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2 years ago

Given mn, find the value of x.

Mathematics

2 answers:

Aleksandr-060686 [28]2 years ago

7 0

Answer:

10x -10

Step-by-step explanation:

this answer is really sure

Ksivusya [100]2 years ago

5 0

Answer:

x = 19

Step-by-step explanation:

The two angles with measures are supplementary.

6x + 9 + 4x - 19 = 180

10x - 10 = 180

10x = 190

x = 19

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30 POINTS HELP PLEASE

Reptile [31]

Answer:

$\large\boxed{a=-3,\ b=-7\ and\ c=3}$

Step-by-step explanation:

$ax^2+bx+c=0$

We have:

$-3a^2=7a-3$ subtract 7a from both sides

$-3a^2-7a=7a-7a-3$

$-3a^2-7a=-3$ add 3 to both sides

$-3a^2-7a+3=-3+3\\\\-3a^2-7a+3=0\to a=-3,\ b=-7, c=3$

8 0

3 years ago

What are you finding when you are asked what is 60% of 175? the part, the whole or the precent

butalik [34]

Answer:

105 is the part.

Step-by-step explanation:

175 is the whole

60% is the percent

x is the part

175 * 0.60 = 105

3 0

2 years ago

What is the exact circumference of a circle with a radius of 15 cm?

Anni [7]

Answer:

Step-by-step explanation:

None of the choices is correct.

Circumference= 2πr = 30π cm

6 0

2 years ago

Round 4.88 to the nearest tenth

Minchanka [31]

4.88 rounded to the nearest tenth is 4.9

7 0

3 years ago

Read 2 more answers

Any athlete who fails the Enormous State University's women's soccer fitness test is automatically dropped from the team. Last y

castortr0y [4]

Using conditional probability, it is found that there is a 0.7873 = 78.73% probability that Mona was justifiably dropped.

Conditional Probability

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.
$P(A \cap B)$ is the probability of both A and B happening.
P(A) is the probability of A happening.

In this problem:

Event A: Fail the test.
Event B: Unfit.

The probability of failing the test is composed by:

46% of 37%(are fit).
100% of 63%(not fit).

Hence:

$P(A) = 0.46(0.37) + 0.63 = 0.8002$

The probability of both failing the test and being unfit is:

$P(A \cap B) = 0.63$

Hence, the conditional probability is:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.63}{0.8002} = 0.7873$

0.7873 = 78.73% probability that Mona was justifiably dropped.

A similar problem is given at brainly.com/question/14398287

3 0

2 years ago