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Fynjy0 [20]
4 years ago
15

What is the value of 2(x+y)² when X=3 and Y=4 ?

Mathematics
2 answers:
Nina [5.8K]4 years ago
8 0

Answer:

{2(3 + 4)}^2 \\ 2 \times 14 \\ 28

Nadya [2.5K]4 years ago
3 0

Answer:

98

Step-by-step explanation:

2 (x + y) ²

= 2 (3 + 4 ) ²

= 2 x 7²

= 2 x 49

=  98

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First find the mean (sum divided by number of values)
4+5+8+10+15=42
42/5=8.4
Then find the difference between each of the numbers and the mean.
4.4, 3.4, 0.4, 1.6, 6.6
Then find the mean of those values.
4.4+3.4+0.4+1.6+6.6=16.4
16.4/5=3.28
Final answer: 3.28
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What is the Inequality and Solution of: The sum of a number and 19 is at least 8.2 ?
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X + 19 >= 8.2

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6 0
3 years ago
A sample size 25 is picked up at random from a population which is normally
Margarita [4]

Answer:

a) P(X < 99) = 0.2033.

b) P(98 < X < 100) = 0.4525

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 normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, 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.

Mean of 100 and variance of 36.

This means that \mu = 100, \sigma = \sqrt{36} = 6

Sample of 25:

This means that n = 25, s = \frac{6}{\sqrt{25}} = 1.2

(a) P(X<99)

This is the pvalue of Z when X = 99. So

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

By the Central Limit Theorem

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

Z = \frac{99 - 100}{1.2}

Z = -0.83

Z = -0.83 has a pvalue of 0.2033. So

P(X < 99) = 0.2033.

b) P(98 < X < 100)

This is the pvalue of Z when X = 100 subtracted by the pvalue of Z when X = 98. So

X = 100

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

Z = \frac{100 - 100}{1.2}

Z = 0

Z = 0 has a pvalue of 0.5

X = 98

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

Z = \frac{98 - 100}{1.2}

Z = -1.67

Z = -1.67 has a pvalue of 0.0475

0.5 - 0.0475 = 0.4525

So

P(98 < X < 100) = 0.4525

6 0
3 years ago
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