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expeople1 [14]
3 years ago
14

When I think of a number, double it, then add seven, I get 25. what number did I think of?

Mathematics
2 answers:
murzikaleks [220]3 years ago
6 0

Answer:

9

Step-by-step explanation:

let the number be x

double it = x+x = 2x

then add seven = 2x+7

i get 25 2x+7= 25

subtract 7 from both sides

2x= 25-7= 18

2x= 18

x= 18/2 = 9

sweet [91]3 years ago
5 0

Answer:

9

Step-by-step explanation:

25-7=18

half of 18 is 9

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A construction company is considering submitting bids for contracts of three different projects. The company estimates that it h
julsineya [31]

Answer:

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

b. E(x) = 0.3

c. S(x)=0.5196

d. E=5,000

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:

P(x)=\frac{3!}{x!(3-x)!}*0.1^{x}*(1-0.1)^{n-x}\\

For binomial distribution:

E(x)=np\\S(x)=\sqrt{np(1-p)}

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

E(x) = np = 3*0.1 = 0.3

Additionally, the standard deviation of the number of bids won is:

S(x)=\sqrt{np(1-p)}=\sqrt{3(0.1)(1-0.1)}=0.5196

Finally, the probability to won 1, 2 or 3 bids is equal to:

P(1)=\frac{3!}{1!(3-1)!}*0.1^{1}*(1-0.1)^{3-1}=0.243\\P(2)=\frac{3!}{2!(3-2)!}*0.1^{2}*(1-0.1)^{3-2}=0.027\\P(3)=\frac{3!}{3!(3-3)!}*0.1^{3}*(1-0.1)^{3-3}=0.001

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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