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katovenus [111]

3 years ago

12

There are 100 hours of labor, 500 lbs of material and 1000 gallons of water available. If the goal is to maximize the total prof

it then the objective function is: (the variables A,B,C & D are the number of widgets of each type produced)
a.min 10A +1268C9D
b.min 10A-15B 7CBD
c.max 10A +15B +7C+8D
d.max A B C D

1 answer:

Black_prince [1.1K]3 years ago

4 0

Answer:

The correct option is option C max 10A +15B +7C+8D

Step-by-step explanation:

As the complete question is not given here thus the complete question is found online and is attached herewith.

From the data the profit for 1 unit of A is 10, B is 15, C is 7 and D is 8, so the profit function is given as 10A+15B+7C+8D. and as profit is to be maximized so the correct option is option C.

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The measure of one angle is 80 degrees. What is the measure of its supplement? A 80 degrees B 100 degrees C 10 degrees D I don't

ArbitrLikvidat [17]

Answer:

B. 100 degrees

Step-by-step explanation:

5 0

3 years ago

How do you write a percent into a fraction or a mixed number in simplest form

love history [14]

To change a percent to a fraction, you remove the percent sign and put the given number over 100. For example, if you are given 78%, it would be 78/100 or 39/50

4 0

3 years ago

An automobile manufacturer has given its car a 46.7 miles/gallon (MPG) rating. An independent testing firm has been contracted t

erastova [34]

Answer:

$z=\frac{46.5-46.7}{\frac{1.1}{\sqrt{150}}}=-2.23$

The p value would be given by:

$p_v =2*P(z$

For this case since th p value is lower than the significance level of0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true mean for this case is significantly different from 46.7 MPG

Step-by-step explanation:

Information given

$\bar X=46.5$ represent the mean

$\sigma=1.1$ represent the population standard deviation

$n=150$ sample size

$\mu_o =46.7$ represent the value to verify

$\alpha=0.05$ represent the significance level for the hypothesis test.

z would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to test if the true mean for this case is 46.7, the system of hypothesis would be:

Null hypothesis: $\mu = 46.7$

Alternative hypothesis: $\mu \neq 46.7$

Since we know the population deviation the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

Replacing we got:

$z=\frac{46.5-46.7}{\frac{1.1}{\sqrt{150}}}=-2.23$

The p value would be given by:

$p_v =2*P(z$

For this case since th p value is lower than the significance level of0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true mean for this case is significantly different from 46.7 MPG

7 0

3 years ago

Suppose an=3an-1 -5an-2 -4an-3. and a4= -7, a5=-36, a6=-85 , find a1 , a2, a3

maria [59]

Answer:

$a_1=4\\\\a_2=0\\\\a_3=3$

Explanation:

The equation is:

$a_n=3a_{n-1}-5a_{n-2}-4a_{n-3}$

and

$a_4=-7, a_5=-36, a_6=-85$

<u>1. Subsittute n = 6 into the equation:</u>

$a_6=3a_{5}-5a_{4}-4a_{3}$

Now subsititute the known values:

$-85=3(-36)-5(-7)-4a_{3}$

You can solve for $a_3$

$-85=-108+35-4a_{3}\\\\4a_3=12\\\\a_3=12/4\\\\a_3=3$

<u />

<u>2. Substitute n = 5 into the equation:</u>

$a_5=3a_{4}-5a_{3}-4a_{2}$

Substitute the known values and solve for $a_2$

$-36=3(-7)-5(3)-4a_{2}\\\\4a_2=-21-15+36=0\\\\a_2=0$

<u>3. Substitute n = 4 into the equation, subsitute the known values and solve for </u> $a_1$ <u />

$a_4=3a_{3}-5a_{2}-4a_{1}$

$-7=3(3)-5(0)-4a_{1}\\\\4a_1=9+7=16\\\\a_1=4$

7 0

3 years ago

Find the product using the distributive property 2oo × 42.3

Greeley [361]

Answer:

Step-by-step explanation:

200 * 42.3 = 200 * (40 + 2.3)

= 200*40 + 200*2.3

= 8000 + 460.0

=8460

4 0

3 years ago