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Delvig [45]
3 years ago
7

Researchers measured the percent body fat and the preferred amount of salt (percent weight/volume) for several children. Here ar

e data for seven children:
Preferred amount of salt x 0.2 0.3 0.4 0.5 0.7 0.9 1.1
Percent body fat y 20 30 22 30 38 23 30
Use your calculator or software: The correlation between percent body fat and preferred amount of salt is about which of the following?
a) τ = 0.08.
b) τ = 0.3.
c) τ = 0.8.
Mathematics
1 answer:
Anon25 [30]3 years ago
4 0

Answer: b) τ = 0.3

Step-by-step explanation:

Given the data :

Amount of salt (x)____% body fat(y)

0.2 _______________20

0.3 _______________30

0.4 _______________22

0.5 _______________30

0.7 _______________38

0.9 _______________23

1.1 ________________30

The correlation Coefficient as obtained from the online pearson correlation Coefficient calculator is 0.3281 = 0.3 (to one decimal place) which implies that a weak positive correlation or relationship exists between the preferred amount of salt taken to the percentage body weight of an individual. This is because the value is positive and closer to 0 than 1. The closer the weaker the degree of correlation. With positive values implying a positive relationship (that is an increase in variable A leads to a corresponding increase in B and vice-versa).

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Answer:

Therefore the rate change of distance between the car and the person at the instant, the car is 24 m from the intersection is 12 m/s.

Step-by-step explanation:

Given that,

A person stand 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 m/s.

From Pythagorean Theorem,

(The distance between car and person)²= (The distance of the car from intersection)²+ (The distance of the person from intersection)²+

Assume that the distance of the car from the intersection and from the person be x and y at any time t respectively.

∴y²= x²+10²

\Rightarrow y=\sqrt{x^2+100}

Differentiating with respect to t

\frac{dy}{dt}=\frac{1}{2\sqrt{x^2+100}}. 2x\frac{dx}{dt}

\Rightarrow \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}. \frac{dx}{dt}

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