Correlation coefficients can test ______ variable(s) at a time. a. two or more b. only one c. one or more d. only two

1 answer:

4 0

Correlation coeffiencients is used to determine the relationships between data, Correlation coefficients can test two or more variable(s) at a time

<h3 /><h3>What are correlation coefficients?</h3>

This is a linear correlation between sets of data. This coefficient can only be between -1 and 1.

Positive values show a positive correlation while a negative value shows negative correlation. A correlation coefficient of zero shows no relationship between variables.

Based on the explanation, we can conclude that Correlation coefficients can test two or more variable(s) at a time

Learn more on correlation here: brainly.com/question/4219149

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6. Antonio is a freelance computer programmer. In his work, he often uses 2n, where n is the number of bits, to find the number

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In general 2^n is the number of pieces of information n bits can store.

(a) Antonio can use the formula 2^n-1 to indicate the maximum size of numbers an n-bit word can store.
(b) for a 16-bit system, the maximum number of keys Millie can collect is 2^(16)-1 = 65536-1=65535
(c) For a 14 byte program, there are 14*8=112 bits.

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2 years ago

A team of 10 players is to be selected from a class of 6 girls and 7 boys. Match each scenario to its probability. You have to d

tankabanditka [31]

The selection of r objects out of n is done in

$C(n, r)= \frac{n!}{r!(n-r)!}$ many ways.

The total number of selections 10 that we can make from 6+7=13 students is

$C(13,10)= \frac{13!}{3!(10)!}= \frac{13*12*11*10!}{3*2*1*10!}= \frac{13*12*11}{3*2}= 286$
thus, the sample space of the experiment is 286

A.
"The probability that a randomly chosen team includes all 6 girls in the class."

total number of group of 10 which include all girls is C(7, 4), because the girls are fixed, and the remaining 4 is to be completed from the 7 boys, which can be done in C(7, 4) many ways.

 $C(7, 4)= \frac{7!}{4!3!}= \frac{7*6*5*4!}{4!*3*2*1}= \frac{7*6*5}{3*2}=35$

P(all 6 girls chosen)=35/286=0.12

B.
"The probability that a randomly chosen team has 3 girls and 7 boys."

with the same logic as in A, the number of groups were all 7 boys are in, is

 $C(6, 3)= \frac{6!}{3!3!}= \frac{6*5*4*3!}{3!3!}= \frac{6*5*4}{3*2*1}=20$

so the probability is 20/286=0.07

C.
"The probability that a randomly chosen team has either 4 or 6 boys."

case 1: the team has 4 boys and 6 girls

this was already calculated in part A, it is 0.12.

case 2, the team has 6 boys and 4 girls.

there C(7, 6)*C(6, 4) ,many ways of doing this, because any selection of the boys which can be done in C(7, 6) ways, can be combined with any selection of the girls.

 $C(7, 6)*C(6, 4)= \frac{7!}{6!1}* \frac{6!}{4!2!} =7*15= 105$

the probability is 105/286=0.367

since case 1 and case 2 are disjoint, that is either one or the other happen, then we add the probabilities:

0.12+0.367=0.487 (approximately = 0.49)

D.
"The probability that a randomly chosen team has 5 girls and 5 boys."

selecting 5 boys and 5 girls can be done in

 $C(7, 5)*C(6,5)= \frac{7!}{5!2} * \frac{6!}{5!1}=21*6=126$

many ways,

so the probability is 126/286=0.44

6 0

3 years ago

Read 2 more answers

OPEN-UP... | GRADE

allochka39001 [22]

Answer:

The answer to your question is: 19 l

Step-by-step explanation:

Data

20 quarts to liters

Process

Rule of three