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Ksenya-84 [330]
3 years ago
6

Can someone help me with this question? i will mark brainliest

Mathematics
1 answer:
zysi [14]3 years ago
3 0
The answer is 0.9
Because the scatter plot is a positive liner relationship the answer would have to be positive but the liner relationship is not perfect so the value would not be 1.
You might be interested in
A+44=B<br> A=1x+76<br> B=-6x+134 <br> What is the value of A?
Rashid [163]
Hi there! A = 78

A + 44 = B
We can plug in the values of A and B into the equation. We then get an equation with only one variable (x), which we can solve.

1x + 76 + 44 = - 6x + 134
Collect terms.

1x + 120 = - 6x + 134
Add 6x to both sides.

7x + 120 = 134
Subtract 120 from both sides

7x = 14
Divide both sides by 7

x = 14 / 7 = 2

A = 1x + 76
Now plug in the value of x we just found

A = 2 + 76 = 78
8 0
3 years ago
If B is the midpoint of AC, AC=CD, AB=3x+4, AC=11x-17, and CE=49, find DE.
Luba_88 [7]

Answer:

11

Step-by-step explanation:

AB = 3x + 4 and AC which is twice of AB is equal to 11x - 17

2 (3x + 4) = 11x - 17

6x + 8 = 11x - 17

8 + 17 = 11x - 6x

25 = 5x

5 = x

AC = CD = 11x - 17 ➡ 11×5 - 17 = 38

CD = 38 and DE = 49 - 38 = 11

6 0
3 years ago
One end of a ramp is raised to the back of a truck 5 feet above the ground. 1 point
sveta [45]

Answer:

38.7°

Step-by-step explanation:

You are basically looking at a triangle with a 90° angle (check picture).

Since we are looking for the ramp's angle and are given the length of the opposite and the hypothenuse, we have to solve the following equation:

sin (\alpha) = opposite / hypothenuse= 5 ft/8ft

which is equivalent to

\alpha = sin^{-1}(5/8)\approx 38.7\deg

6 0
2 years ago
the quadratic function h (t)=-16t^2+150 models a balls height, in feet, over time, in seconds, after it is dropped from a 15 sto
OLEGan [10]

From the equation, h(t)=-16t^2+150. You are given the height of the building which is 15ft and you are asked to find the time it takes to fall in seconds.

h(t)=-16t^2+150

15=-16t^2+150

15 – 150 = -16t^2

-135 = -16t^2

t = 2.9 seconds

4 0
3 years ago
In a 3 hour examination of 320 questions, there are 40 mathematics problems. if twice as much time should be allowed for each ma
xeze [42]
<span>Let the total hour of exam is 3hrs ie 180 minutes.( since answer required in minutes). Let as assume maths questions be in x minutes. Let the assume rest of the question is in y minutes. Given that out of 320 qns 40 is maths qns. therefore, maths qn is =40 and rest of qns is 280. Also given the time of maths qns is twice of rest. They related in the following equation. ie x:y=2:1 ; x/y =2/1 ie x=2y therefore 2x+y= 180 min ; ie 2(2y)+y= 180 ; y=36 therefore resolving for x = 2*36 = 72 time spent for maths qn is 2*x = 2* 72 =144min. rest of maths question is 36 min.</span>
6 0
3 years ago
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