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Tanya [424]
3 years ago
11

Which of the choices below is not a possible correlation coefficient?

Mathematics
1 answer:
fiasKO [112]3 years ago
8 0

Answer:

The condition for r is the following:

-1 \leq r \leq 1

And for this case if we analyze the options the only impossible value is given by:

1.0528

Because this value is higher than 1 and not satisfy the general limits for r

Step-by-step explanation:

The correlation coefficient is a measure of dispersion and is a value between -1 and 1, and is defined as:

r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}

The condition for r is the following:

-1 \leq r \leq 1

And for this case if we analyze the options the only impossible value is given by:

1.0528

Because this value is higher than 1 and not satisfy the general limits for r

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Find the value of n .<br> 6/n = 24/28<br> y = ___.
il63 [147K]

Answer:

n=7

Step-by-step explanation:

Cross multiply, isolate the variable, and divide by the coefficient to solve.

\frac{6}{n}=\frac{24}{28} \\ \\ 24n=168 \\ \\ n=7

Plug back in to check.

\frac{6}{7}=0.857142857 \\ \\ \frac{24}{28} =0.857142857

5 0
3 years ago
Given h(x) = 5x -1, find h(1).
Norma-Jean [14]
H(1) = 4


h(1)= 5 x 1 -1
=5-1
=4
hope that helps <3
5 0
3 years ago
Please help me I don’t know this
inessss [21]

Answer:

28 degrees

Step-by-step explanation:

68 = 2m + 12

56 = 2m

m = 28 degrees

6 0
3 years ago
Height-9ft<br> Width-25ft<br> Volume-7650<br> Length-?
Licemer1 [7]

Step-by-step explanation:

V = h × w × l

7650 = 9 × 25 × l

So,

l = 7650 ÷ 9 ÷ 25

l = 850 ÷ 25

l = 34

For the final, the length is 34.

4 0
3 years ago
The following data gives the speeds (in mph), as measured by radar, of 10 cars traveling north on I-15. 76 72 80 68 76 74 71 78
____ [38]

Answer:

71.123 mph ≤ μ ≤ 77.277 mph

Step-by-step explanation:

Taking into account that the speed of all cars traveling on this highway have a normal distribution and we can only know the mean and the standard deviation of the sample, the confidence interval for the mean is calculated as:

m-t_{a/2,n-1}\frac{s}{\sqrt{n} } ≤ μ ≤ m+t_{a/2,n-1}\frac{s}{\sqrt{n} }

Where m is the mean of the sample, s is the standard deviation of the sample, n is the size of the sample, μ is the mean speed of all cars, and t_{a/2,n-1} is the number for t-student distribution where a/2 is the amount of area in one tail and n-1 are the degrees of freedom.

the mean and the standard deviation of the sample are equal to 74.2 and 5.3083 respectively, the size of the sample is 10, the distribution t- student has 9 degrees of freedom and the value of a is 10%.

So, if we replace m by 74.2, s by 5.3083, n by 10 and t_{0.05,9} by 1.8331, we get that the 90% confidence interval for the mean speed is:

74.2-(1.8331)\frac{5.3083}{\sqrt{10} } ≤ μ ≤ 74.2+(1.8331)\frac{5.3083}{\sqrt{10} }

74.2 - 3.077 ≤ μ ≤ 74.2 + 3.077

71.123 ≤ μ ≤ 77.277

8 0
3 years ago
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