What is the sum when 1/5 is added to 5/2?

What exactly do you mean by *? Like exponent? If so the domain is R= −∞,∞ with its range being [−1,1]. If this wasn’t meant by exponent, please comment and I’ll reply ASAP!

6 0

3 years ago

The diameter of the earth is about 7900 miles. The diameter of Jupiter is about 88,800 miles, approximately 11 times that of the

jeka94

We know that the surface of a sphere is: S = 4 * pi * r^2 The surface of Earth is : Se = 4 * 3.1416 * (7900/2)^2 = 196,066,797.51 And the Jupiter is : Sj = 4 * 3.1416 * (88800/2)^2 = 24,772,840,374.32 So the surface of Jupiter is around 11^2 times bigger than the Earth

8 0

3 years ago

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Fairfield Homes is developing two parcels near Pigeon Fork, Tennessee. In order to test different advertising approaches, it use

evablogger [386]

Answer:

(1) Null Hypothesis, $H_0$ : $\mu_1 = \mu_2$

Alternate Hypothesis, $H_1$ : $\mu_1\neq \mu_2$

(2) Test statistics = -1.48

(3) At the 0.05 significance level, Fair field conclude that the population means are same.

Step-by-step explanation:

Let $\mu_1$ = mean annual family income for 12 people making inquiries at the first development

$\mu_2$ = mean annual family income for 24 people making inquiries at the second development

$s_1$ = standard deviation of annual family income for 12 people making inquiries at the first development

$s_2$ = standard deviation of annual family income for 24 people making inquiries at the second development

$n_1$ = sample of people of first development i.e. 12

$n_2$ = sample of people of second development i.e. 24

(1) Null Hypothesis, $H_0$ : $\mu_1 = \mu_2$ {population means are same}

Alternate Hypothesis, $H_1$ : $\mu_1\neq \mu_2$ {population means are different}

DECISION RULE ;

If the test statistics is less than the critical value of t from table at 5% significance level, then we will accept null hypothesis, $H_0$ .
If the test statistics is more than the critical value of t from table at 5% significance level, then we will reject null hypothesis, $H_0$ .

(2) The test statistics is given by;

$\frac{(X_1bar -X_2bar) - (\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t_n__1+n_2-2$

where, $X_1bar$ = Sample mean income of people at first development

= $153,000

$X_2bar$ = Sample mean income of people at second development = $171,000

$s_1$ = $42,000 and $s_2$ = $30,000

$s_p= \sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2}}$ = $\sqrt{\frac{(12-1)42000^{2}+(24-1)30000^{2} }{12+24-2}}$ = 34344.30

Test statistics = $\frac{(153000 -171000) - 0}{34344.30\sqrt{\frac{1}{12}+\frac{1}{24} } }$ ~ $t_3_4$

= -1.48

(3) At 5% level of significance, t table gives critical value of 2.032 at 34 degree of freedom.Since our test statistics is less than the critical value of t so considering our decision rule, we will accept null hypothesis.

And conclude that population means are same.

4 0

4 years ago