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Brums [2.3K]
3 years ago
15

How many terms are in the expression 2m2n+2mn2-m+3n+9

Mathematics
1 answer:
Travka [436]3 years ago
6 0
Six or Seven. (Sorry if not right, haven’t done this in a while)
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He will have 16 fields 1/4 mile long. 
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Read 2 more answers
The age of United States Presidents on the day of their first inauguration follows a Normal distribution with mean 56 and standa
Talja [164]

Answer:

a) 0.7088 = 70.88% probability that a randomly selected President was less than 60 years old on the day of their first inauguration.

b) The 75th percentile for the age of United States Presidents on the day of inauguration is 61.

c) 0.8643 = 86.43% probability that the average age on the day of their first inauguration for a random sample of 4 United States Presidents exceeds 60 years.

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 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.

The age of United States Presidents on the day of their first inauguration follows a Normal distribution with mean 56 and standard deviation 7.3.

This means that \mu = 56, \sigma = 7.3

(a) (5 points) Compute the probability that a randomly selected President was less than 60 years old on the day of their first inauguration.

This is the pvalue of Z when X = 60. So

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

Z = \frac{60 - 56}{7.3}

Z = 0.55

Z = 0.55 has a pvalue of 0.7088

0.7088 = 70.88% probability that a randomly selected President was less than 60 years old on the day of their first inauguration.

(b) (5 points) Compute the 75th percentile for the age of United States Presidents on the day of inauguration.

This is X when Z has a pvalue of 0.75. So X when Z = 0.675.

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

0.675 = \frac{X - 56}{7.3}

X - 56 = 0.675*7.3

X = 61

The 75th percentile for the age of United States Presidents on the day of inauguration is 61.

(c) (5 points) Compute the probability that the average age on the day of their first inauguration for a random sample of 4 United States Presidents exceeds 60 years.

Now, by the Central Limit Theorem, we have that n = 4, s = \frac{7.3}{\sqrt{4}} = 3.65

This is the pvalue of Z when X = 60. So

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

By the Central Limit Theorem

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

Z = \frac{60 - 56}{3.65}

Z = 1.1

Z = 1.1 has a pvalue of 0.8643

0.8643 = 86.43% probability that the average age on the day of their first inauguration for a random sample of 4 United States Presidents exceeds 60 years.

4 0
3 years ago
What is the measure of c in the parallelogram shown
finlep [7]

Answer: B. 45 degrees

Step-by-step explanation:

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3 years ago
Help can’t seem to figure out the base number?
fredd [130]

9514 1404 393

Answer:

  B = 6.65 in²

Step-by-step explanation:

The base area (B) is the area of the triangular base of the prism. You already know the sides of that triangle are 3, 5, and 7 inches. The figure shows the altitude to the side length 7 is 1.9 inches. The triangle area formula applies:

  A = 1/2bh

  A = 1/2(7 in)(1.9 in) = 6.65 in² . . . . area of one triangular base.

Then the area of one base is B = 6.65 in².

__

The lateral area is the product of the perimeter and the height:

  LA = Ph = 15(12) = 180 . . . . square inches

The total surface area is the sum of the lateral area and the areas of the two bases.

  SA = LA + 2B

  SA = 180 in² + 2(6.65 in²) = 193.3 in²

3 0
3 years ago
Student spend 5 minutes of every 60 minutes at school moving from class to class. Which fraction is equivalent to the number of
Anastaziya [24]
A. 10/120 sorry if I’m wrong
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3 years ago
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