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mamaluj [8]
3 years ago
7

Evaluate 5m2 for m=4

Mathematics
2 answers:
jolli1 [7]3 years ago
4 0
If the question is 5m^2 the answer is 80. And if the question is 5m2 then it is 40
meriva3 years ago
4 0
Okay if you make it 5•4•2 you get 40,
BUT if this was meant to be an exponent as in “5m*2” then it would be 80.
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7 Grade MAth help me and not the pezrson who helped me last time pls
saveliy_v [14]

Answer:

40

Step-by-step explanation:

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Can someone help me with this page
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The first one is A. Number 2 is d. Number 3 is b because 20x20=400 and 21x21=441 and when you add them together you get 841 or 29x29. Number 4 is b because complementary angles are also right angles, they have to add up to 90 degrees. Number 5 is a because every angle is less than 90 degree. Number 6 is 30 because 60+90+x = 180 and when you solve for x you will get 30 degrees.
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Find the value of X.
SpyIntel [72]

Answer:

<h2>5</h2>

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5 0
3 years ago
Read 2 more answers
Can you explain step by step for #15 and #16 I have a test on this tomorrow
RUDIKE [14]

Answer:

15.

\left\{\begin{array}{l}x+y=50\\0.2x+0.1y=6\end{array}\right.

16. 10 ml of 20% saline and 40 ml of 10% saline

Step-by-step explanation:

A chemist takes x ml of 20% saline and y ml of 10% saline. In total, he takes

x + y ml that is 50 ml, so

x + y = 50.

There are  0.2x ml of salt in x ml of 20% saline and 0.1y ml of salt in 10% saline. There are 0.12\cdot 50=6 ml of salt in 50 ml of 12% saline. Thus,

0.2x+0.1y=6

15. We get the system of two equations:

\left\{\begin{array}{l}x+y=50\\0.2x+0.1y=6\end{array}\right.

16. Solve this system. From the first equation:

x=50-y

Substitute it into the second equation:

0.2(50-y)+0.1y=6\\ \\10-0.2y+0.1y=6\\ \\-0.1y=6-10\\ \\-0.1y=-4\\ \\y=40\\ \\x=50-y=50-40=10

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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