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3 years ago

In △ABC, AB = BC = 20 and DE ≈ 9.28. Approximate BD.

Mathematics

1 answer:

Volgvan3 years ago

6 0

Answer:

BD = 5

Step-by-step explanation:

∠A = 60°

Since AB = 20 = BC, Then ∠B = 60° resulting in a equilateral triangle with ∠C = 60°. Then dividing ∠DAE into half gives 4 = angles of 15° yielding each secition of BC being equal to 20/4 = 5. Do i don't think DE ≈ 9.28 cannot occur.

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Answer:

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Step-by-step explanation:

The square root of 121 is 11. There are no other solutions, I believe.

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Simplify the expression: (3 − 7i)(2 − 5i).

Mandarinka [93]

Answer:

Step-by-step explanation:

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3 years ago

×WILL MARK AS BRAINLIEST×<br>Solve the equation for b, d and x

Lera25 [3.4K]

Solve for d:
(3 (a + x))/b = 2 d - 3 c
(3 (a + x))/b = 2 d - 3 c is equivalent to 2 d - 3 c = (3 (a + x))/b:
2 d - 3 c = (3 (a + x))/b
Add 3 c to both sides:
2 d = 3 c + (3 (a + x))/b
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Answer: d = (3 c)/2 + (3 (a + x))/(2 b)

-----------------------------

Solve for x:
(3 (a + x))/b = 2 d - 3 c
Multiply both sides by b/3:
a + x = (2 b d)/3 - b c
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____________________________

Solve for b:
(3 (a + x))/b = 2 d - 3 c
Take the reciprocal of both sides:
b/(3 (a + x)) = 1/(2 d - 3 c)
Multiply both sides by 3 (a + x):
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3 years ago

The number of people arriving at a ballpark is random, with a Poisson distributed arrival. If the mean number of arrivals is 10,

Stella [2.4K]

Answer:

a) 3.47% probability that there will be exactly 15 arrivals.

b) 58.31% probability that there are no more than 10 arrivals.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

If the mean number of arrivals is 10

This means that $\mu = 10$

(a) that there will be exactly 15 arrivals?

This is P(X = 15). So

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 15) = \frac{e^{-10}*(10)^{15}}{(15)!} = 0.0347$

3.47% probability that there will be exactly 15 arrivals.

(b) no more than 10 arrivals?

This is $P(X \leq 10)$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-10}*(10)^{0}}{(0)!} = 0.000045$

$P(X = 1) = \frac{e^{-10}*(10)^{1}}{(1)!} = 0.00045$

$P(X = 2) = \frac{e^{-10}*(10)^{2}}{(2)!} = 0.0023$

$P(X = 3) = \frac{e^{-10}*(10)^{3}}{(3)!} = 0.0076$

$P(X = 4) = \frac{e^{-10}*(10)^{4}}{(4)!} = 0.0189$

$P(X = 5) = \frac{e^{-10}*(10)^{5}}{(5)!} = 0.0378$

$P(X = 6) = \frac{e^{-10}*(10)^{6}}{(6)!} = 0.0631$

$P(X = 7) = \frac{e^{-10}*(10)^{7}}{(7)!} = 0.0901$

$P(X = 8) = \frac{e^{-10}*(10)^{8}}{(8)!} = 0.1126$

$P(X = 9) = \frac{e^{-10}*(10)^{9}}{(9)!} = 0.1251$

$P(X = 10) = \frac{e^{-10}*(10)^{10}}{(10)!} = 0.1251$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.000045 + 0.00045 + 0.0023 + 0.0076 + 0.0189 + 0.0378 + 0.0631 + 0.0901 + 0.1126 + 0.1251 + 0.1251 = 0.5831$

58.31% probability that there are no more than 10 arrivals.

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3 years ago

A commercial builder has a downtown lot with 250 frontage feet on Broadway. The lot is 200’ deep. By code, the builder must allo

Morgarella [4.7K]

Subtract 15 from the depth:

200-15 = 185

Subtract 20 from the width ( 10 from both sides)

250-20 = 230

Area to build on: 185 x 230 = 42,550 square feet.

8 0

4 years ago