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1 year ago

Find the are of the trapezoid

Mathematics

2 answers:

snow_lady [41]1 year ago

6 0

Answer:

Step-by-step explanation:

52√3 ft²

butalik [34]1 year ago

6 0

Answer:

120 ft.

Step-by-step explanation:

I did it by adding 15 and 15, then dividing that by 2, and finally multiplying that by the height, which is 8, to get 120.

Hope this helps :)

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In a complete paragraph, explain how to scope a speech. <br> Please use your own words.

butalik [34]

Answer:

Prepare. You dread public speaking because you don't know what to say.

Psych yourself. ...

Know your audience. ...

Be an expert on your topic. ...

Visualize. ...

Conduct an ocular inspection. ...

Read, watch, and listen to good speakers.

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

TRUE OR FALSE: The solution to the equation 3y – 4 = 6 - 2y is -10.

anygoal [31]

Answer:

false your answer would be 2

Step-by-step explanation:

4 0

2 years ago

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I got -4 but I don't see it on here, did I do something wrong??

Ira Lisetskai [31]

Answer:

it is -4

Step-by-step explanation:

but the answer is not on there so the answer will be D.

5 0

1 year ago

Read 2 more answers

Plz help!! (I just copied and pasted it)

lora16 [44]

Step-by-step explanation:

3d + 4.5g <= 18.

3 possible combinations are:

0 DVD and 4 Video games,

1 DVD and 3 Video games,

2 DVDs and 2 Video games.

5 0

2 years ago

Read 2 more answers

Scott and Letitia are brother and sister. After dinner, they have to do the dishes, with one washing and the other drying. They

ehidna [41]

Answer:

The probability that Scott will wash is 2.5

Step-by-step explanation:

Given

Let the events be: P = Purple and G = Green

$P = 2$

$G = 3$

Required

The probability of Scott washing the dishes

If Scott washes the dishes, then it means he picks two spoons of the same color handle.

So, we have to calculate the probability of picking the same handle. i.e.

$P(Same) = P(G_1\ and\ G_2) + P(P_1\ and\ P_2)$

This gives:

$P(G_1\ and\ G_2) = P(G_1) * P(G_2)$

$P(G_1\ and\ G_2) = \frac{n(G)}{Total} * \frac{n(G)-1}{Total - 1}$

$P(G_1\ and\ G_2) = \frac{3}{5} * \frac{3-1}{5- 1}$

$P(G_1\ and\ G_2) = \frac{3}{5} * \frac{2}{4}$

$P(G_1\ and\ G_2) = \frac{3}{10}$

$P(P_1\ and\ P_2) = P(P_1) * P(P_2)$

$P(P_1\ and\ P_2) = \frac{n(P)}{Total} * \frac{n(P)-1}{Total - 1}$

$P(P_1\ and\ P_2) = \frac{2}{5} * \frac{2-1}{5- 1}$

$P(P_1\ and\ P_2) = \frac{2}{5} * \frac{1}{4}$

$P(P_1\ and\ P_2) = \frac{1}{10}$

<em>Note that: 1 is subtracted because it is a probability without replacement</em>

So, we have:

$P(Same) = P(G_1\ and\ G_2) + P(P_1\ and\ P_2)$

$P(Same) = \frac{3}{10} + \frac{1}{10}$

$P(Same) = \frac{3+1}{10}$

$P(Same) = \frac{4}{10}$

$P(Same) = \frac{2}{5}$

8 0

2 years ago