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IgorLugansk [536]
3 years ago
10

Help plz.....Will give brainlist.

Mathematics
2 answers:
Jlenok [28]3 years ago
6 0

Answer:

50 ft

Step-by-step explanation:

bearhunter [10]3 years ago
3 0

Answer:

Answer            2

Step-by-step explanation:

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3/5 because it can't be 1/6 or 1/2 because they are 50% or smaller. Also it is 2/5 because if you count 20, 40, 60... up to 100 the third number is 50.
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Which expression is equivalent to 2(3x−2y)−4(5y−8x)?
ella [17]
6x-4y-20y+32x
39x-24y
4 0
3 years ago
Pls help me <br><br> Solve for x
Karo-lina-s [1.5K]
8x + 2 + 70 + 60 = 180
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8 0
3 years ago
An automobile manufacturer has discovered that 20% of all the transmissions it installed in a particular style of truck are defe
Hatshy [7]

Answer:

0.148 = 14.8% probability that they will need to order at least one more new transmission

Step-by-step explanation:

For each transmission, there are only two possible outcomes. Either it is defective after a year of use, or it is not. The probability of a transmission being defective is independent of any other transmission. This 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.

20% of all the transmissions it installed in a particular style of truck are defective after a year of use.

This means that p = 0.2

Sold seven trucks:

This means that n = 7

It has two of the new transmissions in stock. What is the probability that they will need to order at least one more new transmission?

This is the probability that at least 3 are defective, that is:

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

In which

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

So

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

P(X = 0) = C_{7,0}.(0.2)^{0}.(0.8)^{7} = 0.2097

P(X = 1) = C_{7,1}.(0.2)^{1}.(0.8)^{6} = 0.3670

P(X = 2) = C_{7,2}.(0.2)^{2}.(0.8)^{5} = 0.2753

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.2097 + 0.3670 + 0.2753 = 0.852

P(X \geq 3) = 1 - P(X < 3) = 1 - 0.852 = 0.148

0.148 = 14.8% probability that they will need to order at least one more new transmission

6 0
3 years ago
Please help; I’m hopeless.
Yuliya22 [10]
Well, you have to find 2 of the same number that add together to make that middle number before the c. Then you have to multiply those 2 identical numbers together to find the value of c!——————————————————————So! For number 10, 7+7 is 14, so 7 and 7 are your two identical numbers.——————————————————————Then you have to multiply them to get c! 7•7=49, so 49 is c.——————————————————————I’ll do 11 for you as well, 12+12=24, and 12•12=144, so 144 is c.
3 0
4 years ago
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