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qaws [65]
2 years ago
12

What is the product of (2p + 7)(3p2 + 4p – 3)?

Mathematics
1 answer:
Nady [450]2 years ago
3 0

Answer:

(2p+7)(3p²+4p-3)

2p(3p²+4p-3)+7(3p²+4p+3)

(6p³+8p²-6p)+ (21p²+28p+21)

6p³+29p²-6p+21

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John got a 76, 68, and 83 on his first three test. What score does he need on his fourth test to achieve an 80 average?
shepuryov [24]
It's 93.
simple calculate it by:
(76+68+83+x)/4=80
then 73*5-(67+75+73+69)=x
where x here is the grade needed to reach an av. of 80. 4 is the number of tests.
*common answer*
3 0
3 years ago
A total of 140 seventh- and eighth-grade boys were asked which sport they preferred, basketball or baseball.
ss7ja [257]

Answer:

The answer is, " About 56% of those who prefer baseball are eight graders"

Step by Step Explanation:

Step 1: Turn 56% to a decimal, you get 0.56

Step 2: Multiply 0.56 by 68, the total amount of people who like baseball

Step 3 : you get 38.08 ( Don't mind the extra .08, just round )

Step 4 : You celebrate with a little victory dance!!!!!

7 0
3 years ago
A substance has a density of 0.70 g/mL, if it has a volume of 3.0 mL, what is its mass?
Naddika [18.5K]

Answer:

<h2>2.1 g</h2>

Step-by-step explanation:

The mass of a substance when given the density and volume can be found by using the formula

mass = Density × volume

From the question we have

mass = 0.70 × 3

We have the final answer as

<h3>2.1 g</h3>

Hope this helps you

4 0
3 years ago
vertices of parallelogram ABCD are A(2, 0), B(4, 3), C(3, 1). How many possibilities are there for the position of the fourth ve
nordsb [41]

Answer:

1 possibility is there for the the position of the fourth vertex.

7 0
3 years ago
The overhead reach distances of adult females are normally distributed with a mean of 197.5 cm197.5 cm and a standard deviation
fiasKO [112]

Answer:

a) 5.37% probability that an individual distance is greater than 210.9 cm

b) 75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c) Because the underlying distribution is normal. We only have to verify the sample size if the underlying population is not normal.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 197.5, \sigma = 8.3

a. Find the probability that an individual distance is greater than 210.9 cm

This is 1 subtracted by the pvalue of Z when X = 210.9. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{210.9 - 197.5}{8.3}

Z = 1.61

Z = 1.61 has a pvalue of 0.9463.

1 - 0.9463 = 0.0537

5.37% probability that an individual distance is greater than 210.9 cm.

b. Find the probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

Now n = 15, s = \frac{8.3}{\sqrt{15}} = 2.14

This probability is 1 subtracted by the pvalue of Z when X = 196. Then

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{196 - 197.5}{2.14}

Z = -0.7

Z = -0.7 has a pvalue of 0.2420.

1 - 0.2420 = 0.7580

75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

The underlying distribution(overhead reach distances of adult females) is normal, which means that the sample size requirement(being at least 30) does not apply.

5 0
3 years ago
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