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Morgarella [4.7K]
4 years ago
14

The amount of profit, p, you earn by selling knives,k, can be determined by:p=200k-500. A) determine the constraints on profit a

nd the constraints on the number of knives sold
B) what happens to your profit as you sell more knives.
C) is it possible to make $14,000 profit? explain.
Mathematics
1 answer:
Marizza181 [45]4 years ago
3 0
The only restraints on profit result from the restriction on the domain which in turn place a restriction on profit.  

0≤k. Knives sold cannot be negative.

So the range of profit or p is:

p=[-500, +oo)

Profit increases as the number of knives sold increases.  $200 more profit is gained by each sale of a knife.

4000=200k-500

4500=200k

k=22.5

Since k is for knives sold, it must be an integer value, so a $4000 profit is not possible.  At 22, p=3900 and at 23, p=4100


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The sum of three consecutive even integers is 36. Let x represent the first integer, x + 2 represent the second integer, and x +
prisoha [69]

Let

x------>  the first even integer

x+2---->the second even integer

x+3---->the third even integer

we know that

x+(x+2)+(x+4)=36 -----> equation that represent the situation

solve for x

Combine like terms in the left side

(x+x+x)+(2+4)=36

3x+6=36

therefore

<u>the answer is the option B</u>

3x+6=36


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What must be true for line EF to not intersect plane L?
aev [14]

Answer:

If line EF is parallel to line CD, it will not intersect plane L.

5 0
1 year ago
Can someone help solve this equation?
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6 0
2 years ago
A computer can be classified as either cutting dash edge or ancient. Suppose that 94​% of computers are classified as ancient. ​
taurus [48]

Answer:

(a) 0.8836

(b) 0.6096

(c) 0.3904

Step-by-step explanation:

We are given that a computer can be classified as either cutting dash edge or ancient. Suppose that 94​% of computers are classified as ancient.

(a) <u>Two computers are chosen at random.</u>

The above situation can be represented through Binomial distribution;

P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....

where, n = number of trials (samples) taken = 2 computers

            r = number of success = both 2

           p = probability of success which in our question is % of computers

                  that are classified as ancient, i.e; 0.94

<em>LET X = Number of computers that are classified as ancient​</em>

So, it means X ~ Binom(n=2, p=0.94)

Now, Probability that both computers are ancient is given by = P(X = 2)

       P(X = 2)  = \binom{2}{2}\times 0.94^{2} \times (1-0.94)^{2-2}

                      = 1 \times 0.94^{2} \times 1

                      = 0.8836

(b) <u>Eight computers are chosen at random.</u>

The above situation can be represented through Binomial distribution;

P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....

where, n = number of trials (samples) taken = 8 computers

            r = number of success = all 8

           p = probability of success which in our question is % of computers

                  that are classified as ancient, i.e; 0.94

<em>LET X = Number of computers that are classified as ancient</em>

So, it means X ~ Binom(n=8, p=0.94)

Now, Probability that all eight computers are ancient is given by = P(X = 8)

       P(X = 8)  = \binom{8}{8}\times 0.94^{8} \times (1-0.94)^{8-8}

                      = 1 \times 0.94^{8} \times 1

                      = 0.6096

(c) <u>Here, also 8 computers are chosen at random.</u>

The above situation can be represented through Binomial distribution;

P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....

where, n = number of trials (samples) taken = 8 computers

            r = number of success = at least one

           p = probability of success which is now the % of computers

                  that are classified as cutting dash edge, i.e; p = (1 - 0.94) = 0.06

<em>LET X = Number of computers classified as cutting dash edge</em>

So, it means X ~ Binom(n=8, p=0.06)

Now, Probability that at least one of eight randomly selected computers is cutting dash edge is given by = P(X \geq 1)

       P(X \geq 1)  = 1 - P(X = 0)

                      =  1 - \binom{8}{0}\times 0.06^{0} \times (1-0.06)^{8-0}

                      = 1 - [1 \times 1 \times 0.94^{8}]

                      = 1 - 0.94^{8} = 0.3904

Here, the probability that at least one of eight randomly selected computers is cutting dash edge​ is 0.3904 or 39.04%.

For any event to be unusual it's probability is very less such that of less than 5%. Since here the probability is 39.04% which is way higher than 5%.

So, it is not unusual that at least one of eight randomly selected computers is cutting dash edge.

7 0
3 years ago
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