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2 years ago

Compute the height (in meters) of a cylinder with a radius of 2m and

Mathematics

1 answer:

Olenka [21]2 years ago

7 0

Answer: 10.98 meters

Step-by-step explanation:

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Question 2 of 4

Alexxandr [17]

Answer:

The time it took for the bottle rocket to reach a height of 200 feet is 60 seconds.

Step-by-step explanation:

The function for the height of the bottle rocket (in feet), <em>t</em> seconds after it is launched is:

$h(t)=1157-16\ t$

Compute the value of <em>t</em> as follows:

$h(t)=1157-16\ t$

$200=1157-16t$

$16t=1157-200\\16t=957\\$

$t=59.8125\\t\approx60$

Thus, the time it took for the bottle rocket to reach a height of 200 feet is 60 seconds.

8 0

2 years ago

How many times will 1/4 go into 2 2/4

Scilla [17]

Well, there are 4 of 1/4 in each whole number. there is two whole numbers, so that makes 8 of 1/4's. Now we add the two in the fraction and we come out with 10. 1/4 can go into 2 2/4 10 times.

7 0

2 years ago

Read 2 more answers

in a 3 digit number, the hundreds digit is four more than the units digit, and the tens digit is 3 less than the sum of the othe

Snowcat [4.5K]

Answer: The number is 480.

Step-by-step explanation:

The question states that in a 3 digit number, the hundreds digit is 4 more than the units digit and the tens digit is twice the hundreds digit.

7 0

2 years ago

Read 2 more answers

PLSSSSS HELP ME I WILL GIVE BRAINLIEST!!!!!!

Leto [7]

Top right is ur answer please give brainliest

3 0

3 years ago

Consider a Poisson distribution with μ = 6.

bearhunter [10]

Answer:

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

b) f(2) = 0.04462

c) f(1) = 0.01487

d) $P(X \geq 2) = 0.93803$

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 successes

e = 2.71828 is the Euler number

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

In this question:

$\mu = 6$

a. Write the appropriate Poisson probability function.

Considering $\mu = 6$

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

b. Compute f (2).

This is P(X = 2). So

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

$P(X = 2) = \frac{e^{-6}*6^{2}}{(2)!} = 0.04462$

So f(2) = 0.04462

c. Compute f (1).

This is P(X = 1). So

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

$P(X = 1) = \frac{e^{-6}*6^{1}}{(1)!} = 0.01487$

So f(1) = 0.01487.

d. Compute P(x≥2)

This is:

$P(X \geq 2) = 1 - P(X < 2)$

In which:

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

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

$P(X = 0) = \frac{e^{-6}*6^{0}}{(0)!} = 0.00248$

$P(X = 1) = \frac{e^{-6}*6^{1}}{(1)!} = 0.01487$

$P(X = 2) = \frac{e^{-6}*6^{2}}{(2)!} = 0.04462$

Then

$P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.00248 + 0.01487 + 0.04462 = 0.06197$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.06197 = 0.93803$

$P(X \geq 2) = 0.93803$

5 0

2 years ago