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3 years ago

53% of what number is 384

Mathematics

2 answers:

fgiga [73]3 years ago

5 0

Answer:

The answer is 384 is 53% of 724.53

TiliK225 [7]3 years ago

3 0

Answer:

The answer is 384 is 53% of 724.53

Step-by-step explanation:

Calculation:

384/53% = 724.53

formula:

value1/% = value2

Hope this helps!

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Based on past experience, the main printer in a university computer centre is operating properly 90% of the time. Suppose inspec

yaroslaw [1]

Answer:

a) 38.74% probability that the main printer is operating properly for exactly 9 inspections.

b) Approximately 100% probability that the main printer is operating properly for at least 3 inspections.

c) The expected number of inspections in which the main printer is operating properly is 9.

Step-by-step explanation:

For each inspection, there are only two possible outcomes. Either it is operating correctly, or it is not. The probability of the printer operating correctly for an inspection is independent of any other inspection, which 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.

Based on past experience, the main printer in a university computer centre is operating properly 90% of the time.

This means that $p = 0.9$

Suppose inspections are made at 10 randomly selected times.

This means that $n = 10$

A) What is the probability that the main printer is operating properly for exactly 9 inspections.

This is $P(X = 9)$ . So

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

$P(X = 9) = C_{10,9}.(0.9)^{9}.(0.1)^{1} = 0.3874$

38.74% probability that the main printer is operating properly for exactly 9 inspections.

B) What is the probability that the main printer is operating properly for at least 3 inspections?

This is:

$P(X \geq 3) = 1 - P(X < 3)$

In which

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

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

$P(X = 0) = C_{10,0}.(0.9)^{0}.(0.1)^{10} \approx 0$

$P(X = 1) = C_{10,1}.(0.9)^{1}.(0.1)^{9} \approx 0$

$P(X = 2) = C_{10,2}.(0.9)^{2}.(0.1)^{8} \approx 0$

Thus:

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0 + 0 + 0 = 0$

Then:

$P(X \geq 3) = 1 - P(X < 3) = 1 - 0 = 1$

Approximately 100% probability that the main printer is operating properly for at least 3 inspections.

C) What is the expected number of inspections in which the main printer is operating properly?

The expected value for the binomial distribution is given by:

$E(X) = np$

In this question:

$E(X) = 10(0.9) = 9$

3 0

3 years ago

For the month of April which option would be better? 1. To be gifted $1,000,000 once only on April 1st. OR 2. To receive a penny

SVETLANKA909090 [29]

Answer:

a.) April has 30 days

If you for option No.2, you would have:

[day, amount($)]

1, 0.01

2, 0.02

3, 0.04

4, 0.08

5, 0.16

6, 0.32

7, 0.64

8, 1.28

9, 2.56

10, 5.12

11, 10.24

12, 20.48

13, 40.96

14, 81.82

15. 163.48

16, 327.68

17, 655.36

18, 1310.72

19, 2621.44

20, 5242.88

21, 10,485.76

22, 20,971.52

23, 41,943.04

24, 83,886.08

25, 167,772.16

26, 335,544.32

27, 671,088.64

28, 1,342,177.28

29, 2,684,354.56

30, 5,368,709.12

Option 2 grants you more than 5 times as much as Option A and is therefore obviously better.

b.) A diagram would show first a slow rise, than a steeper and steeper rise, then would almost growvertically. exponential growth

#coronatime

The most elegant form to describe the given numbers would simply be

$=1*2^x

You start with one, eich doubles after a day (x=1). x is the number of days and how often you multiply by 2

c.) Wasn't sure without calculating, but I guessed Opt.2, because it seemed that one should be tricked into choosing Opt.1

Have a nice day

Brainliest would be appreciated

If there are questions left, feel free to ask them

4 0

3 years ago

WILL AWARD BRAINLIEST!!!

Anna007 [38]

Answer:

A. p(hat) = .139

We divide our sample population by the amount who tested positive. 14851/107109 = .139.

B. 1.62 million

We just multiply the p times the population. 11.69 M * .139 = 1.62 M

C. No

It depends upon the sample method. From what I can tell, I assume all conditions are met and it was not biased.

If it wasn't random, that is a problem, but we aren't given this information.

We can test if it's small enough. It can't be larger than 10% of the population. 107109 * 10 < 11.69 million, so it's small enough.

We can also test if it's large enough. np and nq must be greater than 10. 107100 * .139 > 10, 107100 * .861 > 10.

7 0

4 years ago

Convert 15/125 As A Decimal Using Long Division <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B15%7D%7B125%7D%20" id="TexForm

denpristay [2]

0 . 1 2

125 ) 15.00

125

------

250

-----

The answer is 0.12

7 0

3 years ago

A class of 28 students has 13 boys and 15 girls. What is the ratio of girls to boys in the class? 15:28 13:15 O 13:28 15:13 Prev

uranmaximum [27]

Answer: $\frac{15}{13}$

Step-by-step explanation:

Given

There are 28 students in a class out of which

13 are boys so the remaining 15 are girls

Ratio of girls to boys in the class $=\frac{15}{13}$

8 0

4 years ago