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2 years ago

What percent 591 is of 2364?

2 answers:

7 0

25% because you need to divide 591 by 2364 to get .25 then you move the decimal place over two to the right which gives you 25%!

Novosadov [1.4K]2 years ago

3 0

Since percent means parts out of 100 then x/100=x%

the easy way is to do it is to divide 591 by 2364
591/2364=0.25
0.25/1 multiply by 100/100=25/100=25%

the answer is 25%

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The time taken to deliver a pizza has a uniform probability distribution from 20 minutes to 60 minutes. What is the probability

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Answer:

(1) The probability that the time to deliver a pizza is at least 32 minutes is 0.70.

(2a) The percentage of results more than 45 is 79.67%.

(2b) The percentage of results less than 85 is 91.77%.

(2c) The percentage of results are between 75 and 90 is 15.58%.

(2d) The percentage of results outside the healthy range 20 to 100 is 2.64%.

Step-by-step explanation:

(1)

Let Y = the time taken to deliver a pizza.

The random variable Y follows a Uniform distribution, U (20, 60).

The probability distribution function of a Uniform distribution is:

$f(x)=\left \{ {{\frac{1}{b-a};\ x\in [a, b] } \atop {0};\ otherwise} \right.$

Compute the probability that the time to deliver a pizza is at least 32 minutes as follows:

$P(Y\geq 32)=\int\limits^{60}_{32} {\frac{1}{b-a} } \, dx \\=\frac{1}{60-20} \int\limits^{60}_{32} {1 } \, dx\\=\frac{1}{40}\times[x]^{60}_{32}\\=\frac{1}{40}\times[60-32]\\=0.70$

Thus, the probability that the time to deliver a pizza is at least 32 minutes is 0.70.

(2)

Let X = results of a certain blood test.

It is provided that the random variable X follows a Normal distribution with parameters $\mu = 60$ and $s = 18$ .

The probabilities of a Normal distribution are computed by converting the raw scores to z-scores.

The z-scores follows a Standard normal distribution, N (0, 1).

(a)

Compute the probability that the results are more than 45 as follows:

$P(X>45)=P(\frac{X-\mu}{\sigma}> \frac{45-60}{18})=P(Z>-0.833)=P(Z$

The percentage of results more than 45 is: $0.7967\times100=79.67\%$

Thus, the percentage of results more than 45 is 79.67%.

(b)

Compute the probability that the results are less than 85 as follows:

$P(X$

The percentage of results less than 85 is: $0.9177\times100=91.77\%$

Thus, the percentage of results less than 85 is 91.77%.

(c)

Compute the probability that the results are between 75 and 90 as follows:

$P(75$

The percentage of results are between 75 and 90 is: $0.1558\times100=15.58\%$

Thus, the percentage of results are between 75 and 90 is 15.58%.

(d)

Compute the probability that the results are between 20 and 100 as follows:

$P(20$

Then the probability that the results outside the range 20 to 100 is: $1-0.9736=0.0264$ .

The percentage of results outside the range 20 to 100 is: $0.0264\times100=2.64\%$

Thus, the percentage of results outside the healthy range 20 to 100 is 2.64%.

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3 years ago

Suppose you have a random sample of 50 of these animals. What is the probability that the mean life span of the sample group is

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In what I know of this problem and what you told me I think there is a 1/2 chance!

think about it, if the life span is 15 years old with standard deviation of 3 then that means we can have bats from 12~18 years old and the mean of the 50 numbers can be anything between that depending on the results we got. Then half of values in interval 12~18 is over 15 which gives it a 1/2 probability!

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3 years ago