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FrozenT [24]
3 years ago
13

Find the midpoint of a segment whose endpoints are (9, –7) and (–3, 5).

Mathematics
1 answer:
suter [353]3 years ago
7 0
For this question, you would have to use the midpoint formula. 
(X1 +X2 / 2 , Y1 + Y2 / 2)

In other words, 
(9 + -3 / 2 , -7 +5 / 2)
(6 / 2 , -2 / 2)
(3 , -1)

Your midpoint is (3, -1)
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GarryVolchara [31]
2 bc u just do change in y over the change in x. Start from a point and go down then go to the left.
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According to the histogram, how many shipments had more than 20 light bulbs broken? A) 4 B) 6 C) 7 D) 12
bearhunter [10]

The histogram depicts that the number shipments that had more than 20 light bulbs broken is D. 12.

<h3>What is a histogram?</h3>

A histogram simply means a graphical representation that is used to organize a group of data points into ranges.

From the complete question, the histogram shows that the number shipments that had more than 20 light bulbs broken is 12. This can be interpreted based on the ranges shown in the histogram.

Learn more about histogram on:

brainly.com/question/1581651

6 0
2 years ago
Add (6x^2-3x)+(5x-7)
iragen [17]

No problem.

The answer is 31x.

Hope this helped

7 0
3 years ago
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A local hamburger shop sold a combined total of 389 hamburgers and cheeseburgers on Thursday. There were 61 fever cheeseburgers
allsm [11]

Answer:

  225

Step-by-step explanation:

Let h represent the number of hamburgers sold. Then h-61 is the number of cheeseburgers sold. The combined total is ...

  h +(h -61) = 389

  2h = 450 . . . . . . add 61

  h = 225 . . . . . . . divide by 2

There were 225 hamburgers sold on Thursday.

6 0
3 years ago
A confidence interval (CI) is desired for the true average stray-load loss u (watts) for a certain type of induction motor when
Genrish500 [490]

Answer:

A) CI = (57.12 , 59.48)

B) CI = (57.71 , 58.89)

C) CI = (57.53 , 59.07)

D) n = 239.63

Step-by-step explanation:

a)

given data:

mean, \bar X = 58.3

standard deviation, σ = 3

sample size, n = 25Given CI level is 95%, hence α = 1 - 0.95 = 0.05

α/2 = 0.05/2 = 0.025,

Zc = Z(α/2) = 1.96

ME = Zc * σ \sqrt{n}

ME = 1.96 * 3 \sqrt{25}

ME = 1.18

CI = (\bar X - Zc * s\sqrt{n}  , \barX + Zc * s\sqrt{n})

CI = (58.3 - 1.96 * 3\sqrt{25} , 58.3 + 1.96 * 3\sqrt{25})

CI = (57.12 , 59.48)

b)

Given data:

mean, \bar X = 58.3

standard deviation, σ = 3

sample size, n = 100

Given CI level is 95%, hence α = 1 - 0.95 = 0.05

α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

ME = zc * σ \sqrt{n}

ME = 1.96 * 3\sqrt{100}

ME = 0.59

CI = (\bar X - Zc * s\sqrt{n}  , \barX + Zc * s\sqrt{n})

CI = (58.3 - 1.96 * 3\sqrt{100} , 58.3 + 1.96 * 3\sqrt{100})

CI = (57.71 , 58.89)

c)

sample mean, \bar X = 58.3

sample standard deviation, σ = 3

sample size, n = 100

Given CI level is 99%, hence α = 1 - 0.99 = 0.01

α/2 = 0.01/2 = 0.005, Zc = Z(α/2) = 2.58

ME = Zc * σ \sqrt{n}

ME = 2.58 * 3\sqrt{100}

ME = 0.77

CI = (\bar X - Zc * s\sqrt{n}  , \barX + Zc * s\sqrt{n})

CI = (58.3 - 2.58 * 3\sqrt{100} , 58.3 + 2.58 * 3/\sqrt{100}

CI = (57.53 , 59.07)

D)

Given data:

Significance Level, α = 0.01,

Margin or Error, E = 0.5,

σ = 3

The critical value for α = 0.01 is 2.58.

for calculating population mean we used

n \geq (zc *σ/E)^2

n = (2.58 * 3/0.5)^2

n = 239.63

7 0
3 years ago
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