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Setler79 [48]
3 years ago
12

What is the area of this

Mathematics
1 answer:
Zina [86]3 years ago
8 0
The figure has already been split into 3 shapes, a rectangle, a trapezium, and a triangle. I will show you how to get the area of each, so we can add it together at the end.

Rectangle:
The rectangle is fairly simple, just multiply 9 and 16 together to get 144 in².

Triangle:
Again, the triangle is pretty simple. The area of a triangle is (base × height)/2, so you do (31 × 20)/2 = 620/2 = 310 in².

Trapezium:
To find the area of a trapezium, you first add the two widths, which are 16 and 31, and then you divide it by 2. You then multiply that value by the height. 16 + 31 = 47, and 47 ÷ 2 = 23.5.
The height would be 35 (the total length of the shape) - 20 (the height of the triangle) which is 15, so 23.5 × 15 = 352.5 in².

Now we add them up, so 144 + 310 + 352.5 = 806.5 in², which is your final answer.

I hope this helps!
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** I NEED TO GET THIS RIGHT IM DOING TEST CORRECTION**
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Answer: ~27.5

Step-by-step explanation:

I don't know about all the geometry terms, but since a line is 180°, and SE makes 1 right angle (90°), the other side is also 90° (180-90=90).

This means you have a right triangle with sides 15 and 23 and unknown hypotenuse AS.

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3 years ago
According to a​ survey, 60 ​% of the residents of a city oppose a downtown casino. Of these 60 ​% about 8 out of 10 strongly opp
katrin [286]

Answer:

(a) 0.48

(b) 0.20

(c) it is not unusual for a radomly selected resident to oppose the casino and strongly oppose the​ casino.

Step-by-step explanation:

​(a) Find the probability that a randomly selected resident opposes the casino and strongly opposes the casino. ​

The probability that a radomly selected resident opposes the casino and strongly opposes the cassino is the product of the two probabilities, that a resident opposes the casino and that it strongly opposes the casino (once it is in the first group) as it is shown below.

Use this notation:

  • Probability that a radomly selected resident opposes the casino: P(A)

  • Probability that a resident who opposes the casino strongly opposes it: P(B/A), because it is the probability of event B given the event A

i) Determine the <em>probability that a radomly selected resident opposes the casino</em>, P(A)

Probability = number of favorable outcomes / number of possible outcomes

  • P(A) is <em>given as 60%</em>, which in decimal form is 0.60

ii) Next, determine,the <em>probability that a resident who opposes the casino strongly opposes it</em>, P(B/A):

  • It is given as 8 out of 10 ⇒ P(B/A) = 8/10

iii) You want the probability of both events, which is the joint probability or  intersection: P(A∩B).

So, you can use the definition of conditional probability:

  • P(B/A) = P(A∩B) / P(A)

iv) From which you can solve for P(A∩B)

  • P(A∩B) = P(B/A)×P(A) =  (8/10)×(0.60) = 0.48

(b) Find the probability that a randomly selected resident who opposes the casino does not strongly oppose the casino.

In this case, you just want the complement of the probability that <em>a radomly selected resident who opposes the casino does strongly oppose the casino</em>, which is 1 - P(B/A) = 1 - 8/10 = 1 - 0.8 = 0.2.

​(c) Would it be unusual for a randomly selected resident to oppose the casino and strongly oppose the​ casino?

You are being asked about the joint probability (PA∩B), which you found in the part (a) and it is 0.48.

That is almost 0.50 or half of the population, so you conclude it is not unusual for a radomly selected resident to oppose the casino and strongly oppose the​ casino.

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Answer:

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Step-by-step explanation:

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