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3 years ago

7 (8.5 minus 1.5) + 8 divided by 2

Mathematics

1 answer:

erica [24]3 years ago

5 0

Answer:

eetf fee

Step-by-step explanation:

yyff gf gg gfdf jhdsd tdef gsseig fetygc seeyug. eyyyfvdeetg. dthgd fssergg

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Use the expression 5a – 2b – 3 + 2b – 6a.

ella [17]

Answer:

5a + (-6a) + (-2b) + 2b + (-3) = -a -3 or -(a + 3) so first answer is correct because -(a + 3) is the same as -(3 + a)

Step-by-step explanation:

4 0

3 years ago

After a college football team once again lost a game to their archrival, the alumni association conducted a survey to see if alu

valentina_108 [34]

Answer:

P-value for this hypothesis test is 0.00175.

Step-by-step explanation:

We are given that the alumni association conducted a survey to see if alumni were in favor of firing the coach.

A simple random sample of 100 alumni from the population of all living alumni was taken. Sixty-four of the alumni in the sample were in favor of firing the coach.

Let p = proportion of all living alumni who favored firing the coach

SO, Null Hypothesis, $H_0$ : p = 0.50 {means that the majority of alumni are not in favor of firing the coach}

Alternate Hypothesis, $H_A$ : p > 0.50 {means that the majority of alumni are in favor of firing the coach}

The test statistics that will be used here is One-sample z proportion statistics;

T.S. = $\frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of the alumni in the sample who were in favor of firing the coach = $\frac{64}{100}$ = 0.64

n = sample of alumni = 100

So, test statistics = $\frac{0.64-0.50}{{\sqrt{\frac{0.64(1-0.64)}{100} } } } }$

= 2.92

Now, P-value of the hypothesis test is given by ;

P-value = P(Z > 2.92) = 1 - P(Z $\leq$ 2.92)

= 1 - 0.99825 = 0.00175

Therefore, the P-value for this hypothesis test is 0.00175.

4 0

3 years ago

g People between the ages of 65 and 70 were studied to see if those who exercise more were healthier. One measure of health was

liubo4ka [24]

Answer:

If the person selected had high blood pressure, the probability that he or she did not exercise is P=0.643.

Step-by-step explanation:

We have a group of people between agss of 65 and 70.

They are divided in two groups regarding their amount of exercise: "none", "moderate" and "high".

The response variable is blood pressure, that was categorized as "normal"or "high".

Then, we have the frequencies table as:

Amount of exercise None Moderate High

Blood pressure High 54 23 7

Blood pressure Normal 86 76 32

We have to calculate the probability that, given that the person had high blood pressure, he or she did not exercise.

If the person had high blood pressure, it can be of the any of the 3 groups of exercise amount. There are 54 people with high blood pressure that did not exercise, 23 that did moderate amount of exercise, and 7 that did a high amount.

Then, the probability can be calculated dividing the amount of people that had high blood pressure and did not exercise, by all the people that have high blood pressure (regarding the amount of exercise):

$P(N | HP)=\dfrac{54}{54+23+7}=\dfrac{54}{84}=0.643$