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2 years ago

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The area of the conference table in Mr. Nathan’s office must be no more than 175 ft2. If the length of the table is 18 ft more t

han the width, x, which interval can be the possible widths?

1 answer:

valentina_108 [34]2 years ago

7 0

The answer is  0 < x <span>≤ 7
</span>
First, we know that width = x
Which means that length = x +18

So, the possible equation for the Table's area is

X (X + 18)  ≤ 175

X^2 + 18x - 175  <span>≤ </span>0

Next, we need to calculate is by using complete square method
x^2 + 18x + 81 <span>≤ 175 + 81

(x + 9)^2 </span><span>≤ 256

|x + 9| </span><span>≤ sqrt(256)

|x + 9| </span><span>≤ +-16

-16 </span>≤ x + 9 <span>≤ 16

</span>-16 - 9 ≤ x <span>≤ 16 - 9

</span>-25 ≤ x <span>≤ 7

Since the width couldn't be negative, we can change -25 with 0,

so it become
</span> 0 < x ≤ 7

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Answer:

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b. $E(x) = 0.3$

c. $S(x)=0.5196$

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Step-by-step explanation:

The probability that the company won x bids follows a binomial distribution because we have n identical and independent experiments with a probability p of success and (1-p) of fail.

So, the PMF of X is equal to:

$P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}\\$

Where p is 0.1 and it is the chance of winning. Additionally, n is 3 and it is the number of bids. So the PMF of X is:

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Therefore, the company can expect to win 0.3 bids and it is calculated as:

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So, the expected profit for the company is equal to:

$E=-10,000+50,000(0.243)+100,000(0.027)+150,000(0.001)\\E=5,000$

Because there is a probability of 0.243 to win one bid and it will produce 50,000 of income, there is a probability of 0.027 to win 2 bids and it will produce 100,000 of income and there is a probability of 0.001 to win 3 bids and it will produce 150,000 of income.

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The option that says this is $[1,\infty)$

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