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2 years ago

Two tangents each intersect a circle at opposite

Mathematics

2 answers:

Oksana_A [137]2 years ago

6 0

Answer:

No. They are parallel and parallel lines don't intersect each other.

Step-by-step explanation:

-The two tangents intersect the circle at it's diameters endpoints.

-Tangents are always perpendicular to the radius.

-Since, both tangents meet the circle perpendicularly at the extreme ends of the diameter, they have a parallel relationship.

-Parallel lies don't intersect.

-Hence, the tangents won't intersect at any point since they are parallel.

Monica [59]2 years ago

5 0

Answer:

No, the tangents cannot intersect outside of the circle. When tangents intersect outside of a circle, the measure of the angle they form is one half the difference of the intercepted arcs. So as a result, both intercepted arcs would have to measure 180 degrees which would make two tangents intersecting outside of the circle, improbable.

Step-by-step explanation:

I just did this question on edg ang got it right. I hope this helps somebody!!

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Which problem will have the greater quotient, 376.0 divided by 93 OR 376 divided by 93.01 ? Explain how you know.

emmainna [20.7K]

376/93 = 4.04301075

376/ 93.01 = 4.04257607.

So 376 divided by 93 is greater because the higher the divisor, the lower the quotient.

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2 years ago

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2 years ago

Jeanne wants to start collecting coins and orders a coin collection starter kit. The kit comes with three coins chosen at random

lesya692 [45]

Conditional probability is a measure of the probability of an event given that another event has occurred. If the event of interest is A and the event B is known or assumed to have occurred, "the conditional probability of A given B", or "the probability of A under the condition B", is usually written as P(A|B), or sometimes $P_B(A)$ .

The conditional probability of event A happening, given that event B has happened, written as P(A|B) is given by
$P(A|B)= \frac{P(A \cap B)}{P(B)}$

In the question, we were told that there are three randomly selected coins which can be a nickel, a dime or a quarter.

The probability of selecting one coin is $\frac{1}{3}$

Part A:
To find the probability that all three coins are quarters if the first two envelopes Jeanne opens each contain a quarter, let the event that all three coins are quarters be A and the event that the first two envelopes Jeanne opens each contain a quarter be B.

P(A) means that the first envelope contains a quarter AND the second envelope contains a quarter AND the third envelope contains a quarter.

Thus $P(A)= \frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} = \frac{1}{27}$

P(B) means that the first envelope contains a quarter AND the second envelope contains a quarter

Thus $P(B)= \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$

Therefore, $P(A|B)=\left( \frac{ \frac{1}{27} }{ \frac{1}{9} } \right)= \frac{1}{3}$

Part B:
To find the probability that all three coins are different if the first envelope Jeanne opens contains a dime, let the event that all three coins are different be C and the event that the first envelope Jeanne opens contains a dime be D.

$P(C)= \frac{3}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{6}{27} = \frac{2}{9}$

 $P(D)= \frac{1}{3}$ 

Therefore, $P(C|D)=\left( \frac{ \frac{2}{9} }{ \frac{1}{3} } \right)= \frac{2}{3}$

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2 years ago

Let h(x)=505.5+8e−0.9x . What is h(5) ? rounded to the nearest tenth. any help would be great

Naddik [55]

For this case, the first thing we are going to do is rewrite the function.
We have then:
h (x) = 505.5 + 8 * exp (-0.9 * x)
We evaluate the value of x = 5 in the function.
We have then:
h (5) = 505.5 + 8 * exp (-0.9 * 5)
h (5) = 505.588872
round to the nearest tenth:
h (5) = 505.6
Answer:
the value of h (5) is:
h (5) = 505.6

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2 years ago

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Talulah is an ecologist who studies the change in the penguin population of Antarctica over time. She observed that the populati

Oksana_A [137]

Answer:

P(t) = 27000 * (1/9)^(t/4)

Step-by-step explanation:

This problem can me modelled with an exponencial formula:

P = Po * (1+r)^t

Where P is the final value, Po is the inicial value, r is the rate and t is the amount of time.

In this problem, we have that the inicial population/value is 27000, the rate is -8/9 (negative because the population decays), and the time t is in months, so as the rate is for every 4 months, we use the value (t/4) in the exponencial.

So, our function will be:

P(t) = 27000 * (1-8/9)^(t/4)

P(t) = 27000 * (1/9)^(t/4)

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2 years ago

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