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Alexus [3.1K]
3 years ago
14

Mrs.Helio has 2 3/4 acres of farmland. she will plant corn on 1/4 of this land, potatoes on 1/12 of the land, wheat on 5/8 of th

e land and beans on the rest of the land. Mrs. Hello is thinking about planting less wheatvand more potatoes. if she changes the number of acres used for wheat to 1 1/2, how many acres will be left for the potatoes
Mathematics
1 answer:
Minchanka [31]3 years ago
4 0
2 3/4 acres is the same as 11/4 acres
To find how much of the land is beans, find how much is already taken up, and the rest is beans.
Farmland-Corn-Potatoes-Wheat=Beans
1-(1/4)-(1/12)-(5/8)=1/24

Corn: (1/4)x(11/4)=11/16 acres
Potatoes: (1/12)x(11/4)=11/48 acres
Wheat: (5/8)x(11/4)=55/32 acres or 1 23/32 acres
<span>Beans: (1/24)x(11/4)=11/96 acres

Hope this helps!</span>
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3 years ago
"the distribution of means is the correct comparison distribution when"
alexandr402 [8]

The distribution of means is the correct comparison distribution when

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4 0
3 years ago
Could someone explain this to me?
Ilya [14]
This is a simple differentiation problem. Let's start by taking the derivative of both sides (with respect to x):
5y^4\frac{dy}{dx}+6y^2x+6x^2y\frac{dy}{dx}+20x^3 = 0

Simplify:
\frac{dy}{dx} (5y^4+6x^2y) + 6y^2x+20x^3 = 0

Solve for dy/dx:
\frac{dy}{dx} = \frac{-6y^2x-20x^3}{5y^4+6x^2y}

Now, plug in the given points:
\frac{dy}{dx} = \frac{-6(2)^2(-1)-20(-1)^3}{5(2)^4+6(-1)^2(2)} = \frac{24+20}{80+12}

Further simplification gives:
\frac{dy}{dx}|_{(2,-1)} =\frac{44}{92} =  \frac{11}{23}

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7 0
4 years ago
A company manufactures tires at a cost of $40 each. The following are probabilities of defective tires in a given production run
asambeis [7]

Answer:

The expected cost of the company for a 3000 tires batch is $120255

Step-by-step explanation:

Recall that given a probability of defective tires p, we can model the number of defective tires as a binomial random variable. For 3000 tires, if we have a probability p of having a defective tire, the expected number of defective tires  is 3000p.

Let X be the number of defective tires. We can use the total expectation theorem, as follows: if there are A_1,\dots, A_n events that partition the whole sample space, and we have a random variable X over the sample space, then

E(X) = p(A_1)E(X|A_1) + \dots + p(A_n)E(X|A_n).

So, in this case, we have the following

E(X) = P(p=0\%)E(X|p=0\%)+P(p=1\%)E(X|p=1\%)+P(p=2\%)E(X|p=2\%)+P(p=3\%)E(X|p=3\%) = 0.1\cdot 3000\cdot 0 + 0.3\cdot 3000\cdot 0.01+ 0.4\cdot 3000\cdot 0.02+ 0.2\cdot 3000\cdot 0.03 = 51.

Let Y be the number non defective tires. then X+Y = 3000. So Y = 3000-X. Then E(Y) = 3000-E(X). Then, E(Y) = 2949.

Finally, note that the cost of the batch would be 40Y+45X. Then

E(40Y+45X) = 40E(Y)+45E(X) = 40\cdot 2949+45\cdot 51=120255

3 0
3 years ago
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