1 ~ Q3) According to the Bureau of the Census, 18.1% of the U.S. population lives in the Northeast, 21.9% inn the Mid

vekshin1

Answer:

We can therefore conclude that the geographical distribution of hotline callers could be the same as the U.S population distribution.

Step-by-step explanation:

The null Hypothesis: Geographical distribution of hotline callers could be the same as the U.S. population distribution

Alternative hypothesis: Geographical distribution of hotline callers could not be the same as the U.S. population distribution

The populations considered are the Midwest, South, Northeast, and west.

The number of categories, k = 4

Number of recent calls = 200

Let the number of estimated parameters that must be estimated, m = 0

The degree of freedom is given by the formula:

df = k - 1-m

df = 4 -1 - 0 = 3

Let the significance level be, α = 5% = 0.05

For α = 0.05, and df = 3,

from the chi square distribution table, the critical value = 7.815

Observed and expected frequencies of calls for each of the region:

Northeast

Observed frequency = 39

It contains 18.1% of the US Population

The probability = 0.181

Expected frequency of call = 0.181 * 200 = 36.2

Midwest

Observed frequency = 55

It contains 21.9% of the US Population

The probability = 0.219

Expected frequency of call = 0.219 * 200 =43.8

South

Observed frequency = 60

It contains 36.7% of the US Population

The probability = 0.367

Expected frequency of call = 0.367 * 200 = 73.4

West

Observed frequency = 46

It contains 23.3% of the US Population

The probability = 0.233

Expected frequency of call = 0.233 * 200 = 46

$x^{2} = \sum \frac{(O_{i} - E_{i}) ^{2} }{E_{i} } , i = 1, 2,.........k$

Where $O_{i} =$ observed frequency

$E_{i} =$ Expected frequency

Calculate the test statistic value, x²

$x^{2} = \frac{(39 - 36.2)^{2} }{36.2} + \frac{(55 - 43.8)^{2} }{43.8} + \frac{(60 - 73.4)^{2} }{73.4} + \frac{(46 - 46.6)^{2} }{46.6}$

$x^{2} = 5.535$

Since the test statistic value, x²= 5.535 is less than the critical value = 7.815, the null hypothesis will not be rejected, i.e. it will be accepted. We can therefore conclude that the geographical distribution of hotline callers could be the same as the U.S population distribution.

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