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2 years ago

14

Solve for the variable

1 answer:

Feliz [49]2 years ago

5 0

Step-by-step explanation:

$\frac{20x}{20} = \frac{40}{20} \\ x = 2$

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The question is in the picture.

svetlana [45]

Answer:I figured it out for you. The answer is 8-6*sq 3

Step-by-step explanation:

I used Math way

8 0

2 years ago

12e^5 divided 3e^3 simplified

bonufazy [111]

Answer:

$4 {e}^{2}$

Step-by-step explanation:

$\frac{12 {e}^{5} }{3 {e}^{3} } \\ 12 \div 3 \times {e}^{5 - 3} \\ = 4 {e}^{2}$

hope this helps

brainliest appreciated

good luck! have a nice day!

6 0

3 years ago

6. a. Sixty students in a class took an examination in Physics and Mathematics. If 17 of them passed Physics only, 25 passed in

Ivahew [28]

Let $C$ be the set of all students in the <u>c</u>lassroom.

Let $P$ and $M$ be the sets of students that pass <u>p</u>hysics and <u>m</u>ath, respectively.

We're given

$n(C) = 60$

$n(P \cap M') = 17$

$n(P \cap M) = 25$

$n((P \cup M)') = n(P' \cap M') = 9$

i. We can split up $P$ into subsets of students that pass both physics and math $(P\cap M)$ and those that pass only physics $(P\cap M')$ . These sets are disjoint, so

$n(P) = n(P\cap M) + n(P\cap M') = 25 + 17 = \boxed{42}$

ii. 9 students fails both subjects, so we find

$n(C) = n(P\cup M) + n(P\cup M)' \implies n(P\cup M) = 60 - 9 = 51$

By the inclusion/exclusion principle,

$n(P\cup M) = n(P) + n(M) - n(P\cap M)$

Using the result from part (i), we have

$n(M) = 51 - 42 + 25 = 34$

and so the probability of selecting a student from this set is

$\mathrm{Pr}(M) = \dfrac{34}{60} = \boxed{\dfrac{17}{30}}$

7 0

2 years ago

Whats happenin playas'

Doss [256]

Answer:

nothing, I just want free points!

thx

6 0

3 years ago

Read 2 more answers

PLEASE ANSWER AND EXPLAIN | 98 POINTS

DIA [1.3K]

Answer:

Step-by-step explanation:

R-value can range from -1 to 1 with correlation being strongest at -1 and 1 and weakest at 0.

So the answer is D. -0.9.

3 0

3 years ago

Read 2 more answers