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zvonat [6]
3 years ago
13

F(-2) = 3 . 4(31 + 10)

Mathematics
1 answer:
AnnZ [28]3 years ago
4 0

Answer:

d

Step-by-step explanation:

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Can someone help me pls
creativ13 [48]

Answer:

69

Step-by-step explanation:

caculator

69 hahahahahahaha

5 0
3 years ago
A baby weighed 7.25 lb at birth. At the end of 8 months, the baby weighed 2.5 times its birth weight. How many pounds did the ba
Vesnalui [34]

Answer: 18.125 lbs

Explanation: The baby weighed 18.125 pounds at the end of eight months.

The baby weighed 7.25 lbs at birth. Eight months later, he weighed 2.5 or 2½ times its birth weight. Multiply 7.25 by 2.5  (which equals 18.125).

So now eight months later, he's 18.125 pounds.

4 0
3 years ago
Read 2 more answers
A Candy contains 300 pieces of which 28% are wet how many pieces are red
Dmitrij [34]
I assumed you meant red instead of wet.

To calculate this, multiply 28% (which is the same as .28) times 300.

300 x .28 = 84
8 0
3 years ago
What is the additive inverse of -5a?
vivado [14]
The answer should be the opposite. equaling 0. so 5.

-5 + 5
3 0
3 years ago
A factory worker productivity is normally distributed. one worker produces an average of 75 units per day with a standard deviat
Angelina_Jolie [31]
Let Xi be the random variable representing the number of units the first worker produces in day i.
Define X = X1 + X2 + X3 + X4 + X5 as the random variable representing the number of units the
first worker produces during the entire week. It is easy to prove that X is normally distributed with mean µx = 5·75 = 375 and standard deviation σx = 20√5.
Similarly, define random variables Y1, Y2,...,Y5 representing the number of units produces by
the second worker during each of the five days and define Y = Y1 + Y2 + Y3 + Y4 + Y5. Again, Y is normally distributed with mean µy = 5·65 = 325 and standard deviation σy = 25√5. Of course, we assume that X and Y are independent. The problem asks for P(X > Y ) or in other words for P(X −Y > 0). It is a quite surprising fact that the random variable U = X−Y , the difference between X and Y , is also normally distributed with mean µU = µx−µy = 375−325 = 50 and standard deviation σU, where σ2 U = σ2 x+σ2 y = 400·5+625·5 = 1025·5 = 5125. It follows that σU = √5125. A reference to the above fact can be found online at http://mathworld.wolfram.com/NormalDifferenceDistribution.html.
Now everything reduces to finding P(U > 0) P(U > 0) = P(U −50 √5125 > − 50 √5125)≈ P(Z > −0.69843) ≈ 0.757546   .
5 0
3 years ago
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