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Anna71 [15]
3 years ago
9

200 - pondela

Mathematics
1 answer:
Firdavs [7]3 years ago
7 0

Answer:

add my sc = bmic ava, no spaces

or add my discord = XoXo_ava

Step-by-step explanation:

:)

HOPE I HELPED LOLIOLOL IK IM ANOYING BUT YUH

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Find X using the Pythagorean Theorem
Natalija [7]

Answer:

x = 4.47

Step-by-step explanation:

Formula for hypotenuse:

x =\sqrt{a^2+b^2}

x=\sqrt{4^2+2^2}

x=\sqrt{16+4}

x=\sqrt{20}

x=4.47

8 0
3 years ago
Question 18: please help. I will give brainliest to correct answer.
Veronika [31]

Answer:

3. mean of 2nd is larger

Step-by-step explanation:

mean of 2nd sem 87 > 86.8 1st sem

7 0
4 years ago
Read 2 more answers
Explain how you know that the some of 12.6, 3.1, and 5.4cis greater than 20
ddd [48]
You can add the numbers together so:

 12.6
   3.1
<u>+ 5.4
</u> 21.1 
<u>
</u>Since the sum of the numbers is 21.1, it is greater than 20<u>
</u>
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
Rewrite it as a function and isolate Y. ​
hodyreva [135]

Answer: y= -2x + 12

or f (y ) = 6 - \frac{y}{2}

Step-by-step explanation:

just subtract the 2x to get the y alone

8 0
3 years ago
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