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4 years ago

What is the measure in radians for the central angle of a circle whose radius is 9 cm and intercepted arc length is 7.2 cm

Mathematics

2 answers:

pentagon [3]4 years ago

5 0

Answer:

Ф = 0.8 radians.

Step-by-step explanation:

We have given radius= 9 cm and intercepted arc length= 7.2 cm of a circle.

We have to find the central angle of a circle in radians.

As we know that :

l = rФ where r is a radius, l is a arc length, Ф is the angle of circle.

Ф = l ÷ r

Ф = 7.2cm / 9cm

Ф = 0.8 radians.

Ф = 0.8 radians is the central angle of a circle.

Oksana_A [137]4 years ago

4 0

Answer: 0.8 radians

Step-by-step explanation:

To solve this exercise you must apply the formula shown below:

$S=r\theta$

Where S is the arc lenght, r is the radius of the circle and $\theta$ is the central angle in radians.

Solve for the central angle:

$\theta=\frac{S}{r}$

Now, when you susbtitute the value of the arc length and the radius, you obtain that the central angle is:

$\theta=\frac{7.2cm}{9cm}=0.8radians$

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3 years ago

Hospitals typically require backup generators to provide electricity in the event of a power outage. Assume that emergency backu

JulsSmile [24]

Answer:

a) There is a 10.24% probability that both generators fail during a power outage.

b) There is an 89.76% probability of having a working generator in the event of a power outage, which is not high enough for the hospital.

Step-by-step explanation:

For each emergency backup generator, there are only two possible outcomes. Either they work correctly, or they fail. So we use the binomial 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.

In this problem we have that:

Assume that emergency backup generators fail 32% of the times when they are needed. So they work correctly 100-32 = 68% of the time. So $p = 0.68$

There are two generators, so $n = 2$

a. Find the probability that both generators fail during a power outage

This is $P(X = 0)$

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

$P(X = 0) = C_{2,0}.(0.68)^{0}.(0.32)^{2} = 0.1024$

There is a 10.24% probability that both generators fail during a power outage.

b. Find the probability of having a working generator in the event of a power outage. Is that probability high enough for the hospital? Assume the hospital needs both generators to fail less than 1% of the time when needed.

Either there are no working generators, or there is at least one working generator. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

We want $P(X \geq 1)$ . So

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.1024 = 0.8976$

There is an 89.76% probability of having a working generator in the event of a power outage, which is not high enough for the hospital.

To be high enough for the hospital, this probability should be at least of 99%.

7 0

4 years ago

6/9 in a percentage

motikmotik

Answer:

66.666%

Step-by-step explanation:

6/9=x/100

9x=6x100

9x=600

x=600:9

x=66.666

66.666/100=66.666%

3 0

2 years ago

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faust18 [17]

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aleksandrvk [35]

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