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egoroff_w [7]
3 years ago
15

Help again please...

Mathematics
2 answers:
Tresset [83]3 years ago
5 0

Answer:

C

Step-by-step explanation:

Simplify −4(8−3x)-4(8-3x).

−32+12x≥6x−8-32+12x≥6x-8

Move all terms containing xx to the left side of the inequality.

−32+6x≥−8-32+6x≥-8

Move all terms not containing xx to the right side of the inequality.

6x≥246x≥24

Divide each term by 66 and simplify.

x≥4x≥4

Inequality Form:

x≥4

xeze [42]3 years ago
3 0

Answer:

b

Step-by-step explanation:

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Suppose a geyser has a mean time between irruption’s of 75 minutes. If the interval of time between the eruption is normally dis
lesya [120]

Answer:

(a) The probability that a randomly selected Time interval between irruption is longer than 84 minutes is 0.3264.

(b) The probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is 0.0526.

(c) The probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is 0.0222.

(d) The probability decreases because the variability in the sample mean decreases as we increase the sample size

(e) The population mean may be larger than 75 minutes between irruption.

Step-by-step explanation:

We are given that a geyser has a mean time between irruption of 75 minutes. Also, the interval of time between the eruption is normally distributed with a standard deviation of 20 minutes.

(a) Let X = <u><em>the interval of time between the eruption</em></u>

So, X ~ Normal(\mu=75, \sigma^{2} =20)

The z-score probability distribution for the normal distribution is given by;

                            Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

Now, the probability that a randomly selected Time interval between irruption is longer than 84 minutes is given by = P(X > 84 min)

 

    P(X > 84 min) = P( \frac{X-\mu}{\sigma} > \frac{84-75}{20} ) = P(Z > 0.45) = 1 - P(Z \leq 0.45)

                                                        = 1 - 0.6736 = <u>0.3264</u>

The above probability is calculated by looking at the value of x = 0.45 in the z table which has an area of 0.6736.

(b) Let \bar X = <u><em>sample time intervals between the eruption</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

           n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is given by = P(\bar X > 84 min)

 

    P(\bar X > 84 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{84-75}{\frac{20}{\sqrt{13} } } ) = P(Z > 1.62) = 1 - P(Z \leq 1.62)

                                                        = 1 - 0.9474 = <u>0.0526</u>

The above probability is calculated by looking at the value of x = 1.62 in the z table which has an area of 0.9474.

(c) Let \bar X = <u><em>sample time intervals between the eruption</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

           n = sample of time intervals = 20

Now, the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is given by = P(\bar X > 84 min)

 

    P(\bar X > 84 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{84-75}{\frac{20}{\sqrt{20} } } ) = P(Z > 2.01) = 1 - P(Z \leq 2.01)

                                                        = 1 - 0.9778 = <u>0.0222</u>

The above probability is calculated by looking at the value of x = 2.01 in the z table which has an area of 0.9778.

(d) When increasing the sample size, the probability decreases because the variability in the sample mean decreases as we increase the sample size which we can clearly see in part (b) and (c) of the question.

(e) Since it is clear that the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is very slow(less than 5%0 which means that this is an unusual event. So, we can conclude that the population mean may be larger than 75 minutes between irruption.

8 0
3 years ago
10 POINTS AND BRAINLIEST! PLEASE ANSWER BOTH QUESTIONS!
jok3333 [9.3K]

Segment CF is parallel to segment BE because these segments are side by side and will have the same distance continuously between them. Therefore your answer would be,

Segment BE

∠AOF is the angle that is the supplementary angle to ∠FOD because these are two angles that sum up to 180°. hence, the answer is,

∠AOF

8 0
3 years ago
The line at the bullet roller coaster is moving about 4 feet for every 15 minutes. At this rate, how many hours would it take fo
Brums [2.3K]

Answer:

90

Step-by-step explanation:

4/15 which is our fraction distance/time

our other fraction is 24/x (x is our variable to find our time)

Now we are doing the buultterfly method

set it up

now multiply 4*x= 4x

15*24=360

now divide 4x ÷ 4 to get our variable alone

do the same to the other side

4 ÷ 360= 90

so x= 90 which is our missing number

8 0
3 years ago
Help please :( I can’t seem to get the answer
igomit [66]

Answer:

6 i don't know what the unit is, but its 6 for time

Step-by-step explanation:

It is a two step problem

54=9t

divide both sides by 9 to get the value of t

6=t

I hope that helps you :)

4 0
3 years ago
The difference of the range and the interquartile range of the data set represented by the box plot is
egoroff_w [7]

Answer: sorry but there's no dot plot :/

Step-by-step explanation:

8 0
3 years ago
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