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Sergio039 [100]
3 years ago
14

Which set of numbers can represent the side lengths, in inches, of an acute triangle?

Mathematics
2 answers:
PSYCHO15rus [73]3 years ago
8 0

Answer:

5, 7, 8

Step-by-step explanation:

Serhud [2]3 years ago
5 0

Answer:

An acute triangle is less than 90 degrees. What set of numbers equal less to 90 degrees?

Possible numbers: 30, 12, 16. These equal 58.

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James has $32 and earns $10 per week for his allowance
Leno4ka [110]
The initial value is how much he starts out with, and it says he has $32
So the initial value is B, $32
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I know i have asked a lot of questions but i am in need of help
viva [34]

the answer is 31/12


I hope this helps (;

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3 years ago
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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
1.which side of AFED does AB in AABC correspond to?
erma4kov [3.2K]

Rotate one of them so the right angle is in the same orientation as the other one.

1. AB = DE

2. CB = FE

3. AC = DF

4. Compare the length of two known sides: cb and EF

CB = 3 and EF = 8

8/3 = 2 2/3 scale factor

5. Ab is side de. Multiply the length of ab by the scale factor:

4 x 2 2/3 = 10 2/3

6. FD = sqrt ( 10 2/3^2 + 8^2)

FD = 13 1/3

6 0
3 years ago
What is the probability that at least 2 of 3 randomly selected students are in th3 band
Nutka1998 [239]
The estimated probability would be 0.1... I had this test yesterday. hope this helps
4 0
3 years ago
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