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3 years ago

If d = rt, which of the following equations is correct?

Mathematics

2 answers:

V125BC [204]3 years ago

6 0

The answer is A.r=d/t and C t=d/r

klio [65]3 years ago

5 0

R=d/t and t=d/r are correct I think

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What was the equation of the graph below before it was shifted to the right 1 unit? (equation was g(x)=(x-1.5)^3-(x-1.5))

harina [27]

If graph of the function y=f(x) was shifted to the right 1 unit, then you got new function y=f(x-1).
Reversly, if you have the function $g(x)=(x-1.5)^3-(x-1.5)$ that is obtained by shifting to the right 1 unit, then the initial function was

$g(x+1)=(x+1-1.5)^3-(x+1-1.5), \\ g(x+1)=(x-0.5)^3-(x-0.5)$ .
Answer: Correct choice is D.

3 0

3 years ago

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There is a book sale at the library. The price for each book is $4. Which expression can be used to show how much money the libr

kompoz [17]

Answer:

4x=289

Step-by-step explanation:

this expression could easily help find what the number is by 289/4

5 0

3 years ago

Which is the graph of y=.75x-3

goblinko [34]

Answer:

-225

Step-by-step explanation:

8 0

3 years ago

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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

How do you do this? Thanks for your time and effort!

V125BC [204]

Well first you have to simplify the denominators with x, by multiplying the denominator on the left times the top and bottom of the middle, and vice versa to get 10x/(4x^2-4x)-9(2x-2)/(4x^2-4x)=-1/4 and then you combine the fractions on the left to get 2(9-4x)/(4x^2-4x)=-1/4 and then you cross multiply the fractions to get 8(9-4x)=-4x^2+4x and then simplify to get 72-32x=-4x^2+4x and then 4x^2-36x+72=0 which then we can turn into 4(x-6)(x-3)=0 so x is 6 and 3

3 0

3 years ago