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3 years ago

B is between A and C, AB = 13.7 and BC = 8.3. Find AC. AC = 17 AC = 22 AC = 20 AC = 19

Mathematics

2 answers:

ella [17]3 years ago

6 0

Answer: The required length of AC is 22 units.

Step-by-step explanation: We are given that the point B is between the points A and C, where

$AB=13.7,~~BC=8.3.$

We are to find the length of AC.

The picture of the given situation is shown in the attached figure below.

Since the point B is in between A and C, so we must have

$AC=AB+BC=13.7+8.3=22.$

Thus, the required length of AC is 22 units.

Rom4ik [11]3 years ago

3 0

13.7 + 8.7 = 22
ac = 22

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Answer:

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We divide both sides by 100000 and we got:

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Now we can apply natural logs on both sides;

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And then the value of t would be:

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Step-by-step explanation:

For this case since we know that the interest is compounded continuously, then we can use the following formula:

$A =P e^{rt}$

Where A is the future value, P the present value , r the rate of interest in fraction and t the number of years.

For this case we know that P = 100000 and r =0.12 we want to triplicate this amount and that means $A= 300000$ and we want to find the value for t.

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$3 = e^{0.12 t}$

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$t = \frac{ln(3)}{0.12}= 9.16 years$

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2 years ago

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1 year ago

Cube roots of 1 are +1 and -1 (true and false)

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2 years ago

The city has an average of 13 days of rainfall for April.

zhenek [66]

Using the Poisson distribution, we have that:

There is a 0.0859 = 8.59% probability of having exactly 10 days of precipitation in the month of April.

There is a 0.00022 = 0.022% probability of having less than three days of precipitation in the month of April.

There is a 0.2364 = 23.64% probability of having more than 15 days of precipitation in the month of April.

<h3>What is the Poisson distribution?</h3>

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

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

The parameters are:

x is the number of successes

e = 2.71828 is the Euler number

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

For this problem, the mean is given as follows:

$\mu = 13$

The probability of having exactly 10 days of precipitation in the month of April is P(X = 10), hence:

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

$P(X = 10) = \frac{e^{-13}13^{10}}{(10)!} = 0.0859$

There is a 0.0859 = 8.59% probability of having exactly 10 days of precipitation in the month of April.

The probability of having less than three days of precipitation in the month of April is:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

In which:

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

$P(X = 0) = \frac{e^{-13}13^{0}}{(0)!} \ approx 0$

$P(X = 1) = \frac{e^{-13}13^{1}}{(1)!} = 0.00003$

$P(X = 2) = \frac{e^{-13}13^{2}}{(2)!} = 0.00019$

Then:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0 + 0.00003 + 0.00019 = 0.00022

There is a 0.00022 = 0.022% probability of having less than three days of precipitation in the month of April.

For more than 15 days, the probability is:

P(X > 15) = P(X = 16) + P(X = 17) + ... + P(X = 20)

Applying the formula for each of these values and adding them, we have that P(X > 15) = 0.2364, hence:

There is a 0.2364 = 23.64% probability of having more than 15 days of precipitation in the month of April.

More can be learned about the Poisson distribution at brainly.com/question/13971530

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1 year ago

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Answer:

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