answer plz no explanation needed

Hi there!

A.) Begin by verifying that both endpoints have the same y-value:

g(-1) = 2(-1)² - 4(-1) + 3

Simplify:

g(-1) = 2 + 4 + 3 = 9

g(2) = 2(2)² - 4(2) + 3 = 8 - 8 + 3 = 3

Since the endpoints are not the same, Rolle's theorem does NOT apply.

B.)

Begin by ensuring that the function is continuous.

The function is a polynomial, so it satisfies the conditions of the function being BOTH continuous and differentiable on the given interval (All x-values do as well in this instance). We can proceed to find the values that satisfy the MVT:

[tex]f'(c) = \frac{f(a)-f(b)}{a-b}[/tex]

Begin by finding the average rate of change over the interval:

[tex]\frac{g(2) - g(-1)}{2-(-1)} = \frac{3 - 9 }{2-(-1)} = \frac{-6}{3} = -2[/tex]

Now, Find the derivative of the function:

g(x) = 2x² - 4x + 3

Apply power rule:

g'(x) = 4x - 4

Find the x value in which the derivative equals the AROC:

4x - 4 = -2

Add 4 to both sides:

4x = 2

Divide both sides by 4:

x = 1/2

Solution :

Given data :

Total number of books sold each month= 700

The charts in the display attached below shows the most popular books category by percentages.

Percentage of romance books sold each = 8.5%

Therefore, the number of romance books sold in each month is given by :

[tex]$=8.5 \% \text{ of }\ 700$[/tex]

[tex]$=\frac{8.5}{100}\times 700$[/tex]

= 59.5

≈ 60 books (rounding off)

You can use the fact that total amount is taken as 100%.

The number of romance books in the given Streets Books is 60

Suppose the value of which a thing is expressed in percentage is "a'

Suppose the percent that considered thing is of "a" is b%

Then since percent shows per 100 (since cent means 100), thus we will first divide the whole part in 100 parts and then we multiply it with b so that we collect b items per 100 items(that is exactly what b per cent means).

Thus, that thing in number is [tex]\dfrac{a}{100} \times b[/tex]

Since the total amount of books is 700, and its 8.5% books are romance books, thus we have:

[tex]\text{Number of Romance books} = \dfrac{700}{100} \times 8.5 = 59.5 \approx 60[/tex]

The number of romance books in the given Streets Books is 60

Learn more about percentage here:

https://brainly.com/question/11549320

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Answer:

  (a)  y = −50x + 250

Step-by-step explanation:

In case you don't realize that the graph starts at 250 and decreases by 50 for each increase of 1 in x, you can see if any of the equations match the given points. The only one that does is the first one:

  y = -50x +250

Answer:

(a)  y = −50x + 250

Step-by-step explanation:

Answer:

i think 4 4 kilograms if im wrong sorry

Step-by-step explanation:

Answer:

The sample size necessary is of 168.

Step-by-step explanation:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1 - 0.95}{2} = 0.025[/tex]

Now, we have to find z in the Z-table as such z has a p-value of [tex]1 - \alpha[/tex].

That is z with a pvalue of [tex]1 - 0.025 = 0.975[/tex], so Z = 1.96.

Now, find the margin of error M as such

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

Based on a previous study, arrival delay times have a standard deviation of 39.6 minutes.

This means that [tex]\sigma = 39.6[/tex]

Find the sample size necessary to estimate the mean arrival delay time for all American Airlines flights from Dallas to Sacramento to within 6 minutes with 95% confidence.

This is n for which M = 6. So

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

[tex]6 = 1.96\frac{39.6}{\sqrt{n}}[/tex]

[tex]6\sqrt{n} = 1.96*39.6[/tex]

[tex]\sqrt{n} = \frac{1.96*39.6}{6}[/tex]

[tex](\sqrt{n})^2 = (\frac{1.96*39.6}{6})^2[/tex]

[tex]n = 167.34[/tex]

Rounding up:

The sample size necessary is of 168.

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Answer:

  f = 3 units

Step-by-step explanation:

The ratios of side lengths in this 30°-60°-90° triangle are ...

  1 : √3 : 2

So, the ratio of interest is ...

  1 : √3 = √3 : f

We can see that the numbers in the second ratio are √3 times the numbers in the first ratio, so

  f = √3 × √3 = 3

  f = 3 units

Answer:

y = 7cos bx

Step-by-step explanation:

For a cosine function without pahse shift and vertical shift, but with amplitude given, it will also have period and thus , the formula for the cosine function is;

y = Acos bx

Where;

A is the amplitude

Period = 2π/b

Now, we are told that the amplitude is 7. Thus;

y = 7cos bx

Answer:

5x+20y=425

Step-by-step explanation:

Its 5 bucks for x pairs of skates

Its 20 dollars for y bikes

x+y rentals have to equal 25

all of this is equal to 425. All that is left to do is test with number until the statement is true.

try :

5(5)+(20)(20)=425

x + y do equal 25, and the total is equal to 425.

Answer:

X^2 + 3x - 10=0

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Answer:

  1. see below

  2. g(x) is f(x) translated left 2 and down 3

Step-by-step explanation:

1. The graphs are attached. F(x) is in red; g(x) is in blue.

__

2. The graph of g(x) = f(x -h) +k is the parent function translated by (h, k). Here we have (h, k) = (-2, -3), so g(x) is f(x) translated left 2 and down 3.

Answer:

A sample of 18 is required.

Step-by-step explanation:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1 - 0.92}{2} = 0.04[/tex]

Now, we have to find z in the Z-table as such z has a p-value of [tex]1 - \alpha[/tex].

That is z with a pvalue of [tex]1 - 0.04 = 0.96[/tex], so Z = 1.88.

Now, find the margin of error M as such

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

A previous study indicated that the standard deviation was 2.2 days.

This means that [tex]\sigma = 2.2[/tex]

How large a sample must be selected if the company wants to be 92% confident that the true mean differs from the sample mean by no more than 1 day?

This is n for which M = 1. So

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

[tex]1 = 1.88\frac{2.2}{\sqrt{n}}[/tex]

[tex]\sqrt{n} = 1.88*2.2[/tex]

[tex](\sqrt{n})^2 = (1.88*2.2)^2[/tex]

[tex]n = 17.1[/tex]

Rounding up:

A sample of 18 is required.

Answer:

-11

Step-by-step explanation:

I am going to assume that it is -x^2-5y^3.

-(4^2)-5(-1^3)

-16-5(-1)

-16+5

-11

Answer:

- 11

Step-by-step explanation:

If x = 4,  y = -1

then,

        - x^2 - 5y^3 = - (4)^2 - 5(-1)^3

                            = - 16 + 5

                            = - 11

Answer:

$520.632

Step-by-step explanation:

Answer:

The 1st,Thrid, Fifth Option

Step-by-step explanation:

The first option is true. We can move the orginal square root function to get g(x).

The second option is false. Function g(x) which equals

[tex] \sqrt{x - 3} - 1[/tex]

Domain is all real numbers greater than or equal to 3.

The third option is true. Since minimum point we can get is 0 in a square root function. We have a vertical shift so our new minimum point is

[tex]0 - 1 = - 1[/tex]

We can take the sqr root of 0 so

So all real numbers that are greater than or equal to -1 is true.

The fourth option is false, we need to add 3 instead of subtract 3.

The fifth option is true, we can do that to get back to our original function

Answer:

A) x = 0.

B) f is concave up for (-∞, 0).

C) f is concave down for (0, ∞).

Step-by-step explanation:

We are given the function:

[tex]f(x)=5+12x-x^3[/tex]

A)

We want to find the x-coordinates of all inflection points.

Recall that inflections points (may) occur when the second derivative equals zero. Hence, find the second derivative. The first derivative is given by:

[tex]f'(x) = 12-3x^2[/tex]

And the second:

[tex]f''(x) = -6x[/tex]

Set the second derivative equal to zero:

[tex]0=-6x[/tex]

And solve for x. Hence:

[tex]x=0[/tex]

We must test the solution. In order for it to be an inflection point, the second derivative must change signs before and after. Testing x = -1:

[tex]f''(-1) = 6>0[/tex]

And testing x = 1:

[tex]f''(1) = -6<0[/tex]

Since the signs change for x = 0, x = 0 is indeed an inflection point.

B)

Recall that f is concave up when f''(x) is positive, and f is concave down when f''(x) is negative.

From the testing in Part A, we know that f''(x) is positive for all values less than zero. Hence, f is concave up for all values less than zero. Our interval is:

[tex](-\infty, 0)[/tex]

C)

From Part A, we know that f''(x) is negative for all values greater than zero. So, f is concave down for that interval:

[tex](0, \infty)[/tex]

Answer:

Step-by-step explanation:

yes

Answer:

The projected world population in 2015 was 8,705,121,030 people.

Step-by-step explanation:

Given that the population of the world in 1987 was 5 billion and the annual growth rate was estimated at 2 percent per year, assuming that the world population follows an exponential growth model, to find the projected world population in 2015 the following calculation must be performed :

5,000,000,000 x 1.02 ^ (2015-1987) = X

5,000,000,000 x 1.02 ^ 28 = X

5,000,000,000 x 1.741024 = X

8,705,121,030 = X

Therefore, the projected world population in 2015 was 8,705,121,030 people.

Answer:

1/8

Step-by-step explanation:

Answer: 1/8

Step-by-step explanation:

1/8 + 1/8 = 2/8 = 1/4

Complete Question

The quality control manager at a computer manufacturing company believes that the mean life of a computer is 91 months with a standard deviation of 10 months if he is correct. what is the probability that the mean of a sample of 68 computers would differ from the population mean by less than 2.08 months? Round your answer to four decimal places. Answer How to enter your answer Tables Keypad

Answer:

[tex]P(-1.72<Z<1.72)=0.9146[/tex]

Step-by-step explanation:

From the question we are told that:

Population mean \mu=91

Sample Mean \=x =2.08

Standard Deviation \sigma=10

Sample size n=68

Generally the Probability that The  sample mean  would differ from the population mean

P(|\=x-\mu|<2.08)

From Table

[tex]P(|\=x-\mu|<2.08)=P(|z|<1.72)[/tex]

T Test

[tex]Z=\frac{\=x-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex]

[tex]Z=\frac{2.08}{\frac{10}{\sqrt{68} } }[/tex]

[tex]Z=1.72[/tex]

[tex]P(|\=x-\mu|<2.08)=P(|z|<1.72)[/tex]

[tex]P(-1.72<Z<1.72)[/tex]

Therefore From Table

[tex]P(-1.72<Z<1.72)=0.9146[/tex]

Answer:

The minimum sample size required to create the specified confidence interval is 295.

Step-by-step explanation:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1 - 0.95}{2} = 0.025[/tex]

Now, we have to find z in the Z-table as such z has a p-value of [tex]1 - \alpha[/tex].

That is z with a pvalue of [tex]1 - 0.025 = 0.975[/tex], so Z = 1.96.

Now, find the margin of error M as such

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

Variance of 0.49:

This means that [tex]\sigma = \sqrt{0.49} = 0.7[/tex]

They want to construct a 95% confidence interval with an error of no more than 0.08. What is the minimum sample size required to create the specified confidence interval?

The minimum sample size is n for which M = 0.08. So

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

[tex]0.08 = 1.96\frac{0.7}{\sqrt{n}}[/tex]

[tex]0.08\sqrt{n} = 1.96*0.7[/tex]

[tex]\sqrt{n} = \frac{1.96*0.7}{0.08}[/tex]

[tex](\sqrt{n})^2 = (\frac{1.96*0.7}{0.08})^2[/tex]

[tex]n = 294.1[/tex]

Rounding up:

The minimum sample size required to create the specified confidence interval is 295.

Step-by-step explanation:

Option B

The length of side will be 2m...

hope it helps

Answer:

Step-by-step explanation:

If we are looking for the time(s) that the ball is at a height of 15, we simply sub in a 15 for the height in the position equation and solve for t:

[tex]15=-16t^2+23t+7[/tex] and

[tex]0=-16t^2+23t-8[/tex]

Factor this however you factor a quadratic in class to get

t = .59 seconds and t = .85 seconds.

This means that .59 seconds after the ball was thrown into the air it was 15 feet off the ground. Then the ball reached its max height, gravity took over, and began pulling it back down to earth. The ball passes the height of 15 feet again on its way down after .85 seconds.

Answer:

A significant negative relationship exists between the variables

Step-by-step explanation:

Base on the information given in the question which goes thus : For every unit increase in x, y decreases by 12.8094. The value 12.8094 is the slope which is the rate of change in y variable per unit change in the independent variable. The sign or nature of the slope Coefficient gives an hint about the relationship between the x and y variables. The slope Coefficient in this case is negative and thus we'll have a negative relationship between the x and y variables (an increase in x leads to a corresponding decrease in y). This is a negative association.

Answer:

(a) [tex]H_o:\mu_1 = \mu_2[/tex]     [tex]H_a:\mu_1 > \mu_2[/tex]

(b) [tex]t = 0.74[/tex]

(c) [tex]p =0.2331[/tex]

(d) [tex]CI = (-2.095,4.095)[/tex]

Step-by-step explanation:

Given

[tex]\bar x_1=21,\ s_1=4,\ n_1=12,\\ \bar x_2=20,\ s_2=3,\ n_2=15[/tex]

Solving (a): The hypotheses

The test is right-tailed, means that the alternate hypothesis will contain greater than sign.

So, we have:

[tex]H_o:\mu_1 = \mu_2[/tex]

[tex]H_a:\mu_1 > \mu_2[/tex]

Solving (b); The test statistic (t)

This is calculated as:

[tex]t = \frac{\bar x_1 - \bar x_2}{\sqrt{\frac{s_1^2(n_1 - 1) + s_2^2(n_2 - 1)}{n_1 + n_2 - 2} * (\frac{1}{n_1} + \frac{1}{n_2})}}[/tex]

So, we have:

[tex]t = \frac{21 - 20}{\sqrt{\frac{4^2(12 - 1) + 3^2(15 - 1)}{12 + 15 - 2} * (\frac{1}{12} + \frac{1}{15})}}[/tex]

[tex]t = \frac{1}{\sqrt{\frac{302}{25} * (0.15)}}[/tex]

[tex]t = \frac{1}{\sqrt{12.08 * 0.15}}[/tex]

[tex]t = \frac{1}{\sqrt{1.812}}[/tex]

[tex]t = \frac{1}{1.346}[/tex]

[tex]t = 0.74[/tex]

Solving (c): The P-value

First, we calculate the degrees of freedom

[tex]df = n_1 + n_2 -2[/tex]

[tex]df = 12+15 -2[/tex]

[tex]df = 25[/tex]

Using the t distribution, the p-value is:

[tex]p =TDIST(0.74,25)[/tex]

[tex]p =0.2331[/tex]

Solving (d): The 90% confidence interval

Calculate significance level

[tex]\alpha = 1 - CI[/tex]

[tex]\alpha = 1 - 90\%[/tex]

[tex]\alpha = 0.10[/tex]

Calculate the t value (t*)

[tex]t^* = (\alpha/2,df)[/tex]

[tex]t^* = (0.10/2,25)[/tex]

[tex]t^* = (0.05,25)[/tex]

[tex]t^* = 1.708[/tex]

The confidence interval is calculated using:

[tex]CI = (\bar x - \bar x_2) \± t^* *\sqrt{\frac{s_1^2(n_1 - 1) + s_2^2(n_2 - 1)}{n_1 + n_2 - 2} * (\frac{1}{n_1} + \frac{1}{n_2})}[/tex]

[tex]CI = (21 - 20) \± 1.708 *\sqrt{\frac{4^2(12 - 1) + 3^2(15 - 1)}{12 + 15 - 2} * (\frac{1}{12} + \frac{1}{15})}[/tex]

[tex]CI = 1 \± 1.708 *1.812[/tex]

[tex]CI = 1 \± 3.095[/tex]

Split

[tex]CI = 1 - 3.095 \ or\ 1 + 3.095[/tex]

[tex]CI = -2.095 \ or\ 4.095[/tex]

[tex]CI = (-2.095,4.095)[/tex]

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Answer:

  4/5

Step-by-step explanation:

The wording is ambiguous, as it often is when math expressions are described in English. We assume you intend ...

  [tex]\dfrac{3n+8}{n+7}=\dfrac{4}{3}\\\\3(3n+8)=4(n+7)\qquad\text{multiply by $3(n+7)$}\\\\9n+24=4n+28\qquad\text{eliminate parentheses}\\\\5n=4\qquad\text{subtract $4n+24$}\\\\\boxed{n=\dfrac{4}{5}}\qquad\text{divide by 5}[/tex]

The number is 4/5.

Answer:

A) 21 in²

B) 42 in²

C) 84 in²

D) I) 4 in²

II) 8 in²

III) 16 in²

E) From our calculations, we can see that doubling one part of the dimensions gives an area that is twice the original one while doubling both dimensions gives an area that four times the original one.

Step-by-step explanation:

We are given dimensions of triangle as;

width; w = 3 inches

length; L = 7 inches

A) Area of triangle is;

A = Lw

A = 7 × 3

A = 21 in²

B) If we double the width, then area is;

A = 7 × (2 × 3)

A = 42 in²

Area is twice the original area

C) If we double the width and length, then we have;

Length = 7 × 2 = 14 in

Width = 3 × 2 = 6 in

Area = 14 × 6 = 84 in²

Area is four times the original one

D) Let's try a triangle with base 2 in and height 4 in.

I) formula for area of triangle is;

A = ½ × base × height

A = ½ × 2 × 4

A = 4 in²

II) If we double the width(base) , then area is;

A = ½ × 2 × 2 × 4

A = 8 in²

This is twice the original area.

III) If we double the width(base) and length(height), then we have;

A = ½ × 2 × 2 × 4 × 2

A = 16 in²

This is four times the original area

E) From our calculations, we can see that doubling one part of the dimensions gives an area that is twice the original one while doubling both dimensions gives an area that four times the original one.