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2 years ago

Find sec A and cot B exactly if a = 8 and b = 7.

Mathematics

1 answer:

Shtirlitz [24]2 years ago

8 0

Answer:

sec A= 1.01 and cot B =8.25

Step-by-step explanation:

Given :

sec A and cotB if a =8 and b=7

Now,

= $sec A=\frac{1}{cos A} \\\\\frac{1}{cos 8} \\\\\frac{1}{0.99} \\1.01$

and

$cot B=\frac{cosB}{sin B} \\cotB=\frac{cos 7}{sin7} \\cot B =\frac{0.99}{0.12} \\cot B =8.25$

Therefore, answer will be sec A= 1.01 and cot B =8.25

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Step-by-step explanation:

We are given that the 55 randomly selected political science classes assigned an average of 19.6 pages of essay writing for the course. The standard deviation for these 55 classes was 4.8 pages.

The 50 randomly selected history classes assigned an average of 20.2 pages of essay writing for the course. The standard deviation for these 50 classes was 3.3 pages.

Let $\mu_1$ = mean writing required by political science classes

$\mu_2$ = mean writing required by history classes

SO, Null Hypothesis, $H_0$ : $\mu_1-\mu_2\geq0$ or $\mu_1\geq \mu_2$ {means that political science classes require more or equal writing than history classes}

Alternate Hypothesis, $H_A$ : $\mu_1-\mu_2$ or $\mu_1$ {means that political science classes require less writing than history classes}

The test statistics that will be used here is Two-sample t test statistics as we don't know about the population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t__n__1+_n__2-2$

where, $\bar X_1$ = sample average writing for political science course= 19.6 pages

$\bar X_2$ = sample average writing for history course= 20.2 pages

$s_1$ = sample standard deviation for political science classes = 4.8 pages

$s_2$ = sample standard deviation for history classes = 3.3 pages

$n_1$ = sample of selected political science classes = 55

$n_2$ = sample of selected history classes = 50

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(55-1)\times 4.8^{2}+(50-1)\times 3.3^{2} }{55+50-2} }$ = 4.154

So, the test statistics = $\frac{(19.6-20.2)-(0)}{4.154 \times \sqrt{\frac{1}{55}+\frac{1}{50} } }$ ~ $t_1_0_3$

= -0.739

Now at 0.01 significance level, the t table gives critical value of -2.367 at 103 degree of freedom for left-tailed test. Since our test statistics is more than the critical value of t as -0.739 > -2.367, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that political science classes require more or equal writing than history classes.

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