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Y_Kistochka [10]
3 years ago
14

Find the measures of ZX and ZY.

Mathematics
2 answers:
Ulleksa [173]3 years ago
6 0

mzh=hahsvvrbhegdvvbehe

Lelechka [254]3 years ago
5 0

Answer: 303

Step-by-step explanation:

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Holly rode her bike for 24 miles at 6 miles per hour. She finished her ride at 1:00 p.m.
Bezzdna [24]
Time = distance/speed
Since you want to find the time Holly spent riding, you need to divide her distance (24 miles) by her speed (6 miles/hour) to get the number of hours (4) that she rode. Her starting time added to the time spend riding will give her ending time. One must subtract the riding time from the ending time to find the starting time.

Selection A is appropriate.
8 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
Simplify the expression. Write your answer as a power.
kolbaska11 [484]
(-3/4)^10 please mark as brainiest :)
4 0
3 years ago
Steven investigates the amount of damage to the head gaskets on the trucks in his fleet and finds that the damage index depends
Ierofanga [76]
It's D because it decreases by 2/3 of a point
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3 years ago
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PLEASE HELP!!! I HAVE A TEST!!!
dsp73
PLEASE HELP!!! I HAVE A TEST!!!
121 is 55% of what number?
C. 66.55
4 0
3 years ago
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