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Tomtit [17]
3 years ago
5

Please dont steal I just need a bit of help here I would really appreciate it I will give brainless

Mathematics
1 answer:
Ksenya-84 [330]3 years ago
6 0

Answer:

<h2>7.25</h2>

Step-by-step explanation:

I gave you the answer earlier too!

for more info !

<h2>there s a theorem called MIDPOINT THEOREM </h2>

Which means the 3rd side of the triangle is equal to 1/2 of the line drawn by joining the mid points

<h2>so AB=1/2of YZ</h2>

which means

2x-4 = 1/2×21

thus, 2(2x-4) =21

=> 4x-8 = 21

4x = 29

<h2>x= 29/4 = 7.25</h2>

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I hope this helps you



Area=1/2.b.c.sinA


Area =1/2.4.3.0,99


Area =5,98
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3 years ago
Simplify (2x^2y)^3 (-3xy^2)^2
solniwko [45]
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A population has a mean of 200 and a standard deviation of 50. Suppose a sample of size 100 is selected and x is used to estimat
zmey [24]

Answer:

a) 0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b) 0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

Step-by-step explanation:

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

Normal probability distribution

When the distribution is normal, we use 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 = 200, \sigma = 50, n = 100, s = \frac{50}{\sqrt{100}} = 5

a. What is the probability that the sample mean will be within +/- 5 of the population mean (to 4 decimals)?

This is the pvalue of Z when X = 200 + 5 = 205 subtracted by the pvalue of Z when X = 200 - 5 = 195.

Due to the Central Limit Theorem, Z is:

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

X = 205

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

Z = \frac{205 - 200}{5}

Z = 1

Z = 1 has a pvalue of 0.8413.

X = 195

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

Z = \frac{195 - 200}{5}

Z = -1

Z = -1 has a pvalue of 0.1587.

0.8413 - 0.1587 = 0.6426

0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b. What is the probability that the sample mean will be within +/- 10 of the population mean (to 4 decimals)?

This is the pvalue of Z when X = 210 subtracted by the pvalue of Z when X = 190.

X = 210

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

Z = \frac{210 - 200}{5}

Z = 2

Z = 2 has a pvalue of 0.9772.

X = 195

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

Z = \frac{190 - 200}{5}

Z = -2

Z = -2 has a pvalue of 0.0228.

0.9772 - 0.0228 = 0.9544

0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

7 0
3 years ago
1. Maggie keeps track of daily bird sightings. The table below shows the number and type of swans that were seen in one day. Wha
Vitek1552 [10]

Answer:

hello! i just did this problem and the answer is 12.2%

Step-by-step explanation:

first we need to add all of the swans together, then from there you take the amount of mute swans and divide it by the sum of all the swans! finally, multiply by 100 to convert into a percent and round it to 12.2%

hope this helped plz give me brainliest! <3

8 0
2 years ago
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2. Equation: 2x + 6y = 26
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