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Ivanshal [37]
3 years ago
10

What is the value of x? Enter you answer in the box X= In.

Mathematics
1 answer:
marishachu [46]3 years ago
5 0

Answer:

x = 24

Step-by-step explanation:

The 2 triangles are similar hence ratios of corresponding sides are equal, that is

\frac{15}{5} = 3 = \frac{x}{8} ( multiply by 8 )

x = 8 × 3 = 24

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Find the probability that a point is chosen randomly inside the rectangle is in each given shape...
Arlecino [84]

By taking the quotients between the areas, we see that:

  • a) P = 0.275
  • b) P = 0.85

<h3>How to find the probabilities?</h3>

First we need to find the areas of the 3 shapes.

For the triangle, the area is:

T = 3*5/2 = 7.5

For the blue square, the area is:

S = 3*3 = 9

For the rectangle, the area is:

R = 10*6 = 60

Now, what is the probability that a random point lies on the triangle or in the square?

It is equal to the quotient between the areas of the two shapes and the total area of the rectangle, this is:

P= (7.5 + 9)/60 = 0.275

b) The area of the rectangle that is not the square is:

A = 60 - 9 = 51

Then the probability of not landing on the square is:

P' = 51/60 = 0.85

If you want to learn more about probability:

brainly.com/question/25870256

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6 0
2 years ago
Need help with this math question
Y_Kistochka [10]

Answer:

23%

Step-by-step explanation:

There are 4 male and 3 female freshmen. Thus the total number of freshmen is 7.

On the other hand, we have 14 male students and 16 female students. Thus the total number of students is 30.

If a student is selected at random, the probability that the student is a freshman is;

( 7/30) * 100 = 23.33%

3 0
3 years ago
An architect is standing 250 feet from the base of a building and would like to know the height of the building. If he measures
GREYUIT [131]

see the figure below to better understand the problem

we have that

\begin{gathered} tan55^o=\frac{h}{250}---->\text{ by TOA} \\  \\ solve\text{ for h} \\ h=250*tan55^o \\ h=357.0\text{ ft} \end{gathered}<h2>The answer is 357.0 feet</h2>

8 0
1 year ago
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