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Dahasolnce [82]

2 years ago

5

An Olympic swimmer can swim 50 meters in 21 seconds. How long would it take the swimmer to swim 800 meters if he kept up the sam

e pace?

1 answer:

ICE Princess25 [194]2 years ago

7 0

Answer:

It would take 336 seconds

Step-by-step explanation:

Set up as a proportion:

50 meters/21 seconds = 800 meters/x

Then we cross multiply to find the answer:

336 = x

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There were 3 shots total.

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3 years ago

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nirvana33 [79]

Hello!

$\large\boxed{\text{ slope = -3}}$

Use the slope formula to solve for the slope of the line (m):

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We can plug in the points (0, 4) and (1, 1) to solve:

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3 years ago

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Colt1911 [192]

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2 years ago

A college professor never finishes his lecture before the end of the hour and always finishes his lectures within 2 min after th

Citrus2011 [14]

Answer:

Value of k is $\displaystyle\frac{3}{8}$

Step-by-step explanation:

We are given the following information in the question:

$f(x) = kx^2, 0\leq x \leq 2\\~~~~~~~= 0, \text{ otherwise}$

where x is the time elapses between the end of the hour and the end of the lecture.

We have to find the values of k.

Since, f(x) is the pdf, then,

$\displaystyle\int^\infty_{-\infty} f(x) = 1\\\\\displaystyle\int^2_{0} f(x) = 1\\\\\displaystyle\int^2_{0} kx^2 = 1\\\\k\bigg[\frac{x^3}{3}\bigg]^2_0 = 1\\\\k\times \frac{8}{3} = 1\\\\k = \frac{3}{8}$

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2 years ago

Police estimate that 25% of drivers drive without their seat belts. If they stop 6 drivers at random, find the probability tha

Furkat [3]

Answer:

17.80% probability that all of them are wearing their seat belts.

Step-by-step explanation:

For each driver stopped, there are only two possible outcomes. Either they are wearing their seatbelts, or they are not. The drivers are chosen at random, which mean that the probability of a driver wearing their seatbelts is independent from other drivers. So we use the normal probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Police estimate that 25% of drivers drive without their seat belts.

This means that 75% wear their seatbelts, so $p = 0.75$

If they stop 6 drivers at random, find the probability that all of them are wearing their seat belts.

This is $P(X = 6)$ .

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 6) = C_{6,6}.(0.75)^{6}.(0.25)^{0} = 0.1780$

17.80% probability that all of them are wearing their seat belts.

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3 years ago

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