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Marianna [84]
3 years ago
15

3 • 4/12 to the simplest form

Mathematics
1 answer:
beks73 [17]3 years ago
3 0
9 because you would turn 4 12 in simplest form the times it by 3

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Factor the polynomial 4x^2-12x^2+7x-21 completely
sergejj [24]

Answer:

4x^2-12x^2+7x-21\\\mathrm{Add\:similar\:elements:}\:4x^2-12x^2=-8x^2\\=-8x^2+7x-21

Step-by-step explanation:

6 0
3 years ago
What is 2/2 -+7 need friends anyone
kow [346]

Answer:

-6

Step-by-step explanation:

2/2=1

-+7=-7

1-7=-6

8 0
3 years ago
solve the proportion. the quantity 13 minus b end of quantity divided by 6 equals the quantity 2 minus 0.5 multiplied by b end o
tatyana61 [14]
\frac{13-b}{6}= \frac{2-0.5b}{4} \\ 4(13-b)=6(2-0.5b) \\ 52-4b=12-3b \\ b=40
5 0
3 years ago
The director of admissions at the University of Maryland, University College is concerned about the high cost of textbooks for t
nadezda [96]

Answer:

a. There is evidence that the population mean is above $300.

b. There is no evidence that the population mean is above $300.

c. There is no evidence that the population mean is above $300.

d. The director could ask for cheaper similar books.

Step-by-step explanation:

Let X be the random variable that represents the cost of textbooks. We have observed n = 25 values, \bar{x} = 315.4 and s = 43.20. We suppose that X is normally distributed.

We have the following null and alternative hypothesis

H_{0}: \mu = 300 vs H_{1}: \mu > 300 (upper-tail alternative)

We will use the test statistic

T = \frac{\bar{X}-300}{S/\sqrt{25}} and the observed value is

t_{0} = \frac{315.4 - 300}{43.20/\sqrt{25}} = 1.7824.

If H_{0} is true, then T has a t distribution with n-1 = 24 degrees of freedom.

a. The rejection region is given by RR = {t | t > t_{0.9}} where t_{0.9} = 1.3178 is the 90th quantile of the t distribution with 24 df, so, RR = {t | t > 1.3178}. Because the observed value satisty 1.7824 > 1.3178, there is evidence that the population mean is above $300.

b. If s = 75, then the observed value is t_{0} = \frac{315.4 - 300}{75/\sqrt{25}} = 1.0267. The rejection region for a 0.05 level of significance is RR = {t | t > t_{0.95}} where t_{0.95} = 1.7108 is the 95th quantile of the t distribution with 24 df, so, RR = {t | t > 1.7108}. Because the observed value does not fall inside the rejection region, there is no evidence that the population mean is above $300.

c. If \bar{x} = 305.11 and s = 43.20, the observed value is t_{0} = \frac{305.11 - 300}{43.20/\sqrt{25}} =  0.5914. For RR = {t | t > 1.3178} we have that the observed value does not fall inside RR, therefore, there is no evidence that the population mean is above $300.

d. Because the director of admissions is concerned about the high cost of textbooks, and there is evidence that the population mean of costs is above $300, the director could ask for cheaper similar books.

8 0
3 years ago
Danielle Cecil's driver-rating factor is 1.60 and her car is in age group D and insurance-rating group is 12.
masha68 [24]

The value of the annual premium will be $1134.40

<h3>How to calculate the premium?</h3>

From the information given, the annual base premium will be:

= Liability premium + Collision premium + Comprehensive premium

= $383 + $240 + $86

= $709

The annual premium will be:

= Annual base premium × Driver eating factors

= $709 × 1.60

= $1134.40

Learn more about premium on:

brainly.com/question/25280754

#SPJ1

8 0
2 years ago
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