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elena-s [515]
3 years ago
8

Use the triangle inequality to determine the largest angle in the figure HELP!

Mathematics
2 answers:
Anettt [7]3 years ago
7 0
I think the answer is SI
mestny [16]3 years ago
4 0
Sl hopefully this helps you
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Anyone ?? Know the answers to these both? Due tmr morning
tresset_1 [31]

Answer:

Just use m a t h w a y for the answer

Step-by-step explanation:

Thank me later,or dont thank me at all

6 0
3 years ago
Find the measure Of PR
Misha Larkins [42]

Answer:

18

Step-by-step explanation:

The formula for this is a x b = c x d where a is SR, b is FR, c is QR and d is PR. This will give us the equation 9(16) = 8(5x + 8); 144 = 40x + 64; 80 = 40x; x = 2. Now, plug this into PR which will be 5(2)+8 = 18. This is your answer.

3 0
3 years ago
The mean weight of an adult is 69 kilograms with a variance of 121. If 31 adults are randomly selected, what is the probability
amid [387]

Answer:

0.2236 = 22.36% probability that the sample mean would be greater than 70.5 kilograms.

Step-by-step explanation:

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

Also, important to remember that the standard deviation is the square root of the variance.

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 random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 69, \sigma = \sqrt{121} = 11, n = 31, s = \frac{11}{\sqrt{31}} = 1.97565

What is the probability that the sample mean would be greater than 70.5 kilograms?

This is 1 subtracted by the pvalue of Z when X = 70.5. So

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

By the Central limit theorem

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

Z = \frac{70.5 - 69}{1.97565}

Z = 0.76

Z = 0.76 has a pvalue of 0.7764

1 - 0.7764 = 0.2236

0.2236 = 22.36% probability that the sample mean would be greater than 70.5 kilograms.

8 0
3 years ago
PLEASE HELP !! ILL GIVE BRAINLIEST *EXTRA POINTS*.. <br> IM GIVING 40 POINTS !! DONT SKIP :((.
Fiesta28 [93]

Answer:

y = 1x - 5

Step-by-step explanation:

8 0
3 years ago
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