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weeeeeb [17]
3 years ago
15

HELP ME PLSSSSSSSssssss

Mathematics
1 answer:
castortr0y [4]3 years ago
3 0

Answer:

RoC is 0.75 (3/4); Equation is y=0.75x (y=\frac34 x)

Step-by-step explanation:

You can easily spot that first column, x increases by 1 each line, while the second, y, increases by 0.75.

Rate of change is variation in y divided by the variation in x, or \Delta y \over \Delta x . Replacing with the value we found, we get that the rate of change is \frac{0.75}{1} or \frac34.

To find the equation the easiest way is to go back 4 steps, at the same rate of change, in order to find what's the value for x=0. Going back by 0.75 each step starting from 4, we obtain 2.25, 1.5, 0.75, and finally 0. Our equation becomesy=0.75x

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Vanderbilt Corporation, which utilizes a calendar year as its fiscal year, is the payee on a $12,000, 7%, 8-month note receivabl
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3 0
3 years ago
Each pail of plaster covers 90 square feet of ceiling. What is the least number of plaster you would need to buy to cover the ce
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4 0
3 years ago
IF YOU ANSWER RIGHT ILL GIVE YOU BRAINLTIST
AleksAgata [21]

Answer:

B: 12/25

Step-by-step explanation:

You have the correct answer marked. This is correct because the probability of grabbing a purple marble is 24/50 which if we divide by 2 to simplify is 12/25.

8 0
3 years ago
We are interested in the amount that students study per week. Suppose you collected the following data in hours {4.4, 5.2, 6.4,
olga nikolaevna [1]

Answer:

Step-by-step explanation:

Hello!

The objective is to estimate the average time a student studies per week.

A sample of 8 students was taken and the time they spent studying in one week was recorded.

4.4, 5.2, 6.4, 6.8, 7.1, 7.3, 8.3, 8.4

n= 8

X[bar]= ∑X/n= 53.9/8= 6.7375 ≅ 6.74

S²= 1/(n-1)*[∑X²-(∑X)²/n]= 1/7*[376.75-(53.9²)/8]= 1.94

S= 1.39

Assuming that the variable "weekly time a student spends studying" has a normal distribution, since the sample is small, the statistic to use to perform the estimation is the student's t, the formula for the interval is:

X[bar] ± t_{n-1;1-\alpha /2}* (S/√n)

t_{n-1;1-\alpha /2}= t_{7;0.975}= 2.365

6.74 ± 2.365 * (1.36/√8)

[5.6;7.88]

Using a confidence level of 95% you'd expect that the average time a student spends studying per week is contained by the interval [5.6;7.88]

I hope this helps!

3 0
3 years ago
Read 2 more answers
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