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Pavlova-9 [17]
3 years ago
11

Hellppppp me its so confusing ????!!!!

Mathematics
1 answer:
melomori [17]3 years ago
3 0
I'm not sure but 
A) 3 out of 20
B) 11 out of 100

You might be interested in
Write 3 times the square root of 2 plus 2 times the square root of 3 in simplest form.
lubasha [3.4K]

Answer:

D

Step-by-step explanation:

3\sqrt{2} + 2\sqrt{3}

is in simplest form and cannot be simplified further


8 0
3 years ago
Tommy and his wife went to the store and bought groceries there's some total was $143.50 if the sales tax was 8% what was the to
Nutka1998 [239]

Answer:

154.98

Step-by-step explanation:

* means multiplication

143.50*.08=11.48

143.50+11.48=$154.98

6 0
3 years ago
Read 2 more answers
This exercise illustrates that poor quality can affect schedules and costs. A manufacturing process has 90 customer orders to fi
svp [43]

Answer:

a) 0.0645 = 6.45% probability that the 90 orders can be filled without reordering components.

b) 0.4062 = 40.62%  probability that the 100 orders can be filled without reordering components.

c) 0.9034 = 90.34% probability that the 100 orders can be filled without reordering components

Step-by-step explanation:

For each component, there are only two possible outcomes. Either it is defective, or it is not. The components can be assumed to be independent. This means that 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.

3% of the components are identified as defective

This means that p = 0.03

a. If the manufacturer stocks 90 components, what is the probability that the 90 orders can be filled without reordering components?

0 defective in a set of 90, which is P(X = 0) when n = 90. So

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

P(X = 0) = C_{90,0}.(0.03)^{0}.(0.97)^{90} = 0.0645

0.0645 = 6.45% probability that the 90 orders can be filled without reordering components.

b. If the manufacturer stocks 102 components, what is the probability that the 100 orders can be filled without reordering components?

At most 102 - 100 = 2 defective in a set of 102, so P(X \leq 2) when n = 102

Then

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)

In which

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

P(X = 0) = C_{102,0}.(0.03)^{0}.(0.97)^{102} = 0.0447

P(X = 1) = C_{102,0}.(0.03)^{1}.(0.97)^{101} = 0.1411

P(X = 2) = C_{102,2}.(0.03)^{2}.(0.97)^{100} = 0.2204

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0447 + 0.1411 + 0.2204 = 0.4062

0.4062 = 40.62%  probability that the 100 orders can be filled without reordering components.

c. If the manufacturer stocks 105 components, what is the probability that the 100 orders can be filled without reordering components?

At most 105 - 100 = 5 defective in a set of 105, so P(X \leq 5) when n = 105

Then

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

In which

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

P(X = 0) = C_{105,0}.(0.03)^{0}.(0.97)^{105} = 0.0408

P(X = 1) = C_{105,0}.(0.03)^{1}.(0.97)^{104} = 0.1326

P(X = 2) = C_{105,2}.(0.03)^{2}.(0.97)^{103} = 0.2133

P(X = 3) = C_{105,3}.(0.03)^{3}.(0.97)^{102} = 0.2265

P(X = 4) = C_{105,4}.(0.03)^{4}.(0.97)^{101} = 0.1786

P(X = 5) = C_{105,5}.(0.03)^{5}.(0.97)^{100} = 0.1116

P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0408 + 0.1326 + 0.2133 + 0.2265 + 0.1786 + 0.1116 = 0.9034

0.9034 = 90.34% probability that the 100 orders can be filled without reordering components

3 0
3 years ago
eric plans to wafch 3 movies each month. write an equation to represent the total number of movies n that he will watch in m mon
Solnce55 [7]

Answer:

n = 3m

Step-by-step explanation:

Total movies watched in m months = 3 movies per month times m months

5 0
3 years ago
What is the probability that a point chosen at random in the given figure will be inside the larger triangle and outside the sma
Anna007 [38]
Given:
Area of a triangle = (h*b) / 2

Area of smaller triangle = (4cm * 4cm) / 2 = 16cm² / 2 = 8 cm²
Area of larger triangle = (6cm * 10cm) / 2 = 60cm² / 2 = 30 cm²

30cm² - 8cm² = 22cm²

Probability that the point is inside the large triangle but outside the small triangle:

22cm² / 30cm² → 11/15 → 0.73 or 73%
4 0
2 years ago
Read 2 more answers
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