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Aleksandr-060686 [28]
4 years ago
14

Whats 21 17/12 reduced​

Mathematics
1 answer:
liraira [26]4 years ago
3 0
17/12 = 1 5/12
21 17/12 = 22 5/12
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Convert the following number to percent. 11/3. If needed, round to nearest hundredth.​
muminat

Answer: 27%

Step-by-step explanation:

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What is the multiplicative inverse of -27 and what is the multiplicative inverse for 7/9 and -9/16
liberstina [14]

The multiplicative inverse of a number is the number when you multiply it, it gives you the product of 1.

So for - 27, what number will give us 1 when we multiply it by -27

-27 * -1/27 = 1

So the multiplicative inverse of -27 = -1/27

7/9 * 9/7 = 1

-9/16 * -16/9 = 1

So in essence, just take the number and flip it

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3 0
3 years ago
A food-packaging apparatus underfills 10% of the containers. Find the probability that for any particular 10 containers the numb
Maksim231197 [3]

Answer:

a) P(X = 1) = 0.38742

b) P(X = 3) = 0.05740

c) P(X = 9) = 0.00000

d) P(X \geq 5) = 0.00163

Step-by-step explanation:

For each container, there are only two possible outcomes. Either it is undefilled, or it is not. This means that we can solve this problem using the binomial probability distribution.

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 combinatios 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.

In this problem

There are 10 containers, so n = 10.

A food-packaging apparatus underfills 10% of the containers, so p = 0.1.

a) This is P(X = 1)

P(X = 1) = C_{10,1}.(0.1)^{1}.(0.9)^{9} = 0.38742

b) This is P(X = 3)

P(X = 3) = C_{10,3}.(0.1)^{3}.(0.9)^{7} = 0.05740

c) This is P(X = 9)

P(X = 9) = C_{10,9}.(0.1)^{9}.(0.9)^{1} = 0.00000

d) This is P(X \geq 5).

Either the number is lesser than five, or it is five or larger. The sum of the probabilities of each event is decimal 1. So:

P(X < 5) + P(X \geq 5) = 1

P(X \geq 5) = 1 - P(X < 5)

In which

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

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

P(X = 0) = C_{10,0}.(0.1)^{0}.(0.9)^{10} = 0.34868

P(X = 1) = C_{10,1}.(0.1)^{1}.(0.9)^{9} = 0.38742

P(X = 2) = C_{10,2}.(0.1)^{2}.(0.9)^{8} = 0.1937

P(X = 3) = C_{10,3}.(0.1)^{3}.(0.9)^{7} = 0.05740

P(X = 4) = C_{10,4}.(0.1)^{1}.(0.9)^{9} = 0.38742

So

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.34868 + 0.38742 + 0.19371 + 0.05740 + 0.01116 = 0.99837

Finally

P(X \geq 5) = 1 - P(X < 5) = 1 - 0.99837 = 0.00163

3 0
3 years ago
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