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3 years ago

Solve for j -1/3 = j/4 - 10/3

Mathematics

1 answer:

fgiga [73]3 years ago

6 0

Answer: J=12

Explanation: To make this equation easier to solve, you can make everything a whole number (a.k.a. Remove the fractions). To do so, you can multiply everything in this equation by the Least Common Multiple (LCM) of all the denominators, which is 12. The equation now becomes:

-4=3J-40

Then, solve it using normal algebra:

-4=3J-40
36=3J
J=12

Much easier, right?!

Hope that helps シ

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Which one of the following is always true?a. The coefficient of variation measures variability in a positively skewed data set r

Tom [10]

Answer:

C. The coefficient of variation is a measure of relative dispersion that expresses the standard deviation as a percentage of the mean, for any data on a ratio scale and an interval scale

Step-by-step explanation:

Th Coefficient of Variance is a measure of dispersion that can be calculated using the formula:

$CV = \frac{\sigma_{x} }{\mu_{x} } * 100%$

Where $\sigma{x}$ is the Standard Deviation

and $\mu_{x}$ is the sample mean

From the formula written above, it is shown that the Coefficient of Variation expresses the Standard Deviation as a percentage of the mean.

Coefficient of variation can be used to compare positive as well as negative data on the ratio and interval scale, it is not only used for positive data.

The Interquartile Range is not a measure of central tendency, it is a measure of dispersion.

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3 years ago

Gọi S là tập các giá trị nguyên của tham số m trong khoảng (–20;20) để bất phương

Pie

Answer:

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Step-by-step explanation:

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3 years ago

Ast year, 3,120 students competed in a statewide athletic competition. If each participating school district sent 26 students to

andreyandreev [35.5K]

Answer: 120 schools

Step-by-step explanation:

3,120 students competed in a statewide athletic competition and each participating school district sent 26 students to the athletic competition.

Total number of students= 3120

Total students sent by each school= 26

Number of participating school districts= Total number of students ÷ he students sent by each school

= 3120 / 26

= 120 students

There are 120 participating schools in the competition

3 0

3 years ago

Read 2 more answers

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Alborosie

$\huge\underline\pink{✏ANSWER}$

<h2>Hello and sorry but it's a Gatheted Question Can you fix it Can you? </h2>

Sorry for answering a nonsense

#BrainliestBunch

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3 years ago

According to a 2018 survey by Bankrate, 20% of adults in the United States save nothing for retirement (CNBC website). Suppose t

nalin [4]

Answer:

0.5981 = 59.81% probability that three or less of the selected adults have saved nothing for retirement

Step-by-step explanation:

For each adult, there are only two possible outcomes. Either they save nothing for retirement, or they save something. The probability of an adult saving nothing for retirement is independent of any other adult. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

20% of adults in the United States save nothing for retirement (CNBC website).

This means that $p = 0.2$

Suppose that sixteen adults in the United States are selected randomly.

This means that $n = 16$

What is the probability that three or less of the selected adults have saved nothing for retirement?

This is:

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{16,0}.(0.2)^{0}.(0.8)^{16} = 0.0281$

$P(X = 1) = C_{16,1}.(0.2)^{1}.(0.8)^{15} = 0.1126$

$P(X = 2) = C_{16,2}.(0.2)^{2}.(0.8)^{14} = 0.2111$

$P(X = 3) = C_{16,3}.(0.2)^{3}.(0.8)^{13} = 0.2463$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0281 + 0.1126 + 0.2111 + 0.2463 = 0.5981$

0.5981 = 59.81% probability that three or less of the selected adults have saved nothing for retirement

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3 years ago