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babymother [125]
3 years ago
15

Which quantity is proportional to 45⁄3?

Mathematics
2 answers:
masha68 [24]3 years ago
5 0

Answer:

15/1 is proportional to 45/3

Step-by-step explanation:

helped or?

coldgirl [10]3 years ago
3 0

The fractions that are proportional to 45/3:

90/6

15/1

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The front view of a piece of art is in the shape of a rhombus. The front view of the art has diagonals that are 1.4 yards long a
Zina [86]
1.12 add im an artist so i know trying to help people out
5 0
3 years ago
Read 2 more answers
Question Help Suppose that the lifetimes of light bulbs are approximately normally​ distributed, with a mean of 5656 hours and a
koban [17]

Answer:

a)3.438% of the light bulbs will last more than 6262 hours.

b)11.31% of the light bulbs will last 5252 hours or less.

c) 23.655% of the light bulbs are going to last between 5858 and 6262 hours.

d) 0.12% of the light bulbs will last 4646 hours or less.

Step-by-step explanation:

Normally distributed problems can be solved by 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}

After we find the value of Z, we look into the z-score table and find the equivalent p-value of this score. This is the probability that a score will be LOWER than the value of X.

In this problem, we have that:

The lifetimes of light bulbs are approximately normally​ distributed, with a mean of 5656 hours and a standard deviation of 333.3 hours.

So \mu = 5656, \sigma = 333.3

(a) What proportion of light bulbs will last more than 6262 ​hours?

The pvalue of the z-score of X = 6262 is the proportion of light bulbs that will last less than 6262. Subtracting 100% by this value, we find the proportion of light bulbs that will last more than 6262 hours.

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

Z = \frac{6262 - 5656}{333.3}

Z = 1.82

Z = 1.81 has a pvalue of .96562. This means that 96.562% of the light bulbs are going to last less than 6262 hours. So

P = 100% - 96.562% = 3.438% of the light bulbs will last more than 6262 hours.

​(b) What proportion of light bulbs will last 5252 hours or​ less?

This is the pvalue of the zscore of X = 5252

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

Z = \frac{5252- 5656}{333.3}

Z = -1.21

Z = -1.21 has a pvalue of .1131. This means that 11.31% of the light bulbs will last 5252 hours or less.

(c) What proportion of light bulbs will last between 5858 and 6262 ​hours?

This is the pvalue of the zscore of X = 6262 subtracted by the pvalue of the zscore X = 5858

For X = 6262, we have that Z = 1.81 with a pvalue of .96562.

For X = 5858

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

Z = \frac{5858- 5656}{333.3}

Z = 0.61

Z = 0.61 has a pvalue of .72907.

So, the proportion of light bulbs that will last between 5858 and 6262 hours is

P = .96562 - .72907 = .23655

23.655% of the light bulbs are going to last between 5858 and 6262 hours.

​(d) What is the probability that a randomly selected light bulb lasts less than 4646 ​hours?

This is the pvalue of the zscore of X = 4646

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

Z = \frac{4646- 5656}{333.3}

Z = -3.03

Z = -3.03 has a pvalue of .0012. This means that 0.12% of the light bulbs will last 4646 hours or less.

5 0
3 years ago
PLEASE HELP!!!!!!! A family has two cars. The first car has a fuel efficiency of 35 miles per gallon of gas and the second has a
Marianna [84]

Answer:

The answer to your question is car 1 = 30 gal and car 1 = 20 gal

Step-by-step explanation:

car 1 = a

car 2 = b

Efficiency of car 1 = 35 mi/gal

Efficiency for car 2 = 20 mi/gal

Total distance = 1450

Total gas consumption = 50 gal

Equations

                          35a + 20b = 1450            ------- (I)

                             a   +     b =  50               ------- (II)

Solve by elimination

Multiply equation II by -35

                          35a  + 20b  = 1450

                         -35a  - 35b   = -1750

Simplify

                            0    - 15b    =  -300

Solve for b

                                        b =  -300/-15

Result

                                        b = 20

Substitute b in equation II to find a

                               a +  20  = 50

Solve for a

                               a  = 50 -20

Result

                               a  = 30

4 0
3 years ago
A store sells 4 printers for every computers. Last month the store sells 56 computers, how many printers did they sell?
Kisachek [45]

Answer:

224 printers

Step-by-step explanation:

56computers×4printers per computer=224 printers

3 0
3 years ago
Substitute x and y in the equation
umka2103 [35]
B. y=50x

If you look at the graph, you can see that in each dot the y equals to 50 Times x.
Such at the first dot, the location is at (1,50). 1*50=50
And at the second dot the location is at (2,100). 50*2=100
And at the last dot, the location is at (8,400). 8*50=400
6 0
4 years ago
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