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3 years ago

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A construction company is planning to bid on a building contract. the bid costs the company $1900. the probability that the bid

is accepted is 1/10. if the bid is accepted, the company will make $94000 minus the cost of the bid.
a. what is the expected value in this situation?

1 answer:

Shtirlitz [24]3 years ago

7 0

Not enough information to answer

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Segment wx is shown Explain how you would construct a perpendicular bisector of wx using a compass and a straightedge

n200080 [17]

Answer:

To perpendicular bisector of line segment WX. There are following steps:

1) Draw arcs or circles from points A and B on the both sides of WX.

2) Name the intersection points as W and X.

3) Use the straightedge to draw a line through points W and X.

4) Name the point as O

hence we have construct perpendicular bisector AB of WX which bisects at O.

4 0

3 years ago

The water diet requires you to drink two cups of water every half hour from the time you get up until you go to bed, but otherwi

Elan Coil [88]

Answer:

$7-3.182 \frac{13.638}{\sqrt{4}}= -14.698$

$7+3.182 \frac{13.638}{\sqrt{4}}= 28.698$

Step-by-step explanation:

For this case we have the following info given:

Weight before diet 180 125 240 150

Weight after diet 170 130 215 152

We define the random variable $D = before-after$ and we can calculate the inidividual values:

D: 10, -5, 25, -2

And we can calculate the mean with this formula:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

And the deviation with:

$s= \sqrt{\frac{\sum_{i=1}^n (X_i- \bar X)^2}{n-1}}$

And after replace we got:

$\bar D= 7, s_d = 13.638$

And the confidence interval for this case would be given by:

$\bar D \pm t_{\alpha/2} \frac{s_d}{\sqrt{n}}$

The degrees of freedom are given by:

$df = n-1= 4-1=3$

For the 95% of confidence the value for the significance is $\alpha=0.05$ and the critical value would be $t_{\alpha/2}= 3.182$ . And replacing we got:

$7-3.182 \frac{13.638}{\sqrt{4}}= -14.698$

$7+3.182 \frac{13.638}{\sqrt{4}}= 28.698$

6 0

3 years ago

Sum of three numbers in an as is 33 and the sum of thier cubes is 5049.find the numbers

Alex73 [517]

Sum of three numbers is 33 and sum of their cubes is 5049 then three numbers are 7,11,15 or 15,11,7.

As given,

Let a-d, a, a +d be three numbers

Sum =33

⇒a-d + a +a +d =33

⇒ a =11

Sum of cubes= 5049

⇒ (a-d)³ + a +(a +d)³ =5049

⇒ a³ -d³ -3a²d +3ad² + a³ +a³ +d³ +3a²d +3ad² =5049

⇒3a³ + 6ad² = 5049

⇒ d=± 4

Therefore, sum of three numbers is 33 and sum of their cubes is 5049 then three numbers are 7,11,15 or 15,11,7.

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6 0

1 year ago

Help please!!!!!!!!!! Thank you

andre [41]

2b) stretching
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3 years ago

2/5 as a decimal and percent

lukranit [14]

Decimal: .4
Percent: 40%

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3 years ago