Mathematics College

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Answers

Answer 1

Answer:

a²+13a+40

Step-by-step explanation:

Now the x in the function has been replaced into (a+5) :

(a+5)²+3(a+5) =

(a²+10a+25)+(3a+15) =

a²+13a+40

Hope this helped and have a good day

Related Questions

which of the following are like terms3y^5, 2x^53y^5, 2y^56y^2, 2 z3y^4, 4x^3

Answers

In this case the answer is very simple.

3y^5, 2y^5 are like terms.

Because the variables and their exponents are the same.

Section 5.2-4 Graph the following system of equations and find the solution. Plot the solution on the graph. Enter your answer as (x,y). -2x-3y = 0 x+3y = 3 1 2 3 4 5 -1 -2 -3 -4 -5 1 2 3 4 5 -1 -2 -3 -4 -5 Clear All Draw: LineDot Solution =

Answers

For finding y, we can replace the value of x in any equations.

Solution (-3,2)

Plot graphs

If y varies directly as x, and y is 6 when x is 72, what is the value of y when x is 8?
1/9
2/3
54
96

Answers

Answer:

2/3

Step-by-step explanation:

y = xk

6 = 72K Solve for k Divide both sides by 72

[tex]\frac{1}{12}[/tex] = k

y = xk

y = [tex]\frac{8}{1}[/tex] x [tex]\frac{1}{12}[/tex]

y = [tex]\frac{8}{12}[/tex] I can simplify by dividing the numerator and denominator by 4

y = 2/3

The average score for games played in the NFL is 22 and the standard deviation is 9.3 points. 41 games are randomly selected. Round all answers to 4 decimal places where possible and assume a normal distribution.

a. What is the distribution of ¯x x¯
? ¯xx¯ ~ N( , )
b. What is the distribution of ∑x ? ∑x ~ N ( , )
c. P( ¯x > 19.8214) =
d. Find the 60th percentile for the mean score for this sample size.
e. P(20.6214 < x¯< 23.2262) =
f. Q1 for the x¯distribution =
g. P( ∑x > 829.0774) =

For part c) and e), Is the assumption of normal necessary? NoYes

Answers

Using the normal distribution and the central limit theorem, it is found that:

a) The distribution is: x¯ ~ N(22, 1.45).

b) The distribution is: ∑x ~ N(902, 59.55).

c) P( ¯x > 19.8214) = 0.9332 = 93.32%.

d) The 60th percentile for the mean score for this sample size is of 22.37 points a game.

e) P(20.6214 < x¯< 23.2262) = 0.6312 = 63.12%.

f) Q1 for the x¯distribution = 21 points a game.

g) P( ∑x > 829.0774) = 0.8888 = 88.88%.

Assumption of normality is not necessary, as the sample sizes are greater than 30.

Normal Probability Distribution

The z-score of a measure X of a variable that has mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by the rule presented as follows:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The z-score measures how many standard deviations the measure X is above or below the mean of the distribution, depending if the z-score is positive or negative.From the z-score table, the p-value associated with the z-score is found, and it represents the percentile of the measure X in the distribution.By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].Also by the Central Limit Theorem, for the sum of n instances of a variable, the mean is of [tex]\n\mu[/tex] and the standard deviation is of [tex]\sigma\sqrt{n}[/tex].Finally, by the Central Limit Theorem, assumption of normality is only necessary when the sample size is less than 30.

For a single game, the mean and the standard deviation of the number of points scored are given as follows:

[tex]\mu = 22, \sigma = 9.3[/tex]

For the average of 41 games, the standard error is:

[tex]s = \frac{9.3}{\sqrt{41}} = 1.45[/tex]

Hence the distribution is: x¯ ~ N(22, 1.45).

For the sum of the 41 games, the mean and the standard error are given as follows:

41 x 22 = 902.[tex]s = 9.3\sqrt{41} = 59.55[/tex].

Hence the distribution is: ∑x ~ N(902, 59.55).

In item c, the probability is one subtracted by the p-value of Z when X = 19.8214, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem: