Which equation is the steepest slope? • y=4x • y=-10x • y=-7x • y=3x

Dafna11 [192]

A.The greatest possible sum of a and b can be 47 if the two integers are -1 and 48
B.Yes it is possible. The two integers are-3 and 16.
C. The least possible difference is -49 if the two integers are 1 and -48

4 0

3 years ago

Many, many snails have a one-mile race, and the time it takes for them to finish is approximately normally distributed with mean

Natali [406]

Answer:

The percentage of snails that take more than 60 hours to finish is 4.75%

The relative frequency of snails that take less than 60 hours to finish is .9525

The proportion of snails that take between 60 and 67 hours to finish is 4.52 of 100.

There is a 0% probability that a randomly chosen snail will take more than 76 hours to finish.

To be among the 10% fastest snails, a snail must finish in at most 42.26 hours.

The most typical of 80% of snails that between 42.26 and 57.68 hours to finish.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

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

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.

In this problem, we have that:

Many, many snails have a one-mile race, and the time it takes for them to finish is approximately normally distributed with mean 50 hours and standard deviation 6 hours.

This means that $\mu = 50$ and $\sigma = 6$ .

The percentage of snails that take more than 60 hours to finish is %

The pvalue of the zscore of $X = 60$ is the percentage of snails that take LESS than 60 hours to finish. So the percentage of snails that take more than 60 hours to finish is 100% substracted by this pvalue.

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

$Z = \frac{60 - 50}{6}$

$Z = 1.67$

A Zscore of 1.67 has a pvalue of .9525. This means that there is a 95.25% of the snails take less than 60 hours to finish.

The percentage of snails that take more than 60 hours to finish is 100%-95.25% = 4.75%.

The relative frequency of snails that take less than 60 hours to finish is

The relative frequence off snails that take less than 60 hours to finish is the pvalue of the zscore of X = 60.

In the item above, we find that this value is .9525.

So, the relative frequency of snails that take less than 60 hours to finish is .9525

The proportion of snails that take between 60 and 67 hours to finish is:

This is the pvalue of the zscore of X = 67 subtracted by the pvalue of the zscore of X = 60. So

$X = 67$

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

$Z = \frac{67 - 50}{6}$

$Z = 2.83$

A zscore of 2.83 has a pvalue of .9977.

For X = 60, we have found a Zscore o 1.67 with a pvalue of .9977

So, the percentage of snails that take between 60 and 67 hours to finish is:

$p = .9977 - 0.9525 = .0452$

The proportion of snails that take between 60 and 67 hours to finish is 4.52 of 100.

The probability that a randomly-chosen snail will take more than 76 hours to finish (to four decimal places)

This is 100% subtracted by the pvalue of the Zscore of X = 76.

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

$Z = \frac{76 - 50}{6}$

$Z = 4.33$

The pvalue of Z = 4.33 is 1.

So, there is a 0% probability that a randomly chosen snail will take more than 76 hours to finish.

To be among the 10% fastest snails, a snail must finish in at most hours.

The most hours that a snail must finish is the value of X of the Zscore when p = 0.10.

Z = -1.29 has a pvalue of 0.0985, this is the largest pvalue below 0.1. So what is the value of X when Z = -1.29?

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

$-1.29 = \frac{X - 50}{6}$

$X - 50 = -7.74$

$X = 42.26$

To be among the 10% fastest snails, a snail must finish in at most 42.26 hours.

The most typical 80% of snails take between and hours to finish.

This is from a pvalue of .1 to a pvalue of .9.

When the pvalue is .1, X = 42.26.

A zscore of 1.28 is the largest with a pvalue below .9. So

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

$1.28 = \frac{X - 50}{6}$

$X - 50 = 7.68$

$X = 57.68$

The most typical of 80% of snails that between 42.26 and 57.68 hours to finish.

5 0

3 years ago

What is the area of a sector with a central angle of 255 and a radius of 18.4 m?

Igoryamba

Answer:

The answer is 753.01 m².

Step-by-step explanation:

Given that the formula of sector is θ/360×π×r² where r represents radius of circle. Then, you have to substitute the values into the formula :

$area = \frac{θ}{360} \times \pi \times {r}^{2}$

Let θ = 255°

Let π = 3.14,

Let r = 18.4m,

$area = \frac{255}{360} \times 3.14 \times {18.4}^{2}$

$area = 753.01 {m}^{2} \: (near.hundredth)$

5 0

3 years ago

HELP ME Write and solve a proportion to find y.

Phoenix [80]

Given:

A figure of dilation and the center of dilation is O.

To find:

The value of y.

Solution:

The distance between center and vertex of image (d') is proportional to the distance between center and the corresponding vertex of original figure (d).

$d'\propto d$