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White raven [17]
3 years ago
6

Can y’all help me *********

Mathematics
1 answer:
melisa1 [442]3 years ago
7 0

Answer:

33. The sheep pen is 16 square yards

Step-by-step explanation:

You can divide the rhombus inside into two triangles. Each triangle has a base of 4 yards and height 4 yards.

Area of triangle: A = 0.5*b*h = 0.5(4)(4) = 8

Since there are 2, then 2*8 = 16 square yards.

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-0.8b^2+7.4b+(5.6b-0.2b^2)
puteri [66]

Answer:

what? I don't understand lol

5 0
3 years ago
The 7th term of an A.p is -4. If the common difference is-5, find the 6th term​
Lubov Fominskaja [6]

Answer:

Step-by-step explanation:

a₆ = a₇+(6-7)d = -4 -(-5) =1

7 0
2 years ago
20% of students drive a car. 9 students are randomly are selected and asked if they drive a car. What is the probability that ex
Andre45 [30]

The probability is low.

20% of 9 is 2. Therefore, the probability is low that 5 out of the 9 students drive cars.

6 0
3 years ago
Eight times a number is the same as 30 less than 5 times the number. Find the number?
Charra [1.4K]

Answer:

x = -10

Step-by-step explanation:

We just have to translate the problem correctly

>8 times a number is the same

this tells me that

8x = ...

>as 30 less than 5 times the number.

... = 5x - 30

Now set them equal to eachother

8x = 5x - 30

Now isolate the variable x

8x - 5x = 5x - 5x - 30\\3x = -30\\x = \frac{-30}{3} = -10

6 0
3 years ago
Read 2 more answers
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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