Let x represent the number of pages left to be read in a book consisting of 780 pages.

Evaluate 4 − 0 . 2 5 g + 0 . 5 h 4−0.25g+0.5h4, minus, 0, point, 25, g, plus, 0, point, 5, h when g = 1 0 g=10g, equals, 10 and

abruzzese [7]

Not sure about the weird echo in the question but I think we're being ask to evaluate

4 - 0.25g + 0.5h

when g=10, h=5

That's

4 - 0.25(10) + 0.5(5) = 4 - 2.5 + 2.5 = 4

Answer: 4

6 0

3 years ago

Read 2 more answers

I need help with number 1

Sophie [7]

Changing those words into an equation, we get:
7+3x = 19
Subtract both sides by 7
3x = 12
Divide both sides by 3
x = 4
That number is 4. Have a nice day! :)

6 0

3 years ago

∠A = ___∘ Round your answer to the nearest hundredth.

larisa86 [58]

Answer: 45.58 degrees

Step-by-step explanation:

$\sin A=5/7\\A=sin^{-1}(5/7) \approx 45.58^{\circ}$

6 0

2 years ago

Read 2 more answers

Several programs attempt to address the shortage of qualified teachers by placing uncertified instructors in schools with acute

ss7ja [257]

Answer:

We conclude that the mean scores with uncertified teachers is higher or equal as compared to certified teachers.

Step-by-step explanation:

We are given that reading scores of the students of certified teachers averaged 35.62 points with standard deviation 9.31. The scores of students instructed by uncertified teachers had mean 32.48 points with standard deviation 9.43 points on the same test.

There were 44 students in each group.

Let $\mu_1$ = mean scores with uncertified teachers.

$\mu_2$ = mean scores with certified teachers.

So, Null Hypothesis, $H_0$ : $\mu_1\geq \mu_2$ {means that the mean scores with uncertified teachers is higher or equal as compared to certified teachers}

Alternate Hypothesis, $H_A$ : $\mu_1$ {means that the mean scores with uncertified teachers is lower as compared to certified teachers}

The test statistics that would be used here 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 mean scores of students instructed by uncertified teachers = 32.48 points

$\bar X_2$ = sample mean scores of students instructed by certified teachers = 35.62 points

$s_1$ = sample standard deviation of scores of students instructed by uncertified teachers = 9.43 points

$s_2$ = sample standard deviation of scores of students instructed by certified teachers = 9.31 points

$n_1$ = sample of students under uncertified teachers = 44

$n_2$ = sample of students under certified teachers = 44

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2} +(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(44-1)\times 9.43^{2} +(44-1)\times 9.31^{2} }{44+44-2} }$ = 9.37

So, the test statistics = $\frac{(32.48-35.62)-(0)}{9.37 \times \sqrt{\frac{1}{44} +\frac{1}{44} } }$ ~ $t_8_6$

= -1.572

The value of t test statistics is -1.572.

Since, in the question we are not given the level of significance so we assume it to be 5%. Now, at 5% significance level the t table gives critical values of -1.665 at 86 degree of freedom for left-tailed test.

Since our test statistic is more than the critical values of t as -1.572 > -1.665, 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 the mean scores with uncertified teachers is higher or equal as compared to certified teachers.

7 0

3 years ago

What is the volume of the cylinder shown below?

sveta [45]

Answer:

1008pi

Step-by-step explanation:

the formula for volume of a cylinder is pi r squared h

V equals 144 * 7 * pi

7 0

2 years ago