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3 years ago

PLEASE HELP!

Mathematics

1 answer:

wariber [46]3 years ago

8 0

Answer: The maximum value: P = 6x + 7y - 2 is 84 at (5,8).

Step-by-step explanation:

Here, Object of constraint P(max) = 6x + 7y - 2.

And, the system of constraints,

x≤5, y≤(1/5)x + 7, x≥0, y≥0

Since, x= 5 is a line parallel to y-axis.

And at origin it is giving 0≤5 ( false)

Thus the area of inequality x≤5 does not contain the origin.

Now, x -intercept and y-intercept of line y=(1/5)x + 7 are (-35,0) and (0, 7) respectively.

Also, at (0, 0), 0 ≤(1/5)×0 + 7 (true)

Therefore, inequality y≤(1/5)x + 7 will contain the origin.

Now, x≥0, y≥0 shows the first quadrant.

Thus, we get feasible region ABCD.

In which at A≡(0,7), P = 49.

At B≡(5,8), P= 84,

At C≡(5,0), P= 30,

And, D≡(0,0), P= 0

Therefore at B, P is maximum.

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Suppose that 15% of the fields in a given agricultural area are infested with the sweet potato whitefly. One hundred fields in t

hram777 [196]

Answer:

a. $\mu=15$

b. $\mu=7.8586 \ and \ \mu=22.1414$

c. Choice A- Based on limits above, it is unlikely that we would see x = 45, so it might be possible that the trials are not independent.

Step-by-step explanation:

a.Binomial distribution is defined by the expression

$P(X=k)=C_k^n.p^k.(1-p)^{n-k}$

Let n be the number of trials, $n=100$

and p be the probability of success, $p=15\%$

The mean of a binomial distribution is the probability x sample size.

$\mu=np=100\times0.15=15$

b.Limits within which p is approximately 95%

sd of a binomial distribution is given as: $\sigma=\sqrt npq\\q=1-p$

Therefore, $\sigma=\sqrt(100\times0.015\times0.85)=3.5707$

Use the empirical rule to find the limits. From the rule, approximately 95% of the observations are within to standard deviations from mean.

$sd_1=>\mu-2\sigma=15-2\times3.3507=7.8586\\sd_2=>\mu-2\sigma=15+2\times3.3507=22.1414$

Hence, approximately 95% of the observations are within 7.8586 and 22.1414 (areas of infestation).

c. $x=45$ is not within the limits in b above (7.8586,22.1414). X=45 appears to be a large area of infestation. A.Based on limits above, it is unlikely that we would see x = 45, so it might be possible that the trials are not independent.

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3 years ago

How does the graph of this function compare with the graph of the parent function, y=1/x? It is shifted right 5 units and up 2 u

nasty-shy [4]

Answer:

It is shifted left 5 units and up 2 units from the parent function.

Step-by-step explanation:

Given

$y = \frac{1}{x}$

$y' = \frac{1}{x+5} + 2$

Required

Compare both functions

First, translate y, 5 units left.

The rule is:

$(x,y) \to (x + 5,y)$

So, we have:

$y = \frac{1}{x}$

$y_1 = \frac{1}{x + 5}$

Next, translate y1, 2 units up.

The rule is:

$(x,y) \to (x,y+2)$

So, we have:

$y' = y_1 + 2$

$y' = \frac{1}{x + 5} + 2$

Hence, the transformation is:

5 units left and 2 units up

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3 years ago

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3 years ago

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The equation 4x2 – 24x + 4y2 + 72y = 76 is equivalent to

DerKrebs [107]

Answer:

Option 4 is correct.

The equation $4x^2 -24x + 4y^2 + 72y = 76$ is equivalent to $4(x-3)^2 + 4(y+9)^2 =436$

Step-by-step explanation:'

Given equation: $4x^2 -24x + 4y^2 + 72y = 76$

First group the terms with x and those with y;

$(4x^2-24x)+(4y^2+72y) = 76$

Next, we complete the squares.

We can do this by adding a third term such that the x terms and the y terms are perfect squares.

For this we must either add the same value on the other side of the equation or subtract the same value on the same side so that the equality is maintained.

⇒ $4(x^2-6x) +4(y+18y) = 76$