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bija089 [108]
3 years ago
9

Which of the following values are in the range of the function shown below?

Mathematics
1 answer:
Andrews [41]3 years ago
6 0
I think the answers are 2,1,0
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784/ 93 = 8.43010752688
93 * <span>8.43010752688 = 784</span>

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What is the simpified ratio of 12 cups to 2 pints
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Joseph claims that a scatterplot in which the y-values increase as the x-values increase must have a linear association. Amy cla
My name is Ann [436]

Answer: Choice C

Amy is correct because a nonlinear association could increase along the whole data set, while being steeper in some parts than others. The scatterplot could be linear or nonlinear.

======================================================

Explanation:

Just because the data points trend upward (as you go from left to right), it does not mean the data is linearly associated.

Consider a parabola that goes uphill, or an exponential curve that does the same. Both are nonlinear. If we have points close to or on these nonlinear curves, then we consider the scatterplot to have nonlinear association.

Also, you could have points randomly scattered about that don't fit either of those two functions, or any elementary math function your teacher has discussed so far, and yet the points could trend upward. If the points are not close to the same straight line, then we don't have linear association.

-----------------

In short, if the points all fall on the same line or close to it, then we have linear association. Otherwise, we have nonlinear association of some kind.

Joseph's claim that an increasing trend is not enough evidence to conclude the scatterplot is linear or not.

8 0
3 years ago
A sample of 64 observations has been selected to test whether the population mean is smaller than 15. The sample showed an avera
KengaRu [80]

Answer:

Test statistic = - 0.851063

- 2.520463

Step-by-step explanation:

H0 : μ ≥ 15

H1 : μ < 15

Sample mean, xbar = 14.5

Sample standard deviation, s = 4.7

Sample size = 64

Teat statistic :

(xbar - μ) ÷ (s/√(n))

(14.5 - 15) ÷ (4.7/√(64))

= - 0.851063

The critical value at α = 0.05

Using the T - distribution :

Degree of freedom, df = 64 - 1 = 63

Tcritical(0.05, 63) = 1.6694

Test statistic - critical value

-0.851063 - 1.6694

= - 2.520463

8 0
3 years ago
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