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qwelly [4]
2 years ago
13

PLZZ I NEED HELP if ur good at math help meeee. I NEED HELP

Mathematics
1 answer:
Ilya [14]2 years ago
4 0
It looks like it’s already answered I’m confused about that
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What is the coefficient of the second term of the polynomial? (In other words, what is B?
Scrat [10]

Answer: The coefficient of the second term of the polynomial is B=5

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

7 0
2 years ago
Simplify 4.51 with a power of 0<br><br> 4.51 <br> 1 <br> 0 <br> −1
Bumek [7]

Answer:

1

Step-by-step explanation:

Remember that anything to the power of 0 is equal to 1

4.51^0 = 1

1 is your answer

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√84= √4•√21= final answer is
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3 years ago
The universal set in this diagram is the set of integers from 1 to 15. Place the integers in the correct place in the Venn Diagr
Dmitry_Shevchenko [17]

Answer:

see image

Step-by-step explanation:


7 0
3 years ago
In 2008, data from the Center for Disease Control revealed that 28.5% of all male teenagers, aged 18-19 and attending U.S. colle
m_a_m_a [10]

Answer:

a) H0 : u = 28.5%

H1 : u < 28.5%

b) critical value = - 1.645

c) test statistic Z= - 1.41

d) Fail to reject H0

e) There is not enough evidence to support the professor's claim.

Step-by-step explanation:

Given:

P = 28.5% ≈ 0.285

X = 210

n = 800

p' = \frac{X}{n} = \frac{210}{800} = 0.2625

Level of significance = 0.05

a) The null and alternative hypotheses are:

H0 : u = 28.5%

H1 : u < 28.5%

b) Given a 0.05 significance level.

This is a left tailed test.

The critical value =

-Z_0.05 = -1.645

The critical value = -1.645

c) Calculating the test statistic, we have:

Z = \frac{p' - P}{\sqrt{\frac{P(1-P)}{n}}}

Z = \frac{0.2625 - 0.285}{\sqrt{\frac{28.5(1-28.5)}{800}}}

Z = -1.41

d) Decision:

We fail to reject null hypothesis H0, since Z = -1.41 is not in the rejection region, <1.645

e) There is not enough evidence to support the professor's claim that the proportion of obese male teenagers decreased.

5 0
3 years ago
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