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andrey2020 [161]
1 year ago
8

A study of nutrition in developing countries collected data from the Egyptian village of Nahya. Researchers recorded the mean we

ight (in kilograms) for 170 infants in Nahya each month during their first year of life. A hasty user of statistics enters the data into software and computes the least-squares li~thout looking at the scatterplot first. The result is weight= 4.88 + 0.267 (age).
Required:
Use the residual plot to determine if this linear model is appropriate.

Mathematics
1 answer:
IceJOKER [234]1 year ago
7 0

Answer:

The linear model is not appropriate.

Step-by-step explanation:

In regression, the difference amid the observed-value of the dependent-variable (y) and the predicted-value (\hat y) is known as the residual (e).

e=y-\hat y

A residual plot is a graphical representation of the residuals on the y-axis and the independent variable on the x-axis. If the data-points on the residual plot are randomly spread around the x-axis, a linear regression model is appropriate for the data. And if the residual plots shows a non-random or a U or inverted U pattern, a nonlinear model is more appropriate for the data.

The residual provided shown an inverted U pattern. That is the points are not scattered around he graph.

Thus, the linear model is not appropriate.

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What is the quotient of -8a^8b^-2/ 10a^-4b^-10 in simplified form?
postnew [5]

Here are a few rules with exponents that you need to know:

  1. Dividing exponents of the same base: \frac{x^m}{x^n}=x^{m-n}

For this, divide:

\frac{-8a^8b^{-2}}{10a^{-4}b^{-10}}=\frac{-8}{10}a^{8-(-4)}b^{-2-(-10)}=-\frac{4}{5}a^{12}b^8

<u>Your final answer is -\frac{4}{5}a^{12}b^8</u>

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1 year ago
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Kevin ate 2 slices of cake. Ben ate 1 slice. If Kevin ate 2/6 of the cake and all the slices are the same size, what fraction of
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1/2 of the cake was eaten

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All the slices are the same size

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1 year ago
PI-3.
Fudgin [204]

Answer:

A. $301

B. $721

Step-by-step explanation:

Let $x be the amount of money they raised.

Rowena tried to put the $1 bills into two equal piles and found one left over at the end, then

x=2q_1+1

Polly tried to put the $1 bills into three equal piles and found one left over at the end, then

x=3q_2+1

Frustrated, they tried 4, 5, and 6 equal piles and each time had $1 left over, then

x=4q_3+1\\ \\x=5q_4+1\\ \\x=6q_5+1

Finally Rowena put all the bills evenly into 7 equal piles, and none were left over, then

x=7q_6

This means x-1 is divisible by 2, 3, 4, 5 and 6 without remainder, so

x-1=2\cdot 3\cdot 2\cdot 5n=60n

Hence,

x=60n+1, \ n\in N

The smallest amount of money they could have raised is $301, because

x=60\cdot 5+1=301 is divisible by 7.

Now, the number x=60n+1 should be divisible by 7 and must be greater than 500.

So,

60n+1>500\\ \\60n>499\\ \\n>8

When n = 9,

x=60\cdot 9+1=541 is not divisible by 7.

When n = 10,

x=60\cdot 10+1=601 is not divisible by 7.

When n = 11,

x=60\cdot 11+1=661 is not divisible by 7.

When n = 12,

x=60\cdot 12+1=721 is divisible by 7.

B. The least amount of money they could have raised is $721

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Answer: I do not know honestly

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.7

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