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ra1l [238]
3 years ago
11

What is the area of this trapezoid?​

Mathematics
2 answers:
zubka84 [21]3 years ago
8 0

Answer:

153 units squared

Step-by-step explanation:

Ksivusya [100]3 years ago
8 0

Answer:

The area of given trapezoid =  153 square units

Step-by-step explanation:

Formula:-

Area of trapezoid = h/2[a + b]

h - height of trapezoid

a and b are two parallel  bases of trapezoid

<u>To find the are of trapezoid</u>

here h = 9

a = 24

b = 10

Area =  h/2[a + b]

  = 9/2[24 + 10] = 9/2[34] = 153 square units

Therefore area of given trapezoid =  153 square units

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The answer is given above

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3 years ago
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I don't understand how to do this, can someone please help me?! Thank you! :)
Stella [2.4K]

Answer:

Erm. If you know pie use it with the 3.141  Since its a half circle i believe you need that since thats the equasion with it. I don't really know :P

Step-by-step explanation:

3 0
3 years ago
What is the value of the missing angle?a, 720b, 120c, 128d, 138
Zigmanuir [339]
The sum of all the interior angles in a hexagon is 720°, so if you add up all the angles and then take them away from 720, you'll get your answer:
152 + 85 = 237
237 + 85 = 322
322 + 125 = 447
447 + 135 = 582
So 720 - 582 = 138°, which is your answer. I hope this helps!
3 0
3 years ago
Suppose the number of children in a household has a binomial distribution with parameters n=12n=12 and p=50p=50%. Find the proba
nadya68 [22]

Answer:

a) 20.95% probability of a household having 2 or 5 children.

b) 7.29% probability of a household having 3 or fewer children.

c) 19.37% probability of a household having 8 or more children.

d) 19.37% probability of a household having fewer than 5 children.

e) 92.71% probability of a household having more than 3 children.

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

In this problem, we have that:

n = 12, p = 0.5

(a) 2 or 5 children

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161

P(X = 5) = C_{12,5}.(0.5)^{5}.(0.5)^{7} = 0.1934

p = P(X = 2) + P(X = 5) = 0.0161 + 0.1934 = 0.2095

20.95% probability of a household having 2 or 5 children.

(b) 3 or fewer children

P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{12,0}.(0.5)^{0}.(0.5)^{12} = 0.0002

P(X = 1) = C_{12,1}.(0.5)^{1}.(0.5)^{11} = 0.0029

P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161

P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.0537

P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0002 + 0.0029 + 0.0161 + 0.0537 = 0.0729

7.29% probability of a household having 3 or fewer children.

(c) 8 or more children

P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 8) = C_{12,8}.(0.5)^{8}.(0.5)^{4} = 0.1208

P(X = 9) = C_{12,9}.(0.5)^{9}.(0.5)^{3} = 0.0537

P(X = 10) = C_{12,10}.(0.5)^{10}.(0.5)^{2} = 0.0161

P(X = 11) = C_{12,11}.(0.5)^{11}.(0.5)^{1} = 0.0029

P(X = 12) = C_{12,12}.(0.5)^{12}.(0.5)^{0} = 0.0002

P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.1208 + 0.0537 + 0.0161 + 0.0029 + 0.0002 = 0.1937

19.37% probability of a household having 8 or more children.

(d) fewer than 5 children

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{12,0}.(0.5)^{0}.(0.5)^{12} = 0.0002

P(X = 1) = C_{12,1}.(0.5)^{1}.(0.5)^{11} = 0.0029

P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161

P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.0537

P(X = 4) = C_{12,4}.(0.5)^{4}.(0.5)^{8} = 0.1208

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0002 + 0.0029 + 0.0161 + 0.0537 + 0.1208 = 0.1937

19.37% probability of a household having fewer than 5 children.

(e) more than 3 children

Either a household has 3 or fewer children, or it has more than 3. The sum of these probabilities is 100%.

From b)

7.29% probability of a household having 3 or fewer children.

p + 7.29 = 100

p = 92.71

92.71% probability of a household having more than 3 children.

5 0
3 years ago
In this figure m&lt;3=130 <br> What is m&lt;6 ?
Georgia [21]

Answer: m\angle6=50\°

Step-by-step explanation:

You can observe in the figure two parallel lines that are intersected by a transversal and several angles are formed.

The angles m∠3 and m∠6 are located inside the parallel lines and on one side of the transversal, this angles are known as "Consecutive interior angles" and they are supplementary (which means that they add up 180°).

Therefore, you know that:

m\angle3+m\angle6=180\°

So you can substitute m∠3=130° and solve for m∠6. Then you get:

130\°+m\angle6=180\°\\m\angle6=180\°-130\°\\m\angle6=50\°

4 0
3 years ago
Read 2 more answers
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