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Lana71 [14]
3 years ago
13

Please answer the two questions.

Mathematics
2 answers:
Anarel [89]3 years ago
6 0

Answer:

1. n= 14

2. n=3

Step-by-step explanation:

For the first problem, let's break down the equation. The number being thought about can be represented by <em>n</em>. Then <em>n </em>is increased by 7, or simply put 7+ <em>n. </em>The sum is 21, so to find n you can use the equation 7 + n = 24 and solve for <em>n, </em>which is 14.

For problem two, the same strategy can be used. The number is <em>n, </em>multiplied by 9. So, 9 * <em>n</em> is equal to 27. Solve for n by isolating n, and the answer is 3.

nika2105 [10]3 years ago
4 0

Answer:

1. n=14

2. n=3

Step-by-step explanation:

What you could do is if it says the sum of whatever and it gives you a number in the equation you can set it up.

#1. So for the first one the sum is (21) and a number it gives you is (7). It says <em>increased</em> and it says <em>sum </em>so that means you add.

7+n=21                  *question set up in equation*

-7     -7                   *subtract 7 from both sides to isolate n and get you're answer**

n=14

#2. For the 2nd one it says <em>multiplied</em> and <em>product</em> so you will be using multiplication. It gives you the product (27) and a number in the equation (9). To set it up:

9xn=27                      *question set up in equation*

9/9  =27/9                  *divide 9 by both sides to get answer. You divide bc its the opposite of multiplication**

n=3

<u>For the inverses</u>

If it is addition, subtract

If it is subtraction, add

If it is multiplication, divide

If it is division, multiply

<u>For context clues to set up equation</u>

Sum--> addition

Difference--> subtraction

Quotient--> division

Product--> multiplication


Hope it helps!

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The mean of a population is 74 and the standard deviation is 15. The shape of the population is unknown. Determine the probabili
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a) 0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.

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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.

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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.

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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.

The mean of a population is 74 and the standard deviation is 15.

This means that \mu = 74, \sigma = 15

Question a:

Sample of 36 means that n = 36, s = \frac{15}{\sqrt{36}} = 2.5

This probability is 1 subtracted by the pvalue of Z when X = 78. So

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

By the Central Limit Theorem

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

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Z = 1.6

Z = 1.6 has a pvalue of 0.9452

1 - 0.9452 = 0.0548

0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.

Question b:

Sample of 150 means that n = 150, s = \frac{15}{\sqrt{150}} = 1.2247

This probability is the pvalue of Z when X = 77 subtracted by the pvalue of Z when X = 71. So

X = 77

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

Z = \frac{77 - 74}{1.2274}

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Z = 2.45 has a pvalue of 0.9929

X = 71

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

Z = \frac{71 - 74}{1.2274}

Z = -2.45

Z = -2.45 has a pvalue of 0.0071

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c. A random sample of size 219 yielding a sample mean of less than 74.2

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Z = \frac{74.2 - 74}{1.0136}

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Z = 0.2 has a pvalue of 0.5793

0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2

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