Solve the problem.

6y + 4 – 3y = -5 + y + 17

Luda [366]

Answer:

Step-by-step explanation:

6y+4-3y=-5+y+17

6y-3y-y= -5+17-4

2y=8

y=4

5 0

3 years ago

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The distribution of lifetimes of a particular brand of car tires has a mean of 51,200 miles and a standard deviation of 8,200 mi

Orlov [11]

Answer:

a) 0.277 = 27.7% probability that a randomly selected tyre lasts between 55,000 and 65,000 miles.

b) 0.348 = 34.8% probability that a randomly selected tyre lasts less than 48,000 miles.

c) 0.892 = 89.2% probability that a randomly selected tyre lasts at least 41,000 miles.

d) 0.778 = 77.8% probability that a randomly selected tyre has a lifetime that is within 10,000 miles of the mean

Step-by-step explanation:

Problems of normally distributed distributions are 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:

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

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

In this question, we have that:

$\mu = 51200, \sigma = 8200$

Probabilities:

A) Between 55,000 and 65,000 miles

This is the pvalue of Z when X = 65000 subtracted by the pvalue of Z when X = 55000. So

X = 65000

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

$Z = \frac{65000 - 51200}{8200}$

$Z = 1.68$

$Z = 1.68$ has a pvalue of 0.954

X = 55000

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

$Z = \frac{55000 - 51200}{8200}$

$Z = 0.46$

$Z = 0.46$ has a pvalue of 0.677

0.954 - 0.677 = 0.277

0.277 = 27.7% probability that a randomly selected tyre lasts between 55,000 and 65,000 miles.

B) Less than 48,000 miles

This is the pvalue of Z when X = 48000. So

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

$Z = \frac{48000 - 51200}{8200}$

$Z = -0.39$

$Z = -0.39$ has a pvalue of 0.348

0.348 = 34.8% probability that a randomly selected tyre lasts less than 48,000 miles.

C) At least 41,000 miles

This is 1 subtracted by the pvalue of Z when X = 41,000. So

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

$Z = \frac{41000 - 51200}{8200}$

$Z = -1.24$

$Z = -1.24$ has a pvalue of 0.108

1 - 0.108 = 0.892

0.892 = 89.2% probability that a randomly selected tyre lasts at least 41,000 miles.

D) A lifetime that is within 10,000 miles of the mean

This is the pvalue of Z when X = 51200 + 10000 = 61200 subtracted by the pvalue of Z when X = 51200 - 10000 = 412000. So

X = 61200

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

$Z = \frac{61200 - 51200}{8200}$

$Z = 1.22$

$Z = 1.22$ has a pvalue of 0.889

X = 41200

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

$Z = \frac{41200 - 51200}{8200}$

$Z = -1.22$

$Z = -1.22$ has a pvalue of 0.111

0.889 - 0.111 = 0.778

0.778 = 77.8% probability that a randomly selected tyre has a lifetime that is within 10,000 miles of the mean

4 0

3 years ago

A standard fair die is rolled one time. What is the probability that it lands on a

VLD [36.1K]

Answer:

The die is assumed to have six sides labeled

1

,

2

,

3

,

4

,

5

,

6

The numbers greater than 5 are

6

.

The numbers less than 3 are

2

,

1

.

So the numbers greater than 5 OR less than 3 are the union of the two sets or 6, and 2,1. The probability of rolling one of these numbers is

3

6

=

1

2

or

50%

.

Step-by-step explanation:

Sorry its sloppy

8 0

3 years ago

Reynaldo drew a rectangular design that was 6 in. wide and 8in. long. He used a copy machine to enlarge the design so that the w

son4ous [18]

Answer:

$13\frac{1}{3}\ in$

Step-by-step explanation:

we know that

The dilation is a non-rigid transformation that produces similar figures

If two figures are similar, then the ratio of its corresponding sides is proportional and this ratio is called the scale factor

step 1

Find the scale factor

Let

z ----> the scale factor

$z=\frac{10}{6}$ ----> ratio of corresponding sides

simplify

$z=\frac{5}{3}$

step 2

Find the length of the enlarged design (L)

Multiply the length of the original design by the scale factor

so

$L=\frac{5}{3}(8)=\frac{40}{3}\ in$

Convert to mixed number

$\frac{40}{3}\ in=\frac{39}{3}+\frac{1}{3}=13\frac{1}{3}\ in$

7 0

4 years ago

A square sticky note has sides that are 7 centimeters long. What is the sticky note's area?

Minchanka [31]

Answer:

side- 7

area of square - side x side

7 x 7 - 49² cm

8 0

3 years ago

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