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SVEN [57.7K]
2 years ago
9

What percent of 70 equals 21

Mathematics
2 answers:
Sedbober [7]2 years ago
8 0
.3=30%



Is your answers
Misha Larkins [42]2 years ago
4 0

Answer:

30% of 70 = 21

Step-by-step explanation:

21/70 = .3

.3=30%

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Zolol [24]
The answer is A. k/6 + 8 - 2/k
8 0
3 years ago
Use a t-distribution to answer this question. Assume the samples are random samples from distributions that are reasonably norma
Dimas [21]

Using the t-distribution, it is found that the proportion in a t-distribution less than −1.4 if the samples have sizes n1=30 and n2=40 is of 0.083.

<h3>How to find a proportion in a t-distribution?</h3>

The proportion is found using a calculator, with three inputs:

  • The tail of the test, if it is left, right, or two-tailed.
  • The test statistic.
  • The amount of degrees of freedom.

In this problem, we have a left-tailed proportion, as we want the proportion that is less than a value, with test statistic t = -1.4 and 30 + 40 - 2 = 68 df, hence the proportion is of 0.083.

More can be learned about the t-distribution at brainly.com/question/16162795

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8 0
2 years ago
Please can you help me with this
Gwar [14]

Answer:

The answer would be 0

Step-by-step explanation:

Because halfway of 4, -2 is 0

7 0
3 years ago
Read 2 more answers
What is -3(2x-3)=-15
Oksanka [162]

Answer:

18

Step-by-step explanation:

-3(2x-3)

-3x(-6)

-3x-6=18 because the negatives cancel each other out.

7 0
3 years ago
Read 2 more answers
A new sample of 225 employed adults is chosen. Find the probability that less than 7.1% of the individuals in this sample hold m
arsen [322]

Answer:

The probability that less than 7.1% of the individuals in this sample hold multiple jobs is 0.0043.

Step-by-step explanation:

Let <em>X</em> = number of individuals in the United States who held multiple jobs.

The probability that an individual holds multiple jobs is, <em>p</em> = 0.13.

The sample of employed individuals selected is of size, <em>n</em> = 225.

An individual holding multiple jobs is independent of the others.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 225 and <em>p</em> = 0.13.

But since the sample size is too large Normal approximation to Binomial can be used to define the distribution of proportion <em>p</em>.

Conditions of Normal approximation to Binomial are:

  • np ≥ 10
  • n (1 - p) ≥ 10

Check the conditions as follows:

np=225\times 0.13=29.25>10\\n(1-p)=225\times (1-0.13)=195.75>10

The distribution of the proportion of individuals who hold multiple jobs is,

p\sim N(p, \frac{p(1-p)}{n})

Compute the probability that less than 7.1% of the individuals in this sample hold multiple jobs as follows:

P(p

*Use a <em>z</em>-table.

Thus, the probability that less than 7.1% of the individuals in this sample hold multiple jobs is 0.0043.

7 0
3 years ago
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