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4 years ago

What value of x makes this equation true? -2x+3=-15

Mathematics

2 answers:

Dafna1 [17]4 years ago

7 0

Answer: 9

Step-by-step explanation:

-2x+3=-15(move 3 to the other side, it turn -3, -15-3=-18

-3 -3

-2x=-18(Divide -2 to the other side, -18•/•-2=9; if both number negative, the answer will be positive)

X=9

zhannawk [14.2K]4 years ago

5 0

Answer:

x = 9

Step-by-step explanation:

-2x + 3 = -15

-2x = -15 - 3

-2x = -18

x = 9

Hope this helps! :)

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What is the measure of angle PMN? (Image included)

yarga [219]

A quadrilateral QNML is a trapezoid, therefore |∡QNM| + |∡NML| = 180°

|∡NML| = 180° - |∡QNM| ⇒ |∡NML| = 180° - 82° = 98°

|∡NML| = |∡NMP| + |∡PML| ⇒ |∡NMP| = |∡NML| - |∡PML|

|∡NMP| = 98° - 30° = 68°

Answer: |∡NMP| = 68° .

5 0

4 years ago

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

Mkey [24]

Answer:

(a) Probability that 2 or fewer will withdraw is 0.2061.

(b) Probability that exactly 4 will withdraw is 0.2182.

(c) Probability that more than 3 will withdraw is 0.5886.

(d) The expected number of withdrawals is 4.

Step-by-step explanation:

We are given that a university found that 20% of its students withdraw without completing the introductory statistics course.

Assume that 20 students registered for the course.

The above situation can be represented through binomial distribution;

$P(X =r) = \binom{n}{r} \times p^{r} \times (1-p)^{n-r};x=0,1,2,3,......$

where, n = number of trials (samples) taken = 20 students

r = number of success

p = probability of success which in our question is probability

that students withdraw without completing the introductory

statistics course, i.e; p = 20%

Let X = Number of students withdraw without completing the introductory statistics course

So, X ~ Binom(n = 20 , p = 0.20)

(a) Probability that 2 or fewer will withdraw is given by = P(X $\leq$ 2)

P(X $\leq$ 2) = P(X = 0) + P(X = 1) + P(X = 2)

= $\binom{20}{0} \times 0.20^{0} \times (1-0.20)^{20-0}+ \binom{20}{1} \times 0.20^{1} \times (1-0.20)^{20-1}+ \binom{20}{2} \times 0.20^{2} \times (1-0.20)^{20-2}$

= $1 \times1 \times 0.80^{20}+ 20 \times 0.20^{1} \times 0.80^{19}+ 190\times 0.20^{2} \times 0.80^{18}$

= 0.2061

(b) Probability that exactly 4 will withdraw is given by = P(X = 4)

P(X = 4) = $\binom{20}{4} \times 0.20^{4} \times (1-0.20)^{20-4}$

= $4845\times 0.20^{4} \times 0.80^{16}$

= 0.2182

(c) Probability that more than 3 will withdraw is given by = P(X > 3)

P(X > 3) = 1 - P(X $\leq$ 3) = 1 - P(X = 0) - P(X = 1) - P(X = 2) - P(X = 3)

= $1-(\binom{20}{0} \times 0.20^{0} \times (1-0.20)^{20-0}+ \binom{20}{1} \times 0.20^{1} \times (1-0.20)^{20-1}+ \binom{20}{2} \times 0.20^{2} \times (1-0.20)^{20-2}+\binom{20}{3} \times 0.20^{3} \times (1-0.20)^{20-3})$