Please Help Quickly!!!<br><br> If lim x--> 4 f(x) = 5

At Bruno’s Video World, the regular price for a DVD is d dollars. How many DVDs can be purchased for x dollars when the DVDs are

zimovet [89]

Only a so many dvd can be pars

8 0

4 years ago

Solve:<br> 3(2x-1) = 9(x+3)

lions [1.4K]

Answer:

x=-10

Explanation:

6x-3=9x+27(distributive property)

6x-9x=27+3(move like terms to one side)

-3x=30(combine like terms)

x=-10(divide both sides by -3)

3 0

3 years ago

Read 2 more answers

Finn changes his mind and, from now on, decides to take the normal route to work everyday. On any given day, the time (in minute

Margaret [11]

Answer:

The 33rd percentile of the time it takes Finn to get to work on any given day is 31.04 minutes.

There is a 61.92% probability that Finn took more than 40.5 minutes to get to work on the first day or more than 38.5 minutes to get to work on the second day.

Step-by-step explanation:

This can be solved by the the z-score formula:

On a normaly distributed set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a value X is given by:

$Z = \frac{X - \mu}{\sigma}$

Each z-score value has an equivalent p-value, that represents the percentile that the value X is:

The problem states that:

Mean = 35, so $\mu = 35$

Variance = 81. The standard deviation is the square root of the variance, so $\sigma = \sqrt{81} = 9$ .

Find the 33rd percentile of the time it takes Finn to get to work on any given day. Do not include any units in your answer.

Looking at the z-score table, $z = -0.44$ has a pvalue of 0.333. So what is the value of X when $z = -0.44$ .

$Z = \frac{X - \mu}{\sigma}$

$-0.44 = \frac{X - 35}{9}$

$X - 35 = -3.96$

$X = 31.04$

The 33rd percentile of the time it takes Finn to get to work on any given day is 31.04 minutes.

Over the next 2 days, find the probability that Finn took more than 40.5 minutes to get to work on the first day or more than 38.5 minutes to get to work on the second day.

$P = P_{1} + P_{2}$

$P_{1}$ is the probability that Finn took more than 40.5 minutes to get to work on the first day. The first step to solve this problem is finding the z-value of $X = 40.5$ .

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{40.5 - 35}{9}$

$Z = 0.61$

$Z = 0.61$ has a pvalue of 0.7291. This means that the probability that it took LESS than 40.5 minutes for Finn to get to work is 72.91%. The probability that it took more than 40.5 minutes if $P_{1} = 100% - 72.91% = 27.09% = 0.2709$

$P_{2}$ is the probability that Finn took more than 38.5 minutes to get to work on the second day. Sine the probabilities are independent, we can solve it the same way we did for the first day, we find the z-score of

$X = 38.5$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{38.5 - 35}{9}$

$Z = 0.39$

$Z = 0.39$ has a pvalue of 0.6517. This means that the probability that it took LESS than 38.5 minutes for Finn to get to work is 65.17%. The probability that it took more than 38 minutes if $P_{1} = 100% - 65.17% = 34.83% = 0.3483$

So:

$P = P_{1} + P_{2} = 0.2709 + 0.3483 = 0.6192$

There is a 61.92% probability that Finn took more than 40.5 minutes to get to work on the first day or more than 38.5 minutes to get to work on the second day.

7 0

4 years ago

Quadrilateral ABCD has vertices A(-2,4), B(-3.-3), C(-5,-2), D(-5,2). Which of the following are the coordinates of A’ after a d

Gala2k [10]

<h2>Answer:</h2>

$\boxed{A'(-5,10)}$

<h2>Step-by-step explanation:</h2>

Dilation is when you stretch something by the same amount in two perpendicular directions. In other words, dilation changes size, not overall shape. Dilation factors greater than one makes the shape greater while dilation factors less than one makes the shape smaller. The scale factor in this problem is:

$2\frac{1}{2}$

Which is a mixed fraction. This can be transformed into an improper fraction as follows:

$2\frac{1}{2}=2+\frac{1}{2}=\frac{4+1}{2}=\frac{5}{2}$

So the scale factor is $\frac{5}{2}$ . Since the center of dilation is the origin, we just need to multiply each vertex of the quadrilateral by the scale factor. Thus, for A we have:

$A'(-2\times \frac{5}{2},4\times \frac{5}{2})=\boxed{A'(-5,10)}$

3 0

4 years ago

How many different 11-letter arrangements can be made using the letters in the word Firecracker?

zhenek [66]

Answer:

Total number of arrangements = 1,663,200

Step-by-step explanation:

Given:

11 - letter

FIRECRACKER

F = 1

I = 1

R = 3

E = 2

C = 2

A =1

K = 1

Find:

Total number of arrangements = ?

Computation:

Note: Repeated letter will be avoid.

$Number\ of\ arrangements = \frac{!11}{!3 \times !2 \times !2}\\\\Number\ of\ arrangements = \frac{!11}{!3 \times !2 \times !2} \\\\Number\ of\ arrangements = \frac{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3}{3 \times 2 \times 2 } \\\\Number\ of\ arrangements = 1,663,200$

Total number of arrangements = 1,663,200

6 0

4 years ago

Please Help Quickly!!! If lim x--> 4 f(x) = 5