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yulyashka [42]
3 years ago
11

G(x) = 35x^2– 48x – 27; 7x + 3 how do i solve this using long division?

Mathematics
1 answer:
Rina8888 [55]3 years ago
6 0

35<em>x</em>² = 7<em>x</em> • 5<em>x</em>, and

5<em>x</em> (7<em>x</em> + 3) = 35<em>x</em>² + 15<em>x</em>

Subtract this from the dividend to get an initial remainder of

(35<em>x</em>² - 48<em>x</em> - 27) - (35<em>x</em>² + 15<em>x</em>) = -63<em>x</em> - 27

-63<em>x</em> = 7<em>x</em> • (-9), and

-9 (7<em>x</em> + 3) = -63<em>x</em> - 27

Subtract this from the previous remainder to get a new one of

(-63<em>x</em> - 27) - (-63<em>x</em> - 27) = 0

and we're done.

Now just gather the terms in bold (and the remainder, but since it's 0 we leave it out). So we have

(35<em>x</em>² - 48<em>x</em> - 27) / (7<em>x</em> + 3) = 5<em>x</em> - (63<em>x</em> + 27) / (7<em>x</em> + 3)

(35<em>x</em>² - 48<em>x</em> - 27) / (7<em>x</em> + 3) = 5<em>x</em> - 9

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i:

The appropriate null hypothesis is H_0: p \geq 0.2

The appropriate alternative hypothesis is H_1: p < 0.2

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ii:

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Step-by-step explanation:

Question i:

The device will be withdrawn if fewer than 20% of all of the museum’s visitors make use of it.

At the null hypothesis, we test if the proportion is of at least 20%, that is:

H_0: p \geq 0.2

At the alternative hypothesis, we test if the proportion is less than 20%, that is:

H_1: p < 0.2

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

0.2 is tested at the null hypothesis:

This means that \mu = 0.2, \sigma = \sqrt{0.2*0.8} = \sqrt{0.16} = 0.4.

The device will be withdrawn if fewer than 20% of all of the museum’s visitors make use of it. Of a random sample of 100 visitors, 15 chose to use the device.

This means that n = 100, X = \frac{15}{100} = 0.15

Test statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{0.15 - 0.20}{\frac{0.4}{\sqrt{100}}}

z = -1.25

P-value of the test and decision:

The p-value of the test is the probability of finding a sample proportion below 0.15, which is the p-value of z = -1.25.

Looking at the z-table, z = -1.25 has a p-value of 0.1057.

The p-value of the test is 0.1057 > 0.05, which means that there is not sufficient evidence that fewer than 20% of the museum visitors make use of the device, and so, it should not be withdrawn.

Question ii:

The p-value of the test is 0.1057

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