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agasfer [191]
3 years ago
12

The Wellbuilt Company produces two types of wood chippers, economy and deluxe. The deluxe model requires 3 hours to assemble and

1/2 hour to paint, and the economy model requires 2 hours to assemble and 1 hour to paint. The maximum number of assembly hours available is 24 per day, and the maximum number of painting hours available is 8 per day. If the profit on the deluxe model is $98 per unit and the profit on the economy model is $72 per unit, how many units of each model will maximize profit? deluxe units economy units?

Mathematics
1 answer:
seropon [69]3 years ago
4 0

Answer:

The maximum profit is reached with 4 deluxe units and 6 economy units.

Step-by-step explanation:

This is a linear programming problem.

We have to optimize a function (maximize profits). This function is given by:

P=98D+72E

being D: number of deluxe units, and E: number of economy units.

The restrictions are:

- Assembly hours: 3D+2E\leq24

- Paint hours: 0.5D+1E\leq8

Also, both quantities have to be positive:

D\geq 0\\\\E\geq0

We can solve graphically, but we can evaluate the points (D,E) where 2 or more restrictions are saturated (we know that one of this points we will have the maximum profit)

(8;0) \rightarrow  P=98*8+72*0= 784\\\\(0;8) \rightarrow P= 98*0+72*8=576\\\\(4;6) \rightarrow P=98*4+72*6=824

The maximum profit is reached with 4 deluxe units and 6 economy units.

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Determine whether the following probability experiment represents a binomial experiment and explain the reason for your answer.
allochka39001 [22]

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No. See explanation below.

Step-by-step explanation:

Since the cards are being selected <u>without replacement,</u> every time we select a card, <u>the probability varies</u> (since there is one less card) and therefore, the probability doesn't remain the same for every trial and therefore, the probability of success changes for every trial.

It is because of this that this probability experiment doesn't represent a binomial experiment.

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4 years ago
Question Help Suppose that the lifetimes of light bulbs are approximately normally​ distributed, with a mean of 5656 hours and a
koban [17]

Answer:

a)3.438% of the light bulbs will last more than 6262 hours.

b)11.31% of the light bulbs will last 5252 hours or less.

c) 23.655% of the light bulbs are going to last between 5858 and 6262 hours.

d) 0.12% of the light bulbs will last 4646 hours or less.

Step-by-step explanation:

Normally distributed problems can be solved by the z-score formula:

On a normaly distributed set with mean \mu and standard deviation \sigma, the z-score of a value X is given by:

Z = \frac{X - \mu}{\sigma}

After we find the value of Z, we look into the z-score table and find the equivalent p-value of this score. This is the probability that a score will be LOWER than the value of X.

In this problem, we have that:

The lifetimes of light bulbs are approximately normally​ distributed, with a mean of 5656 hours and a standard deviation of 333.3 hours.

So \mu = 5656, \sigma = 333.3

(a) What proportion of light bulbs will last more than 6262 ​hours?

The pvalue of the z-score of X = 6262 is the proportion of light bulbs that will last less than 6262. Subtracting 100% by this value, we find the proportion of light bulbs that will last more than 6262 hours.

Z = \frac{X - \mu}{\sigma}

Z = \frac{6262 - 5656}{333.3}

Z = 1.82

Z = 1.81 has a pvalue of .96562. This means that 96.562% of the light bulbs are going to last less than 6262 hours. So

P = 100% - 96.562% = 3.438% of the light bulbs will last more than 6262 hours.

​(b) What proportion of light bulbs will last 5252 hours or​ less?

This is the pvalue of the zscore of X = 5252

Z = \frac{X - \mu}{\sigma}

Z = \frac{5252- 5656}{333.3}

Z = -1.21

Z = -1.21 has a pvalue of .1131. This means that 11.31% of the light bulbs will last 5252 hours or less.

(c) What proportion of light bulbs will last between 5858 and 6262 ​hours?

This is the pvalue of the zscore of X = 6262 subtracted by the pvalue of the zscore X = 5858

For X = 6262, we have that Z = 1.81 with a pvalue of .96562.

For X = 5858

Z = \frac{X - \mu}{\sigma}

Z = \frac{5858- 5656}{333.3}

Z = 0.61

Z = 0.61 has a pvalue of .72907.

So, the proportion of light bulbs that will last between 5858 and 6262 hours is

P = .96562 - .72907 = .23655

23.655% of the light bulbs are going to last between 5858 and 6262 hours.

​(d) What is the probability that a randomly selected light bulb lasts less than 4646 ​hours?

This is the pvalue of the zscore of X = 4646

Z = \frac{X - \mu}{\sigma}

Z = \frac{4646- 5656}{333.3}

Z = -3.03

Z = -3.03 has a pvalue of .0012. This means that 0.12% of the light bulbs will last 4646 hours or less.

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3 years ago
What are the zeros of g(x)= x³ + 6x² - 9x-54?
Veseljchak [2.6K]

Answer:

3,-3

Step-by-step explanation:

g(3)= (3)³+6(3)²-9(3)-54

= 27+54-27-54

=81-81

= 0

g(-3) = (-3)³+6(-3)²-9(-3)-54

= -27+54+27-54

= 27-27

= 0

For more information comment me !

STAY CONNECTED

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2 years ago
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