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4 years ago

Which of these is a member of the sample space when you flip a coin twice and then roll a die?

Mathematics

2 answers:

dalvyx [7]4 years ago

8 0

I think the answer will be Heads-Tails-3. Hope it help!

Reptile [31]4 years ago

3 0

I believe it's Heads-Tails-3

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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.

5 0

3 years ago

If segment DF is an angle bisector, solve for x

klasskru [66]

Answer:

x = 19

Step-by-step explanation:

Since DF is an angle bisector then it divides the opposite sides into segments that are proportional to the other 2 sides, that is

$\frac{DE}{DG}$ = $\frac{EF}{GF}$ , substitute values

$\frac{25}{22.5}$ = $\frac{x-9}{9}$ ( cross- multiply )

22.5(x - 9) = 225 ( divide both sides by 22.5 )

x - 9 = 10 ( add 9 to both sides )

x = 19

8 0

3 years ago

Please help me

yKpoI14uk [10]

9514 1404 393

Answer:

A) 5x+12 = -12x-12

D) 5x+12 = -5x-12

Step-by-step explanation:

If you subtract the right side expression from both sides, you will get an equation with something equal to zero. If the 'something' has a variable in it, there is exactly one solution.

A: (5x+12) -(-12x-12) = 17x+24 = 0 . . . one solution

B: (5x+12) -(5x-5) = 17 = 0 . . . . no solutions

C: (5x+12)-(5x+12) = 0 = 0 . . . . infinite solutions

D: (5x+12) -(-5x-12) = 10x +24 = 0 . . . one solution

6 0

3 years ago

Subtract. 7 1/5 − (−3/5)

mojhsa [17]

This would just be 7 1/5 + 3/5 since the negative sign gets distributed and makes (-3/5) into a positive number.

From there you have basic math and you get 7 4/5.

5 0

3 years ago

Read 2 more answers

Juan invest $3700 in a simple interest account at a rate of 4% for 15 years

OleMash [197]

Question:

Juan Invest $3700 In A Simple Interest Account At A Rate Of 4% For 15 Years. How Much Money Will Be In The Account After 15 Years?

Answer:

There will be $ 5920 in account after 15 years

Solution:

The simple interest is given by formula:

$S.I = \frac{p \times n \times r }{100}$

Where,

p is the principal

n is number of years

r is rate of interest

From given,

p = 3700

r = 4 %

t = 15 years

Therefore,

$S.I = \frac{3700 \times 4 \times 15 }{100}\\\\S.I = 37 \times 4 \times 15\\\\S.I = 2220$

How Much Money Will Be In The Account After 15 Years?

Total money = principal + simple interest

Total money = 3700 + 2220

Total money = 5920

Thus there will be $ 5920 in account after 15 years

8 0

4 years ago