H(x) = −(x − 5)2 + 3g(x) = 4 cos(2x − π) − 2
f(x)

x y
0 −5
1 0
2 3
3 4
4 3
5 0
6 −5
Which function has the largest maximum?

Answers:
 f(x)
 g(x)
 h(x) 
All three functions have the same maximum value.

Answers

Answer 1
Answer:

Answer:

The correct option is 1. The function f(x) has the largest maximum.

Step-by-step explanation:

The vertex form of a parabola is

y=a(x-h)^2+k

Where, (h,k) is vertex.

The given functions is

h(x)=-(x-5)^2+3

Here, a=-1, h=-5 and k=3. Since the value of a is negative, therefore it is an downward parabola and vertex is the point of maxima.

Thus the maximum value of the function h(x) is 3.

g(x)=4\cos (2x-\pi)-2

The value of cosine function lies between -1 to 1.

-1\leq \cos (2x-\pi)\leq 1

Multiply 4 on each side.

-4\leq 4\cos (2x-\pi)\leq 4

Subtract 2 from each side.

-4-2\leq 4\cos (2x-\pi)-2\leq 4-2

-6\leq 4\cos (2x-\pi)-2\leq 2

Therefore the maximum value of the function g(x) is 2.

From the given table it is clear that the maximum value of the function f(x) is 4 at x=3.

Since the function f(x) has the largest maximum, therefore the correct option is 1.

Answer 2
Answer: Max(f(x)) = f(3) = 4

Max(g(x)) = g(π/2) = 2

Max(h(x)) = h(5) = 3

Of these values, f(3) = 4 is the largest.


The function with the largest maximum is f(x).

Related Questions

QuestionIf $500 is borrowed with an interest of 21.0% compounded monthly, what is the total amount of money needed to pay itback in 1 year? Round your answer to the nearest dollar. Do not round at any other point in the solving process; only roundyour final answer.
If f(x) = x/2 +8, what is f(x) when x=10?
Find he least common multiple of 16, 21, and 24
What is the following sum?5(3squareroot x) +9 (3squareroot x)
Select the correct answer. Solve the following equation by completing the square. 1/4x^2 + x + 1/4 = 0

Suppose you are asked to choose a whole number between 1 and 13, inclusive. (a) What is the probability that it is odd? (b) What is the probability that it is even? (c) What is the probability that it is a multiple of 3?

Answers

a. this depends on if 1 and 13 are included. If so, it would be 7 of 13. If not, 5 of 11
b. same thing, depends on if 1 and 13 are included. If so, 6 of 13. If not, 6 of 11
c. again, depends. If so, 4 of 13. If not, 4 of 11

Julio is planning activities for a trip that lasts from Thursday to Sunday. He is interested in visiting a museum, hiking the trails, attending the food festival, riding a jet ski, exploring a famous park, or going shopping. each activity takes a full day, and he will not do any activity twice. Julio wants to either visit a museum or explore the park on Thursday, and he is definitely attending the food festival on Saturday. How many different ways can he plan activities for his trip?A)24
B)30
C)40
D)120

Answers

Thursday: 2 choices {museum, park}

Friday: 4 choices {shopping, jet ski, hiking, and the choice not chosen for Thu}

Saturday: 1 choice {food}

Sunday: 3 choices {the remaining 3 after Friday's choice}

The total number of ways these choices can be arranged is 2×4×1×3 = 24

The apprpropriate selection is A) 24.

If $75 is invested at an interest rate of 8% per year and is compounded monthly, how much money is in the account in 15 years?

Answers

The amount of money is $248.02.

Principal amount = P=75

Interest rate = r = 8% = 0.08

Number of years = t = 15

Number of times compounded in a year = n = 12

A = Amount after t years.

After 15 years there will be:

A=P\left(1+(r)/(n)\right)^(nt)\nA=75\left(1+(0.08)/(12)\right)^(\left(12\cdot15\right))\nA=248.019110806\nA \approx 248.02

Learn more: brainly.com/question/10241002

Answer:

$248.03

Step-by-step explanation:

The formula you use for this is as follows:

A(t)=P(1+(r)/(n))^(nt)

where A(t) is the amount after the compounding is done, P is the initial amount invested, r is the interest rate in decimal form, n is the number of times the compounding is done per year, and t is the time in years.  Using that information and filling in our equation gives us this:

A(t)=75(1+(.08)/(12))^((12)(15))

which simplifies down to

A(t)=75(1+.0066667)^(180)

which simplifies further to

A(t)=75(3.307118585)

which multiplies to $248.0338938.  Round to the nearest cent to get your answer.

A hypothesis regarding the weight of newborn infants at a community hospital is that the mean is 6.6 pounds. A sample of seven infants is randomly selected, and their weights at birth are recorded as:
9.0, 7.3, 6.0, 8.8, 6.8, 8.4, and 6.6 pounds.
If Alpha = 0.05,
1. What is the critical t-value?
2. What is the decision for a statistically significant change in average weights at birth at the 5% level of significance?

Answers

Answer:

1. Critical value t=±2.447

2. The null hypothesis is failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the birth weight significantly differs from 6.6 lbs.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the birth weight significantly differs from 6.6 lbs.

Then, the null and alternative hypothesis are:

H_0: \mu=6.6\n\nH_a:\mu\neq 6.6

The significance level is 0.05.

The sample has a size n=7.

The sample mean is M=7.56.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=1.18.

The estimated standard error of the mean is computed using the formula:

s_M=(s)/(√(n))=(1.18)/(√(7))=0.446

Then, we can calculate the t-statistic as:

t=(M-\mu)/(s/√(n))=(7.56-6.6)/(0.446)=(0.96)/(0.446)=2.152

The degrees of freedom for this sample size are:

df=n-1=7-1=6

For a two-tailed test with 5% level of significance and 6 degrees of freedom, the critical value for t is ±2.447.

As the test statistic t=2.152 is under 2.447 and over -2.447, it falls in the acceptance region, so the effect is not significant. The null hypothesis is failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the birth weight significantly differs from 6.6 lbs.

Sample mean and standard deviation calculations:

M=(1)/(n)\sum_(i=1)^n\,x_i\n\n\nM=(1)/(7)(9+7.3+6+. . .+6.6)\n\n\nM=(52.9)/(7)\n\n\nM=7.56\n\n\ns=\sqrt{(1)/(n-1)\sum_(i=1)^n\,(x_i-M)^2}\n\n\ns=\sqrt{(1)/(6)((9-7.56)^2+(7.3-7.56)^2+(6-7.56)^2+. . . +(6.6-7.56)^2)}\n\n\ns=\sqrt{(8.32)/(6)}\n\n\ns=√(1.39)=1.18\n\n\n

Select all that justify the following statement.2•1/2=1
commutative - addition
inverse - addition
associative - multiplication
symmetric
commutative - multiplication
associative - addition
inverse - multiplication

Answers

Answer:

G: Inverse multiplication

Step-by-step explanation:

Statement is 2•1/2=1

Now, what this means is that when we multiply a number by it's inverse form, the result will be 1.

This means that this statement denotes an inverse multiplication.

Thus, the last option is the correct answer

What is the least What is the least common multiple of four and 12

Answers

Answer: LCM(4,12) = 12

Step-by-step explanation:

This of this as singing all the multiples of 4

4, 8, 12.

4 + 4 + 4.

Answer:

2

Step-by-step explanation:

Other Questions
Consider the following analogy: You are a hiring manager for a large company. For every job applicant, you must decide whether to hire the applicant based on your assessment of whether he or she will be an asset to the company. Suppose your null hypothesis is that the applicant will not be an asset to the company. As in hypothesis testing, there are four possible outcomes of your decision: (1) You do not hire the applicant when the applicant will not be an asset to the company, (2) you hire the applicant when the applicant will not be an asset to the company, (3) you do not hire the applicant when the applicant will be an asset to the company, and (4) you hire the applicant when the applicant will be an asset to the company. 1. Which of the following outcomes corresponds to a Type I error? A. You hire the applicant when the applicant will not be an asset to the company. B. You do not hire the applicant when the applicant will be an asset to the company. C. You do not hire the applicant when the applicant will not be an asset to the company. D. You hire the applicant when the applicant will be an asset to the company. 2. Which of the following outcomes corresponds to a Type II error? A. You hire the applicant when the applicant will not be an asset to the company. B. You hire the applicant when the applicant will be an asset to the company. C. You do not hire the applicant when the applicant will be an asset to the company. D. You do not hire the applicant when the applicant will not be an asset to the company. As a hiring manager, the worst error you can make is to hire the applicant when the applicant will not be an asset to the company. The probability that you make this error, in our hypothesis testing analogy, is described by:________.