Given (x) = -3x+2, evaluate f(5)

Solve H = fraction with p minus a in the numerator and 3 times w in the denominator for p. SOLVE FOR P

omeli [17]

D. p = fraction H over the quantity 3 times w + a

5 0

2 years ago

Do political science classes require less writing than history classes? The 55 randomly selected political science classes assig

tia_tia [17]

Answer:

No, political science classes does not require less writing than history classes.

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.

8 0

3 years ago

Identify the surface whose equation is given. rho2(sin2(φ) sin2(θ) + cos2(φ)) = 49

Paha777 [63]

Answer:

The surface is a cylindrical surface with radius 7 units

Step-by-step explanation:

The equation is properly written as:

$\rho^{2} (sin^{2} \phi sin^{2} \theta + cos^{2} \phi) = 49$

The above equation takes the form $y^{2} + z^{2} = r^{2}$

Where $y^{2} = \rho^{2} sin^{2} \phi sin^{2} \theta$

$y = \rho sin \phi sin \theta$

and $z^{2} = \rho^{2}cos^{2} \phi$

$z = \rho cos \phi$

$r^{2} = 49$

r = 7 units

The surface is a cylindrical surface with radius 7 units

4 0

2 years ago

Order 34 x 10^2,1.2x10^7,8.11×10-^3 and 435 from least to greatest

FromTheMoon [43]

Answer:

The ascending order is $8.11\times 10^{-3},435, 34\times 10^2, 1.2\times 10^7$

Step-by-step explanation:

First of all find the exact value of each number.

$34\times 10^2=34\times 100=3400$

$1.2\times 10^7=1.2\times 10000000=12000000$

$8.11\times 10^{-3}=\frac{8.11}{1000}=0.00811$

The fourth number is 435.

Using the above calculation we can easily arrange these numbers form least to greatest.

The ascending order is

$0.00811, 435, 3400, 12000000$

It can be written as

$8.11\times 10^{-3},435, 34\times 10^2, 1.2\times 10^7$ .

4 0

3 years ago

What is the probability of rolling a multiple of 5

givi [52]

Answer:

since you didn't give us a range, the possibility is infinity as there is no specified number of sides, and that answer doesn't really help. please be more specific in your future questions.

if the die has 6 sides, the probability of rolling a multiple of 5 is 1/6.

5 0

2 years ago