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2 years ago

You ride your bicycle 40 meters. How many complete revolutions does the front wheel make?

Mathematics

1 answer:

Delicious77 [7]2 years ago

7 0

First, we convert the given radius of the wheel to meters giving us an answre of 0.325 m. Then, we calculate for the circumference.
C = 2πrr
Substituting,
C = 2π(0.325 m) = 2.04 m
Then, we have a road that is 40 m long, the number of complete revolutions is,
n = 40/2.04 m = 20

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Raj’s bathtub is clogged and is draining at a rate of 1. 5 gallons of water per minute. The table shows that the amount of water

Morgarella [4.7K]

Answer:

Draining rate is at 1.5 gallons per minute

That would be y = 1.5g / x.

From the function or the equation, here is the answer.

The range would be all real numbers such that 0 ≤ y ≤ 40.

Step-by-step explanation:

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2 years ago

The daily consumption of milk in a house is 3.25 litres. How much milk will be consumed in 30 days?

leonid [27]

Answer:

Milk consumed in 30 days = (30 days)(3.25 litres/day) = 97.5 litres

Step-by-step explanation:

You are given a rate at which the milk is consumed (3.25 litres per 1 day). This is just multiplied by the 30 days to get your answer.

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2 years ago

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The average blood alcohol concentration (bac) of eight male subjects was measured after consumption of 15 ml of ethanol (corresp

Neporo4naja [7]

Answer:

After 5 minutes, the BAC was increasing at the rate of 0.0137 mg/mL a minute.

Step-by-step explanation:

The average blood alcohol concentration (bac) is modeled by the following function.

$c(t) = 0.0225te^{-0.0467t}$

In which t is measured in minuted.

How rapidly was the BAC increasing after 5 minutes?

This is c'(t) when t = 5.

Using the derivative of the product.

Derivative of the product:

$c(t) = f(t)*g(t)$

$c'(t) = f'(t)*g(t) + f(t)*g'(t)$

In which problem:

$f(t) = 0.0225t$

$g(t) = e^{-0.0467t}$

$c'(t) = 0.0225e^{-0.0467t} - 0.0225*0.0467*te^{-0.0467t}$

$c'(5) = 0.0225e^{-0.0467*5} - 0.0225*0.0467*5e^{-0.0467*5} = 0.0137$

After 5 minutes, the BAC was increasing at the rate of 0.0137 mg/mL a minute.

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2 years ago

See You Later Based on a Harris Interactive poll, 20% of adults believe in reincarnation. Assume that six adults are randomly se

REY [17]

Answer:

a) There is a 0.15% probability that exactly five of the selected adults believe in reincarnation.

b) 0.0064% probability that all of the selected adults believe in reincarnation.

c) There is a 0.1564% probability that at least five of the selected adults believe in reincarnation.

d) Since $P(X \geq 5) < 0.05$ , 5 is a significantly high number of adults who believe in reincarnation in this sample.

Step-by-step explanation:

For each of the adults selected, there are only two possible outcomes. Either they believe in reincarnation, or they do not. So we use the binomial probability distribution to solve this question.

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.

In this problem we have that:

$n = 6, p = 0.2$

a. What is the probability that exactly five of the selected adults believe in reincarnation?

This is P(X = 5).

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

$P(X = 5) = C_{6,5}.(0.2)^{5}.(0.8)^{1} = 0.0015$

There is a 0.15% probability that exactly five of the selected adults believe in reincarnation.

b. What is the probability that all of the selected adults believe in reincarnation?

This is P(X = 6).

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

$P(X = 6) = C_{6,6}.(0.2)^{6}.(0.8)^{0} = 0.000064$

There is a 0.0064% probability that all of the selected adults believe in reincarnation.

c. What is the probability that at least five of the selected adults believe in reincarnation?

This is

$P(X \geq 5) = P(X = 5) + P(X = 6) = 0.0015 + 0.000064 = 0.001564$

There is a 0.1564% probability that at least five of the selected adults believe in reincarnation.

d. If six adults are randomly selected, is five a significantly high number who believe in reincarnation?

5 is significantly high if $P(X \geq 5) < 0.05$

We have that

$P(X \geq 5) = P(X = 5) + P(X = 6) = 0.0015 + 0.000064 = 0.001564 < 0.05$

Since $P(X \geq 5) < 0.05$ , 5 is a significantly high number of adults who believe in reincarnation in this sample.

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2 years ago

Kari is flying a kite. She releases 50 feet of string. What is the approximate difference in the height of the kite when the str

nirvana33 [79]

Sinα=h/L where h=height, L=string length...

h=Lsinα so

h(25°)=50sin25≈21.1ft

h(45°)=50sin45≈35.4ft

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2 years ago

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