All categories

3 years ago

1/71 in multiplication inverse

Mathematics

1 answer:

Cloud [144]3 years ago

8 0

Answer:

Step-by-step explanation:

1/71 * ? = 1

1/71 * x = 1 [consider 'x']

x = 1/(1/71)

x = (1/1)/(1/71) [1 = (1/1)]

x = (1/1)* (71/1)

x = 71.

Hope you're satisfied.

You might be interested in

What is the value of 0.52 +<br> 3<br> 5

Mrrafil [7]

Answer: 35.52

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

What is the answer???<br> P-6+2=-4+10

miskamm [114]

Answer:p=6-2-4+10

P=4-4+10

P=10

Step-by-step explanation:

3 0

3 years ago

What is the area of the hexagon shown below

GalinKa [24]

For you to do area it is length times width.

7 0

3 years ago

Katie's swim team is having an end-of-season pizza party. When the party started, there were 50 slices of pizza. Now there are 2

ss7ja [257]

Answer:

i think its c im not sure so it might be wrong im so sorry if its wrong

Step-by-step explanation:

6 0

3 years ago

Refer to the Trowbridge Manufacturing example in Problem 2-35. The quality control inspection proce- dure is to select 6 items,

Ivanshal [37]

Answer:

77.64% probability that there will be 0 or 1 defects in a sample of 6.

Step-by-step explanation:

For each item, there are only two possible outcomes. Either it is defective, or it is not. The probability of an item being defective is independent of other items. So we use the binomial probability distribution 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.

The true proportion of defects is 0.15

This means that $p = 0.15$

Sample of 6:

This means that $n = 6$

What is the probability that there will be 0 or 1 defects in a sample of 6?

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

In which

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

$P(X = 0) = C_{6,0}.(0.15)^{0}.(0.85)^{6} = 0.3771$

$P(X = 1) = C_{6,1}.(0.15)^{1}.(0.85)^{5} = 0.3993$

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.3771 + 0.3993 = 0.7764$

77.64% probability that there will be 0 or 1 defects in a sample of 6.

5 0

3 years ago