Please help 20 points

5x = 24

Step-by-step explanation:

[tex] - 7 + 5x - 2 = 15[/tex]

Combine like terms

[tex]5x = 15 + 7 + 2[/tex]

Further Workings :

Simplify

[tex]5x = 24[/tex]

[tex] \frac{5x}{5} = \frac{24}{5} \\ \\ x = 4.8[/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.

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:

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]

Answer:

True

Step-by-step explanation:

When covariance matrices of corresponding class are identical and diagonal matrix and their class probability is same then the class is normally distributed and its discriminant functions are linear.

Answer:

a. [tex]f(t) = (0.99)^{t} 100%\\[/tex]

b. The x - intercept represents the time at which the battery life drops to 0 %.

c. The y - intercept represents the initial battery life.

Step-by-step explanation:

a. Construct an equation using f(t) and t that represents the relationship between

"f(t)" the battery capacity of a typical smartphone after "t" months and "L" the

number of months the smartphone has been in normal use.

Let the initial battery life be I.

After one month, it decreases by 1%. So, our new value, I' = initial value - decrease = I - 1%I = 99%I.

After two months, it decreases by 1% to our new value I". So, our new value, I" = initial value - decrease = I' - 1%I' = 99%I' = 99%(99%)I = (99%)²I

After three months, it decreases by 1% to our new value I'". So, our new value, I"' = initial value - decrease = I" - 1%I" = 99%I" = 99%(99%)²I = (99%)³I

We see a pattern here.

After t months, f(t)

[tex]f(t) = (0.99)^{t} 100%\\[/tex]

b. What does the horizontal intercept represent in context? (The x-intercept)

The x - intercept represents the time at which the battery life drops to 0 %.

c. What does the vertical intercept represent in context? (The y-intercept)

The y - intercept represents the initial battery life.

Answer:

4(3x-10)

Step-by-step explanation:

12x-40 = (4*3)x-(4*10) = 4(3x-10). The GCF is 4.

9514 1404 393

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) 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:

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:

i think 4 4 kilograms if im wrong sorry

Step-by-step explanation:

Answer:

X^2 + 3x - 10=0

Answer:

Step-by-step explanation:

yes

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.

Answer:

Chisquare statistic

Step-by-step explanation:

The most appropriate test to use here is the Chisquare test as it isemployed when testing the relationship between two variables. Both the T statistic and z statistics are the used for testing difference in means. However, in the analysis above, we are given two variables with each having its own levels. Hence, the Chisquare test is the most appropriate in this kind of situation a it measures if a difference exists between the two variables. It tests if they are related or independent of on one another

Answer:

1. 55 degree

2. 50 degree

3. 88 degree

4. 50 degree

Step-by-step explanation:

1.

Angle AE interior 180-120 = 60

Angle CD interior 180-112 = 68

x = 180- (60+68) = 180-128 = 52 degree

2.

interior 180 -120= 40

interior 180- 110 = 90

x = 180 - (40+90) = 180-130 = 50 degree

3.

exterior 90-52= 38

exterior 90-40 = 50

x = 180-(52+40)= 180-92 = 88 degree

4.

interior 180- (45+50) = 180-95 = 85 degree

interior adjoining triangle 180 - 85 = 95

all angles add up to 180

interior 180- (35 + 95)= 180-130 = 50 degree

x = 50 degree

Answer:

$520.632

Step-by-step explanation:

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:

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.

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

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:

1/8

Step-by-step explanation:

Answer: 1/8

Step-by-step explanation:

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

Answer:

The confidence interval is [tex](85 - \frac{14.81}{\sqrt{n}},85 + \frac{14.81}{\sqrt{n}})[/tex], in which n is the size of the sample.

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.

[tex]M = 1.645\frac{9}{\sqrt{n}} = \frac{14.81}{\sqrt{n}}[/tex]

The lower end of the interval is the sample mean subtracted by M. So it is [tex]85 - \frac{14.81}{\sqrt{n}}[/tex]

The upper end of the interval is the sample mean added to M. So it is [tex]85 + \frac{14.81}{\sqrt{n}}[/tex]

The confidence interval is [tex](85 - \frac{14.81}{\sqrt{n}},85 + \frac{14.81}{\sqrt{n}})[/tex], in which n is the size of the sample.

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]