Juan and Lizette rented a car for one week to drive from Phoenix to Boise. The car rental rate was $250 per week and $0.20 per mile. By the most direct route, the drive is 926 miles. How much did they spend on the rental car?

Answer:

-192

Step-by-step explanation:

it is a geometric progression

r=-4

Answer:

m = 4

n = 5

Step-by-step explanation:

[tex]m + 8 = 3m\\\\m - 3m = - 8\\\\-2m = - 8\\\\m = 4[/tex]

[tex]2n - 1 = 9 \\\\2n = 9 + 1\\\\2n = 10\\\\n = 5[/tex]

Answer:

Step-by-step explanation:

Polynomial f(x) has the following conditions: zeros of -4 (multiplicity 3), 1 (multiplicity 1), and with f(0) = 320.

The first part zeros of -4 means (x+4) and multiplicity 3 means (x+4)^3.

The second part zeros of 1 means (x-1) and multiplicity 1 means (x-1).

The third part f(0) = 320 means substituting x=0 into (x+4)^3*(x-1)*k =320

(0+4)^3*(0-1)*k = 320

-64k = 320

k = -5

Combining all three conditions, f(x)

= -5(x+4)^3*(x-1)

= -5(x^3 + 3*4*x^2 + 3*4*4*x + 4^3)(x-1)

= -5(x^4 + 12x^3 + 48x^2 + 64x - x^3 - 12x^2 - 48x - 64)

= -5(x^4 + 11x^3 + 36x^2 + 16x -64)

= -5x^3 -55x^3 - 180x^2 - 80x + 320

Answer:

Step-by-step explanation:

-4 is a root for 3 times and 1 is root for once

so (x+4)^3 * (x-1) is part of f(x)

the constant term there is 4^3*(-1)=-64

so there is a multiplier of 320/-64=-5

f(x) = -5 * (x+4)^3 * (x-1)

Answer:

3b^2 + 2b -8

Step-by-step explanation:

* means multiply

^ means exponent

3b * b = 3b^2

3b * 2 = 6b

-4 * b = -4b

-4 * 2 = -8

3b^2 + 6b -4b -8

3b^2 + 2b -8

Answer:

By using a pair of compasses and a ruler you can draw all angles

Answer:

v See below. v

Step-by-step explanation:

LM = MN

11x - 21 = 8x + 15

[tex]3x-21=15\\3x=36\\[/tex]

x = 12

LM = 11(12) - 21 = 132 - 21 = 111

MN = 8(12) + 15 = 96 + 15 = 111

LN = 111 + 111 = 222

Answer:

The next number is going to be 21

Answer:

19

Step-by-step explanation:

4 even number

3,5,7 odd numbers

14 even

17, 19, 21 even

Answer:

Direct variation

[tex]k = 2.5[/tex]

Step-by-step explanation:

Given

The attached table

Required

The type of variation

First, we check for direct variation using:

[tex]k = \frac{y}{x}[/tex]

Pick corresponding points on the table

[tex](x,y) = (-2,-5)[/tex]

So:

[tex]k = \frac{-5}{-2} = 2.5[/tex]

[tex](x,y) = (4,10)[/tex]

So:

[tex]k = \frac{10}{4} = 2.5[/tex]

[tex](x,y) = (6,15)[/tex]

So:

[tex]k = \frac{15}{6} = 2.5[/tex]

Hence, the table shows direct variation with [tex]k = 2.5[/tex]

Answer:

Below.

Step-by-step explanation:

22-8 = 14x2 = 28.

Answer:

AB is perpendicular to [GH] and GH is [A]

Step-by-step explanation:

Answer: A.

Step-by-step explanation:

Hope this helps!

Answer:

the answer is 50 but I don't know if

Answer:

The first one (90, 90) is supplimentary, the next two (54, 36. and 45, 45) are complimentary, and the last two are supplimentary.

Step-by-step explanation:

A complimentary angle is two angles that add up to 90, and supplimentary is two angles that add up to 180! :)

Answer:

1st picture at the top would be a supplementary angle because a supplementary angles always add to 180 degrees.

the 54 and 36 one is a complementary angle

the 45 and 45 would be complementary angle

the last two on the bottom would both be supplementary angles.

High hopes-

Barry

Answer:

[tex] \large{ \tt{❃ \: S \: O \: L \: U \: T \: I \: O \: N : }}[/tex]

[tex] \large{ \tt{❉ \: m \: \angle \:SRQ = m \: \angle \: SRJ\: + \: m \: \angle \:JRQ}}[/tex]

[tex] \large{ \tt{⟼ \: 166 \degree = 110 \degree + m \: \angle \: JRQ}}[/tex]

[tex] \large{ \tt{⟼ \: 166 \degree - 110 \degree = m \: \angle \: JRQ}}[/tex]

[tex] \boxed{ \large{ \tt{⟼ \: 56 \degree = m \: \angle \: JRQ}}}[/tex]

Answer:

a) {3,5}{3,10}{5,10}

b) [tex]P(A)=\frac{1}{3}[/tex]

c) [tex]P(B)=\frac{2}{3}[/tex]

d) [tex]P(C)=\frac{1}{3}[/tex]

e) [tex]P(A and C)=0[/tex]

f) [tex]P(A or B)=1[/tex]

g) [tex]P(B and C)=\frac{1}{3}[/tex]

h) [tex]P(A or C)=\frac{2}{3}[/tex]

i) [tex]P(C given B)=\frac{1}{2}[/tex]

j) [tex]P(C given A)=0[/tex]

k) [tex]P(not B)=\frac{1}{3}[/tex]

l) [tex]P(not C)=\frac{2}{3}[/tex]

Yes, events A and B are mutually exclusive. Because the results can either be even or odd, not both. No, events B and C are not mutually exclusive because the result can be both, odd and prime.

Step-by-step explanation:

a)

In order to solve part a of the problem, we need to find the possible outcomes, in this case, the possible outcomes are:

{3,5}{3,10} and {5,10}

We could think of the oppsite order, for example {5,3}{10,3}{10,5} but these are basically the same as the previous outcomes, so we will just take three outcomes in our sample space. We can think of it as drawing the two chips at the same time.

b)

Now the probability of the sum of the chips to be even. There is only one outcome where the sum of the chips is even, {3,5} since 3+5=8 the other outcomes will give us an odd number, so:

[tex]P=\frac{#desired}{#possible}[/tex]

[tex]P(A)=\frac{1}{3}[/tex]

c) For the probability of the sum of the chips to be odd, there are two outcomes where the sum of the chips is odd, {3,10} since 3+10=13 and {5,10} since 5+10=15 the other outcomes will give us an even number, so:

[tex]P(B)=\frac{2}{3}[/tex]

d) The probability of the sum of the chips is prime. There is only one outcome where the sum of the chips is prime, {3,10} since 3+10=13 the other outcomes will give us non prime results, so:

[tex]P(C)=\frac{1}{3}[/tex]

e) The probability of the sum of the chips to be even and prime. There are no results where we can get an even and prime number, since the only even and prime number there is is number 2 and no outcome will give us that number, so:

P(A and C)=0

f) The probability of the sum of the chips is even or odd. We can either get even or odd results, so no matter what outcome we get, we will get an odd or even result so:

[tex]P(A or B)=1[/tex]

g) The probability of the sum of the chips is odd and prime. There is only one outcome where the sum of the chips is odd and prime, {3,10} since 3+10=13 the other outcomes will give us non prime results, so:

[tex]P(B and C)=\frac{1}{3}[/tex]

h) The probability of the sum of the chips is even or prime. There are two outcomes where the sum of the chips is even or prime, {3,10} since 3+10=13 and {3,5} since 3+5=8 so:

[tex]P(A or C)=\frac{2}{3}[/tex]

i) The probability of the sum of the chips is prime given that the sum of the chips is odd. There are two possible results where the sum of the chips is odd {3,10} and {5,10} and only one of those results is even, {3,10}, so

[tex]P(C given B)=\frac{1}{2}[/tex]

j) The probability of the sum of the chips is prime given that the sum of the chips is even. There is only one possible even result: {3,5} but that result isn't prime, so

[tex]P(C given A)=0[/tex]

k) The probability of the sum of the chips is not odd. There is only one outcome where the sum of the chips is not odd (even), {3,5} so:

[tex]P(not B)=\frac{1}{3}[/tex]

l) The probability of the sum of the chips is not prime. There are two outcomes where the sum of the chips is not prime, {3,5} and {5,10} so:

[tex]P(not C)=\frac{2}{3}[/tex]

Are events A and B mutually exclusive?

Yes, events A and B are mutually exclusive.

Why or why not?

Because the results can either be even or odd, not both.

Are events B and C mutually exclusive?

No, events B and C are not mutually exclusive.

Why or Why not?

Because the result can be both, odd and prime.

Answer:

150 dollars. if I am wrong correct me

Answer:

C and D

Step-by-step explanation:

15 to 30 galons at $9.95 to $21.00

the minimum amount can be found by calculating the minimum amount sold at a minimum price 15*9.95 = $149.25

the maximum amount can be found by calculating the maximum amount sold at a maximum price 30*21 = $630

there are 2 choices that are between 149.25 and 630, C, and D

Answer:

The answer is the third one below

Answer:

a) By the Central Limit Theorem, it has an approximately normal shape, with mean(center) 0.9 and standard deviation(spread) 0.035.

b) Average numbers of children between 0.83 and 0.97 are reasonably likely in the sample.

c) 0.0021 = 0.21% probability that the average number of children per family in the sample will be 0.8 or less

d) 0.9958 = 99.58% probability that the average number of children per family in the sample will be between 0.8 and 1.0

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

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

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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean 0.9 and standard deviation 1.1.

This means that [tex]\mu = 0.9, \sigma = 1.1[/tex]

Suppose a television network selects a random sample of 1000 families in the United States for a survey on TV viewing habits.

This means that [tex]n = 1000, s = \frac{1.1}{\sqrt{1000}} = 0.035[/tex]

a. Describe (as shape, center and spread) the sampling distribution of the possible values of the average number of children per family.

By the Central Limit Theorem, it has an approximately normal shape, with mean(center) 0.9 and standard deviation(spread) 0.035.

b. What average numbers of children are reasonably likely in the sample?

By the Empirical Rule, 95% of the sample is within 2 standard deviations of the mean, so:

0.9 - 2*0.035 = 0.83

0.9 + 2*0.035 =  0.97

Average numbers of children between 0.83 and 0.97 are reasonably likely in the sample.

c. What is the probability that the average number of children per family in the sample will be 0.8 or less?

This is the p-value of Z when X = 0.8. So

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

By the Central Limit Theorem

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

[tex]Z = \frac{0.8 - 0.9}{0.035}[/tex]

[tex]Z = -2.86[/tex]

[tex]Z = -2.86[/tex] has a p-value of 0.0021

0.0021 = 0.21% probability that the average number of children per family in the sample will be 0.8 or less.

d. What is the probability that the average number of children per family in the sample will be between 0.8 and 1.0?

p-value of Z when X = 1 subtracted by the p-value of Z when X = 0.8.

X = 1

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

[tex]Z = \frac{1 - 0.9}{0.035}[/tex]

[tex]Z = 2.86[/tex]

[tex]Z = 2.86[/tex] has a p-value of 0.9979

X = 0.8

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

[tex]Z = \frac{0.8 - 0.9}{0.035}[/tex]

[tex]Z = -2.86[/tex]

[tex]Z = -2.86[/tex] has a p-value of 0.0021

0.9979 - 0.0021 = 0.9958

0.9958 = 99.58% probability that the average number of children per family in the sample will be between 0.8 and 1.0

Answer:

Step-by-step explanation:

[tex]h(t)=-16t^2+96t\\\\h(t)=-t(16t-96)[/tex]

[tex]96=2^5*3\\\\16=2^4\\\\h(t)=-t(2^5*3*t-2^4)=-2^4t(2^1*3*t-1)\\\\h(t)=-16t(6t-1)[/tex]

the b) part is easy do it!

Answer:

0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

Gestation periods:

1) 0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.

2) 0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.

3) 0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.

4) 0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.

Step-by-step explanation:

To solve these questions, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

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

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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The height, X, of all 3-year-old females is approximately normally distributed with mean 38.72 inches and standard deviation 3.17 inches.

This means that [tex]\mu = 38.72, \sigma = 3.17[/tex]

Sample of 10:

This means that [tex]n = 10, s = \frac{3.17}{\sqrt{10}}[/tex]

Compute the probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

This is 1 subtracted by the p-value of Z when X = 40. So

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

By the Central Limit Theorem

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

[tex]Z = \frac{40 - 38.72}{\frac{3.17}{\sqrt{10}}}[/tex]

[tex]Z = 1.28[/tex]

[tex]Z = 1.28[/tex] has a p-value of 0.8997

1 - 0.8997 = 0.1003

0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

Gestation periods:

[tex]\mu = 266, \sigma = 16[/tex]

1. What is the probability a randomly selected pregnancy lasts less than 260 days?

This is the p-value of Z when X = 260. So

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

[tex]Z = \frac{260 -  266}{16}[/tex]

[tex]Z = -0.375[/tex]

[tex]Z = -0.375[/tex] has a p-value of 0.3539.

0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.

2. What is the probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less?

Now [tex]n = 20[/tex], so:

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

[tex]Z = \frac{260 - 266}{\frac{16}{\sqrt{20}}}[/tex]

[tex]Z = -1.68[/tex]

[tex]Z = -1.68[/tex] has a p-value of 0.0465.

0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.

3. What is the probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less?

Now [tex]n = 50[/tex], so:

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

[tex]Z = \frac{260 - 266}{\frac{16}{\sqrt{50}}}[/tex]

[tex]Z = -2.65[/tex]

[tex]Z = -2.65[/tex] has a p-value of 0.0040.

0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.

4. What is the probability a random sample of size 15 will have a mean gestation period within 10 days of the mean?

Sample of size 15 means that [tex]n = 15[/tex]. This probability is the p-value of Z when X = 276 subtracted by the p-value of Z when X = 256.

X = 276

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

[tex]Z = \frac{276 - 266}{\frac{16}{\sqrt{15}}}[/tex]

[tex]Z = 2.42[/tex]

[tex]Z = 2.42[/tex] has a p-value of 0.9922.

X = 256

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

[tex]Z = \frac{256 - 266}{\frac{16}{\sqrt{15}}}[/tex]

[tex]Z = -2.42[/tex]

[tex]Z = -2.42[/tex] has a p-value of 0.0078.

0.9922 - 0.0078 = 0.9844

0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.

Answer:

This question seems incorrect.

Kindly take a look again and re-state it properly to enable me give the most accurate answer.

Thank you

Answer:

B. 0.41

Step-by-step explanation:

Binomial experiment:

In a binomial experiment, the probability of the success on each trial is always the same.

The probability of success on trial 4 is 0.41.

This means that the probability of success on trial 8, and all the other 10 trials, is of 0.41, and thus the correct answer is given by option B.

Answer:

40 degrees un edge

Step-by-step explanation:

Answer:

The person above me got this correct, so the answer to this is 40! I just did the Unit Test and got a 100%!

Answer:

[tex]y = \frac{5ct}{3b}[/tex]

Step-by-step explanation:

[tex]\frac{5t}{4y} =\frac{3b}{4c}[/tex]

1. start by multiplying y to both sides:

y × [tex]\frac{5t}{4y} =\frac{3b}{4c}[/tex] × y

[tex]\frac{5t}{4} =\frac{3b}{4c}y[/tex]

2. divide both sides by [tex]\frac{3b}{4c}[/tex]

[tex]\frac{5t}{4}/\frac{3b}{4c} =\frac{3b}{4c}y/\frac{3b}{4c}[/tex]

[tex]y = \frac{5ct}{3b}[/tex]

Hi there!  

»»————-　★　————-««

I believe your answer is:  

[tex]V=1000\text{mm}^3[/tex]

»»————-　★　————-««  

Here’s why:  

⸻⸻⸻⸻

I am assuming by the infomation given that the figure is a cube.

⸻⸻⸻⸻

[tex]\boxed{\text{Finding the volume of the cube...}}\\\\S = 10mm; V= s^3\\--------------\\\rightarrow V = 10^3\\\\\rightarrow V = 10 * 10 * 10\\\\\rightarrow \boxed{V=1000\text{mm}^3}[/tex]

⸻⸻⸻⸻

»»————-　★　————-««  

Hope this helps you. I apologize if it’s incorrect.  

Answer:

They are both 42 cm

Step-by-step explanation:

Answer:

Range: { 0,3,5,7,9}

Step-by-step explanation:

The range is the values that y takes

Range: { 0,3,5,7,9}

Now we have to find,

The range of the table of values,

→ Range = ?

Then the range will be the numbers that is in the Y column.

→ Range = ?

→ Range = (value that Y takes)

→ Range = 0,3,5,7,9

Therefore, the range is 0,3,5,7,9.

Answer:

x=9

Step-by-step explanation:

Hello!

∫[(x+1)/(x-1)dx

∫t+2/t dt

∫t/t + 2/t dt

∫1 + 2/t dt

∫1dt + ∫2/t dt

∫t + 2In (|t|)

x - 1 + 2In (|x-1|)

x + 2In (|x-1|) + C, C ∈ R

Good luck! :)