MATHEMATICS COLLEGE

Evaluate: 2! + (8 - 3)!

Answers

Answer 1

Answer:

Answer:

122

Step-by-step explanation:

2! = 2

(8-3)! =

(5)! = 120

2+120 =122

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WRITE THE expression in complete factored form 2y^2(p-4)-7(p-4)

Answers

For this case we have the following expression:
2y ^ 2 (p-4) -7 (p-4)
The first thing you should observe is the similar terms in both parts of the expression.
We note that the term (p-4) is repeated in both parts of the expression.
Therefore, by doing common factor (p-4) we have:
(p-4) (2y ^ 2-7)
Answer:
The expression in complete factored form is:
(p-4) (2y ^ 2-7)

p - 4 is common to the 2 parts so we have

(2y^2 - 7)(p - 4) Answer

In determining automobile-mileage ratings, it was found that the mpg (X) for a certain model is normally distributed, with a mean of 33 mpg and a standard deviation of 1.7 mpg. Find the following:__________. a. P(X<30)
b. P(28c. P(X>35)
d. P(X>31)
e. the mileage rating that the upper 5% of cars achieve.

Answers

The upper 5% of cars have a mileage rating of 35.805 mpg

What is z score?

Z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:

z = (raw score - mean) / standard deviation

Given; mean of 33 mpg and a standard deviation of 1.7

a) For < 30:

z = (30 - 33)/1.7 = -1.76

P(x < 30) = P(z < -1.76) = 1 - 0.8413 = 0.0392

b) For < 28:

z = (28 - 33)/1.7 = -2.94

P(x < 28) = P(z < -2.94) = 0.0016

c) For > 35:

z = (35 - 33)/1.7 = 1.18

P(x > 35) = P(z > 1.18) = 1 - P(z < 1.18) = 1 - 0.8810 = 0.119

d) For > 31:

z = (31 - 33)/1.7 = -1.18

P(x > 31) = P(z > -1.18) = 1 - P(z < -1.18) = 0.8810

e) The upper 5% of cars achieve have a z score of 1.65, hence:

1.65 = (x - 33)/1.7

x = 35.805 mpg

The upper 5% of cars have a mileage rating of 35.805 mpg

Find out more on z score at: brainly.com/question/25638875

Answer:

a) P(X < 30) = 0.0392.

b) P(28 < X < 32) = 0.2760

c) P(X > 35) = 0.1190

d) P(X > 31) = 0.8810

e) At least 35.7965 mpg

Step-by-step explanation:

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:

$Z = (X - \mu)/(\sigma)$

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.

In this question, we have that:

$\mu = 33, \sigma = 1.7$

a. P(X<30)

This is the pvalue of Z when X = 30. So

$Z = (X - \mu)/(\sigma)$

$Z = (30 - 33)/(1.7)$

$Z = -1.76$

$Z = -1.76$ has a pvalue of 0.0392.

Then

P(X < 30) = 0.0392.

b) P(28 < X < 32)

This is the pvalue of Z when X = 32 subtracted by the pvalue of Z when X = 28. So

X = 32

$Z = (X - \mu)/(\sigma)$

$Z = (32 - 33)/(1.7)$

$Z = -0.59$

$Z = -0.59$ has a pvalue of 0.2776.

X = 28

$Z = (X - \mu)/(\sigma)$

$Z = (28 - 33)/(1.7)$

$Z = -2.94$

$Z = -2.94$ has a pvalue of 0.0016.

0.2776 - 0.0016 = 0.2760.

P(28 < X < 32) = 0.2760

c) P(X>35)

This is 1 subtracted by the pvalue of Z when X = 35. So

$Z = (X - \mu)/(\sigma)$

$Z = (35 - 33)/(1.7)$

$Z = 1.18$

$Z = 1.18$ has a pvalue of 0.8810.

1 - 0.8810 = 0.1190

P(X > 35) = 0.1190

d. P(X>31)

This is 1 subtracted by the pvalue of Z when X = 31. So

$Z = (X - \mu)/(\sigma)$

$Z = (31 - 33)/(1.7)$

$Z = -1.18$

$Z = -1.18$ has a pvalue of 0.1190.

1 - 0.1190 = 0.8810

P(X > 31) = 0.8810

e. the mileage rating that the upper 5% of cars achieve.

At least the 95th percentile.

The 95th percentile is X when Z has a pvalue of 0.95. So it is X when Z = 1.645. Then

$Z = (X - \mu)/(\sigma)$

$1.645 = (X - 33)/(1.7)$

$X - 33 = 1.645*1.7$

$X = 35.7965$

At least 35.7965 mpg

4. A recipe says that 2 3/5 cups of flour are needed to make 1 batch of biscuits. How many cups of flour are needed to make 2 batches of biscuits? O 43/5 O 51/5 O 26/10 O 6

Answers

the answer is the second option, 5 1/5

Sixteen college freshmen were asked to record the number of alcoholic drinks they typically consume in a week. Here are their data: 2, 4, 6, 0, 1, 10, 9, 0, 6, 3, 6, 8, 5, 4, 6, 2. What is the variance of the number of alcoholic drinks consumed per week?a.3.26
b.2.96
c.12.25
d.8.75

Answers

Solution: The correct option is d. 8.75

Explanation:

The formula for variance is:

$Variance =\frac{\sum(x-\bar{x})^(2)}{n}$

First we need to find the mean $\bar{x}$ of the given data:

$\bar{x}=(2+4+6+0+1+10+9+0+6+3+6+8+5+4+6+2)/(16)=(72)/(16) =4.5$

Now let's find $\sum(x-\bar{x})^(2)$ , please have a look at the attached picture:

$\therefore Variance = (140)/(16)=8.75$

During the soccer season, Cathy made 27 of the 54 goals she attempted. Karla made 18 of the 45 goals she attempted. Patty made 34% of the goals she attempted. List the athletes in order of their goal-scoring percentage from least to greatest.