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3 years ago

PLEASE HELP QUESTION ATTACHED

Mathematics

2 answers:

PSYCHO15rus [73]3 years ago

8 0

(6,3) the one selected is correct

goldfiish [28.3K]3 years ago

8 0

Answer:

(6,3)

Step-by-step explanation:

x + 3y = 15

4x + 2y = 30

This system of equations can either be solved by elimination method or substitution method or both

We are going to use both method to solve this system of education

First we are gong to use elimination method to find the value of y

x + 3y = 15 -------------(1)

4x + 2y = 30 -----------(2)

Using the elimination method, to eliminate x, we want to make coefficients of x the same, to do that we multiply equation (1) by 4

4x + 12y =60 --------------(3)

subtract equation (2) from equation (3)

(4x-4x =0 12y - 2y = 10y 60 -30=30 )

We will have;

10y = 30

Divide 10 by both-side of the equation

10y/10 = 30/10

y = 3

We will use substitution method to get the value of x,

We will now substitute y=3 in equation (1)

x + 3y = 15

x + 3(3) = 15

x + 9 = 15

Subtract 9 from both-side of the equation

x +9 - 9 = 15 -9

x = 6

The solution to the system of the equations is x=6 and y =3

(6, 3)

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Kaylib’s eye-level height is 48 ft above sea level, and addison’s eye-level height is 85 and one-third ft above sea level. how m

GalinKa [24]

The addison see to the horizon at 2 root 2mi.

We have given that,Kaylib’s eye-level height is 48 ft above sea level, and addison’s eye-level height is 85 and one-third ft above sea level.

We have to find the how much farther can addison see to the horizon

<h3>Which equation we get from the given condition?</h3>

$d=\sqrt{\frac{3h}{2} }$

Where, we have

d- the distance they can see in thousands

h- their eye-level height in feet

For Kaylib

$d=\sqrt{\frac{3\times 48}{2} }\\\\d=\sqrt{{3(24)} }\\\\\\d=\sqrt{72}\\\\d=\sqrt{36\times 2}\\\\\\d=6\sqrt{2}....(1)$

For Addison h=85(1/3)

$d=\sqrt{\frac{3\times 85\frac{1}{3} }{2} }\\d\sqrt{\frac{256}{2} } \\d=\sqrt{128} \\d=8\sqrt{2} .....(2)$

Subtracting both distances we get

$8\sqrt{2}-6\sqrt{2} =2\sqrt{2}$

Therefore, the addison see to the horizon at 2 root 2mi.

To learn more about the eye level visit:

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3 years ago

A variety of two types of snack packs are delivered to a store. The box plots compare the number of calories in each snack pack

adelina 88 [10]

OPTIONS:

A. The interquartile range of the trail mix data is greater than the range of the cracker data.

B. The value 70 is an outlier in the trail mix data.

C. The upper quartile of the trail mix data is equal to the maximum value of the cracker data.

D. The number of calories in the packs of trail mix have a greater variation than the number of calories in the packs of crackers.

Answer:

Step-by-step explanation:

With the given information about how the box plot looks like, let's examine each option to see if they are true or not.

Option A: "The interquartile range of the trail mix data is greater than the range of the cracker data."

The interquartile range of trail mix data = 105 - 90 = 15

Range of cracker data = 100 - 70 = 30

Option A is NOT TRUE.

Option B: "The value 70 is an outlier in the trail mix data."

This is NOT TRUE. There are not outliers as 70 is the minimum value if the ranges of the data set for the trail mix.

Option C: "The upper quartile of the trail mix data is equal to the maximum value of the cracker data."

Upper quartile of the trail mix data = 105

Max value of cracker data = 100

This statement is NOT TRUE.

Option D: "The number of calories in the packs of trail mix have a greater variation than the number of calories in the packs of crackers."

The greater the range value, the greater the variation. Thus,

Range value of the trail mix data = 115 - 70 = 45

Range value of the cracker data = 100 - 70 = 30

This is statement is correct because trail mix data have a greater range value, hence, it has a greater variation in the number of calories.

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PSYCHO15rus [73]

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2 years ago

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