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Oduvanchick [21]
4 years ago
8

Solve the inequality:

Mathematics
1 answer:
olya-2409 [2.1K]4 years ago
4 0

Answer:

8 - 2x <4 gives us x>2

|x-8| >22 gives us (-inf, -14) and (30,inf)

|x+5| < 4 gives us (-9,-1)

Step-by-step explanation:

Inequalities differs from equalities in that they return a range of values for x and not a single value (case of equalities)

The first one

8 - 2x <4 if you subtract every side 8, you will get -2x<-4

Next divide every side with -2, since its a negative number you have to change the orientation of the inequaiity, thus we have x>4/2, x>2

The second

|x-8| >22, it has to be treated as this. -22>x-2>22 which can be treated as two single inequalities, -22>x-8, and the other, x-8>22. The fist gives you x<-14, and the other x>30, two non intercepting ranges, so your range will be (-inf, -14) and (30, inf)

The third

|x+5| < 4 if you apply the same steps from above, you will have the following  -4<x+5<4 This results in two intercepting ranges, giving you the range (-9,-1)

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A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and
pav-90 [236]

Answer:

ans=13.59%

Step-by-step explanation:

The 68-95-99.7 rule states that, when X is an observation from a random bell-shaped (normally distributed) value with mean \mu and standard deviation \sigma, we have these following probabilities

Pr(\mu - \sigma \leq X \leq \mu + \sigma) = 0.6827

Pr(\mu - 2\sigma \leq X \leq \mu + 2\sigma) = 0.9545

Pr(\mu - 3\sigma \leq X \leq \mu + 3\sigma) = 0.9973

In our problem, we have that:

The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 53 months and a standard deviation of 11 months

So \mu = 53, \sigma = 11

So:

Pr(53-11 \leq X \leq 53+11) = 0.6827

Pr(53 - 22 \leq X \leq 53 + 22) = 0.9545

Pr(53 - 33 \leq X \leq 53 + 33) = 0.9973

-----------

Pr(42 \leq X \leq 64) = 0.6827

Pr(31 \leq X \leq 75) = 0.9545

Pr(20 \leq X \leq 86) = 0.9973

-----

What is the approximate percentage of cars that remain in service between 64 and 75 months?

Between 64 and 75 minutes is between one and two standard deviations above the mean.

We have Pr(31 \leq X \leq 75) = 0.9545 = 0.9545 subtracted by Pr(42 \leq X \leq 64) = 0.6827 is the percentage of cars that remain in service between one and two standard deviation, both above and below the mean.

To find just the percentage above the mean, we divide this value by 2

So:

P = {0.9545 - 0.6827}{2} = 0.1359

The approximate percentage of cars that remain in service between 64 and 75 months is 13.59%.

4 0
4 years ago
Solve for x -3x+6=2x-24<br><br>-30<br><br>-6 <br><br>6<br><br>30
Svetach [21]

-3x + 6 = 2x - 24.

Add 3x to both sides.

6 = 5x - 24.

Add 24 to both sides.

30 = 5x.

Divide it all by 5.

<em><u>x = 6.</u></em>

8 0
4 years ago
A regular hexagon measures 3x+5 units on each side. What is the perimeter in the simplest form?
vichka [17]
If it is hexagon, 6 sides
1 side = 3x+5
6 sides = 6(3x+5)

perimeter
18x + 30
4 0
4 years ago
WILL GIVE BRAINLIEST!!
charle [14.2K]

Answer:

Total Cost in a day is . Profit function is . And the revenue function is . In a day, bookstore must sell at least 15 books, because of 6n-90=0, in order to make profit.

Step-by-step explanation:

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6 0
3 years ago
In one lottery, a player wins the jackpot by matching all five numbers drawn from white balls (1 through 41) and matching the nu
puteri [66]

Answer:

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Step-by-step explanation:

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So the probability of winning the jackpot is 1 / 2,787,760,560.

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