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3 years ago

7

A kite is flying 11 ft off the ground. Its line is pulled taut and casts a 9 - ft shadow. Find the length of the line. If necess

ary, round your answer to the nearest tenth.

1 answer:

jolli1 [7]3 years ago

8 0

I believe the answer would be 6.3 feet if you use Pythagorean theorem

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A scientist illegally introduces new species of honey bees in their labs. The growth of the population of this new species is mo

OleMash [197]

The answer A because 2 weeks equal 14 days and so you divide 14 by 450 and get 30

7 0

3 years ago

Waiting on the platform, a commuter hears an announcement that the train is running five minutes late. He assumes the arrival ti

natima [27]

Answer:

D. 91%

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Less than 15 minutes.

Event B: Less than 10 minutes.

We are given the following probability distribution:

$f(T = t) = \frac{3}{5}(\frac{5}{t})^4, t \geq 5$

Simplifying:

$f(T = t) = \frac{3*5^4}{5t^4} = \frac{375}{t^4}$

Probability of arriving in less than 15 minutes:

Integral of the distribution from 5 to 15. So

$P(A) = \int_{5}^{15} = \frac{375}{t^4}$

Integral of $\frac{1}{t^4} = t^{-4}$ is $\frac{t^{-3}}{-3} = -\frac{1}{3t^3}$

Then

$\int \frac{375}{t^4} dt = -\frac{125}{t^3}$

Applying the limits, by the Fundamental Theorem of Calculus:

At $t = 15$ , $f(15) = -\frac{125}{15^3} = -\frac{1}{27}$

At $t = 5$ , $f(5) = -\frac{125}{5^3} = -1$

Then

$P(A) = -\frac{1}{27} + 1 = -\frac{1}{27} + \frac{27}{27} = \frac{26}{27}$

Probability of arriving in less than 15 minutes and less than 10 minutes.

The intersection of these events is less than 10 minutes, so:

$P(B) = \int_{5}^{10} = \frac{375}{t^4}$

We already have the integral, so just apply the limits:

At $t = 10$ , $f(10) = -\frac{125}{10^3} = -\frac{1}{8}$

At $t = 5$ , $f(5) = -\frac{125}{5^3} = -1$

Then

$P(A \cap B) = -\frac{1}{8} + 1 = -\frac{1}{8} + \frac{8}{8} = \frac{7}{8}$

If given the train arrived in less than 15 minutes, what is the probability it arrived in less than 10 minutes?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{7}{8}}{\frac{26}{27}} = 0.9087$

Thus 90.87%, approximately 91%, and the correct answer is given by option D.

3 0

3 years ago

Mr. Summerhill went to the grocery store to buy candied apples for the next RFF. He noticed that the candied apples were $0.20 p

iren [92.7K]

Answer:

Your question cannot be answered with any assurance of being accurate because there is insufficient information. If one were to answer your question, exactly as asked, without interpolating any assumptions, then the answer could be zero because you only stated that he bought the apples but did not state that he left the store or that he left the store with the apples he bought. He could have gone directly to customer service and requested a refund for all the apples, or for all but the three which were just right, or a substitution of some other item for those apples which were bruised or rotten or for all of them, or he could have given the apples to someone else in the store or he could have been in the store with a friend or acquaintance and the friend left the store with the apples and Jared left the store empty handed. Perhaps Jared owns a pet pig and the pig won’t care if the apples are rotten or bruised and Jared bought them at a steep discount, because of their condition, as a treat for his pig. In which case, kudos to Jared for buying his pig a treat.

7 0

2 years ago

In circle A, ∠BAE ≅ ∠DAE. Circle A is shown. Line segments A B, A E, and A D are radii. Lines are drawn from point B to point E

xeze [42]

Answer:

The missing figure is attached down

The length of BE is 27 units ⇒ 3rd answer

Step-by-step explanation:

In circle A:

∠BAE ≅ ∠DAE
Line segments A B, A E, and A D are radii
Lines are drawn from point B to point E and from point E to point D to form secants B E and E D
The length of B E is 3 x minus 24 and the length of E D is x + 10

We need to find the length of BE

∵ AB and AD are radii in circle A

∴ AB ≅ AD

In Δs EAB and EAD

∵ ∠BAE ≅ ∠DAE ⇒ given

∵ AB = AD ⇒ proved

∵ EA = EA ⇒ common side in the two triangles

- Two triangles have two corresponding sides equal and the

including angles between them are equal, then the two

triangles are congruent by SAS postulate of congruence

∴ Δ EAB ≅ Δ EAD ⇒ SAS postulate of congruence

By using the result of congruence

∴ EB ≅ ED

∵ EB = 3 x - 24

∵ ED = x + 10

- Equate the two expressions to find x

∴ 3 x - 24 = x + 10

- Add 24 to both sides

∴ 3 x = x + 34

- Subtract x from both sides

∴ 2 x = 34

- Divide both sides by 2

∴ x = 17

Substitute the value of x in the expression of the length of BE to find its length

∵ BE = 3 x - 24

∵ x = 17

∴ BE = 3(17) - 24

∴ BE = 51 - 24

∴ BE = 27

The length of BE is 27 units

8 0

3 years ago

Read 2 more answers

Leo's bank balances at the end of months 1, 2, and 3 are $1,700.00, $1,734.00, and $1,768.68, respectively. The balances form a

quester [9]

Answer:

i think it$1,991.82. this is what i got.

Step-by-step explanation:

4 0

3 years ago