Use the distributive property to evaluate 4(2x – 1) when x = 5.<br> O4<br> O 39<br> O 32<br> O 36

babunello [35]

If the functions are inverses, then f(g(x)) = x.

A.

$\displaystyle f(g(x))=f\left(\frac{3x+1}{3}\right)=\frac{1}{\frac{3x+1}{x}-3}\\\\=\frac{1}{\left(\frac{3x+1-3x}{x}\right)}=\frac{1}{\left(\frac{1}{x}\right)}=x$

The functions are inverses of each other.

B. The domain of f(x) = x≠3. The domain of g(x) is x≠0.

The domain of f(g(x)) is (-∞, 0) ∪ (0, ∞).

The domain of g(f(x)) is (-∞, 3) ∪ (3, ∞).

4 0

3 years ago

Factor the trinomial: x^2-x-12 x 2 − x − 12

aliya0001 [1]

----------------------------

7 0

3 years ago

Read 2 more answers

What is 378 times by 9

Korvikt [17]

The answer for 378 x 9 = 3402

7 0

4 years ago

Please help me with this question please

zheka24 [161]

*im sorry but I cant see the bottom number but I'll consider it as x

tan 45° = 3/x

x = 3/tan45

x is 3

hope this helps :)

5 0

4 years ago

A university found that of its students withdraw without completing the introductory statistics course. Assume that students reg

polet [3.4K]

Answer:

A university found that 30% of its students withdraw without completing the introductory statistics course. Assume that 20 students registered for the course.

a. Compute the probability that 2 or fewer will withdraw (to 4 decimals).

= 0.0355

b. Compute the probability that exactly 4 will withdraw (to 4 decimals).

= 0.1304

c. Compute the probability that more than 3 will withdraw (to 4 decimals).

= 0.8929

d. Compute the expected number of withdrawals.

= 6

Step-by-step explanation:

This is a binomial problem and the formula for binomial is:

$P(X = x) = nCx p^{x} q^{n - x}$

a) Compute the probability that 2 or fewer will withdraw

First we need to determine, given 2 students from the 20. Which is the probability of those 2 to withdraw and all others to complete the course. This is given by:

$P(X = x) = nCx p^{x} q^{n - x}\\P(X = 2) = 20C2(0.3)^2(0.7)^{18}\\P(X = 2) =190 * 0.09 * 0.001628413597\\P(X = 2) = 0.027845872524$

$P(X = x) = nCx p^{x} q^{n - x}\\P(X = 1) = 20C1(0.3)^1(0.7)^{19}\\P(X = 1) =20 * 0.3 * 0.001139889518\\P(X = 1) = 0.006839337111$

$P(X = x) = nCx p^{x} q^{n - x}\\P(X = 0) = 20C0(0.3)^0(0.7)^{20}\\P(X = 0) =1 * 1 * 0.000797922662\\P(X = 0) = 0.000797922662$

Finally, the probability that 2 or fewer students will withdraw is

$P(X = 2) + P(X = 1) + P(X = 0) \\= 0.027845872524 + 0.006839337111 + 0.000797922662\\= 0.035483132297\\= 0.0355$

b) Compute the probability that exactly 4 will withdraw.

$P(X = x) = nCx p^{x} q^{n - x}\\P(X = 4) = 20C4(0.3)^4(0.7)^{16}\\P(X = 4) = 4845 * 0.0081 * 0.003323293056\\P(X = 4) = 0.130420974373\\P(X = 4) = 0.1304$

c) Compute the probability that more than 3 will withdraw

First we will compute the probability that exactly 3 students withdraw, which is given by

$P(X = x) = nCx p^{x} q^{n - x}\\P(X = 3) = 20C3(0.3)^3(0.7)^{17}\\P(X = 3) = 1140 * 0.027 * 0.002326305139\\P(X = 3) = 0.071603672205\\P(X = 3) = 0.0716$

Then, using a) we have that the probability that 3 or fewer students withdraw is 0.0355+0.0716=0.1071. Therefore the probability that more than 3 will withdraw is 1 - 0.1071=0.8929

d) Compute the expected number of withdrawals.

E(X) = 3/10 * 20 = 6

Expected number of withdrawals is the 30% of 20 which is 6.

5 0

4 years ago

Use the distributive property to evaluate 4(2x – 1) when x = 5. O4 O 39 O 32 O 36