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3 years ago

The sum of the lengths of the two bases of a trapezoid is 22 cm and its area is 66 cm2.

Mathematics

1 answer:

ludmilkaskok [199]3 years ago

8 0

Answer:

h = 6 cm

Step-by-step explanation:

Given that,

The area of a trapezoid, A = 66 cm²

The sum of the lengths of the two bases of a trapezoid is 22 cm.

We need to find the height of the trapezoid.

The area of a trapezium is given by :

$A=\dfrac{1}{2}\times (\text{sum of parallel sides})\times \text{height}$

Substitute the values,

$h=\dfrac{2A}{\text{sum of parallel sides}}\\\\h=\dfrac{2\times 66}{22}\\\\h=6\ cm$

So, the height of the trapezoid 6 cm.

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In a bag with 8 red, 2 white, 4 yellow, and 7 green marbles, what is the ratio of GREEN to WHITE marbles? Write your answer in c

LenKa [72]

The ratio between green to white 7:2

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3 years ago

Jeremiah makes a recipe that calls for 1 and 1/2 cups of flour and 3/4 stick of butter if Jeremiah uses 3 sticks of butter how m

levacccp [35]

We know that
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the answer is
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3 years ago

Simplify your answer should contain only positive exponents (3x^2)^4

S_A_V [24]

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Step-by-step explanation:

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3 years ago

A computer chess game and a human chess champion are evenly matched. They play ten games. Find probabilities for the following e

AleksAgata [21]

Answer:

(a) The probability that each of them win five games is 0.2461.

(b) The probability that the computer wins seven games is 0.1172.

Step-by-step explanation:

Let the random variable X = the human wins a chess game.

The probability that the human wins a game is, P (X) = p = 0.50.

The number of games played by the computer and the human is, n = 10.

The random variable $X\sim Bin(n = 10,p=0.50)$

The probability distribution of the Binomial random variable X is:

$P (X=x)={n\choose x}p^{x}(1-p)^{n-x}={10\choose x}(0.50)^{x}(1-0.50)^{10-x}$

(a)

If both the computer and the human wins five games each then the probability of the human winning 5 games is:

$P (X=5)={10\choose 5}(0.50)^{5}(1-0.50)^{10-5}\\=252\times 0.03125\times0.03125\\=0.246094\\\approx 0.2461$

Thus, the probability that each of them win five games is 0.2461.

(b)

If the computer wins 7 games then the number of games won by the human is, 10 - 7 = 3.

P (Computer winning 7 games) = P (Human winning 3 games)

The probability that the human wins 3 games is:

$P (X=3)={10\choose 3}(0.50)^{3}(1-0.50)^{10-3}\\=120\times 0.125\times0.0078125\\=0.1171875\\\approx 0.1172$

Thus, the probability that the computer wins seven games is 0.1172.

(c)

Compute the probability that the human wins at least 7 games as follows:

$P(X\geq 7)=P(X=7)+P(X=8)+P(X=9)+P(X=10)\\={10\choose 7}(0.50)^{7}(1-0.50)^{10-7}+{10\choose 8}(0.50)^{8}(1-0.50)^{10-8}\\+{10\choose 9}(0.50)^{9}(1-0.50)^{10-9}+{10\choose 10}(0.50)^{10}(1-0.50)^{10-10}\\=0.1172+0.044+0.009+0.0009\\=0.1711$

Thus, the probability that the human wins at least 7 games is 0.1711.

5 0

3 years ago

If you're good at trig please help meeeee Show full working out ;)

Trava [24]

We break down the area into two triangles - since we're given all their side lengths, we use Heron's formula: $[area]=\sqrt{s(s-a)(s-b)(s-c)},$ where $s=\frac{a+b+c}{2}$

$[ABD]=\sqrt{(265)(115)(85)(65)}\approx 12975.92$

$[CBD]=\sqrt{(225)(125)(75)(25)}\approx 7261.84$

$[ABCD]=[ABD]+[CBD]\approx\boxed{20238\,\text{m}^2}.$

7 0

3 years ago