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3 years ago

Which equation is equivalent to 5 - 2x = 4?

2 answers:

5 0

-2x = -1
x = 1/2
Any equation that returns x=1/2 is a solution to this problem. For example, x= (3•4)/6 can be simplified so that x=1/2

Tanya [424]3 years ago

4 0

Hey there!

$\bold{Simplify}$ both sides of your equation
$\bold{-2x+5=4}$
$\bold{Subtract}$ by the number $\bold{5}$
$\bold{Like\downarrow}$
$\bold{-2x+5-5=4-5}$
$\bold{Cancel\rightarrow5-5}$ because it equals $\bold{Keep\rightarrow4-5}$ because it helps us solve for our answer
$\bold{4-5=-1}$
$\bold{We\ get:-2x=-1}$
$\bold{Divide}$ by the $\bold{-2}$ on each of your sides
$\bold{Like\downarrow}$
$\bold{\frac{-2x}{-2}=\frac{-1}{-2}}$
$\bold{Cancel\rightarrow\frac{-2x}{-2}}$ because it gives us the result of $\bold{Keep\rightarrow\frac{-1}{-2}}$ because it gives us the answer
$\boxed{\boxed{\bold{Answer:x=\frac{1}{2}}}}\checkmark$

Good luck on your assignment and enjoy your day!

~ $\frak{LoveYourselfFirst:)}$

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

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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.

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