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2 years ago

If my diameter is 20.6ft how would i get my radius,area,circumference?

Mathematics

2 answers:

zheka24 [161]2 years ago

6 0

Step-by-step explanation:

Diameter = 20.6

$radius = \frac{diameter}{ = 2} = \frac{20.6}{2} = 10.3 \: ft \\ \\ c = \pi \:d = 3.14 \times 20.6 = 64.68 \: ft \\$

sveta [45]2 years ago

4 0

Step-by-step explanation:

Radius is half of the diameter

Circumference is 2*pi*radius

Area is pi*radius^2

So r (radius)= 10.3

C (circumference)=64.72

and A (area)=333.29

feel free to ask more questions :)

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2 years ago

A scientist estimated that a mixture would need 6 milliliters of a chemical to balance. The actual amount needed was 7 millilite

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2 years ago

5^x + 5^x+1= 150<br> Find x<br> Help me pls i need soon<br> Tksss

raketka [301]

Answer:

$x=2$

Step-by-step explanation:

So we have the equation:

$5^x+5^{x+1}=150$

First, separate the second term:

$5^x+(5\cdot5^x)=150$

Let u equal 5ˣ. So:

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Combine like terms:

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2 years ago

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Rudiy27

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Step-by-step explanation:

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2 years ago

Read 2 more answers

1. A report from the Secretary of Health and Human Services stated that 70% of single-vehicle traffic fatalities that occur at n

Nuetrik [128]

Using the binomial distribution, it is found that there is a 0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

For each fatality, there are only two possible outcomes, either it involved an intoxicated driver, or it did not. The probability of a fatality involving an intoxicated driver is independent of any other fatality, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

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

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

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

70% of fatalities involve an intoxicated driver, hence $p = 0.7$ .

A sample of 15 fatalities is taken, hence $n = 15$ .

The probability is:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)$

Hence

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

$P(X = 10) = C_{15,10}.(0.7)^{10}.(0.3)^{5} = 0.2061$

$P(X = 11) = C_{15,11}.(0.7)^{11}.(0.3)^{4} = 0.2186$

$P(X = 12) = C_{15,12}.(0.7)^{12}.(0.3)^{3} = 0.1700$

$P(X = 13) = C_{15,13}.(0.7)^{13}.(0.3)^{2} = 0.0916$

$P(X = 14) = C_{15,14}.(0.7)^{14}.(0.3)^{1} = 0.0305$

$P(X = 15) = C_{15,15}.(0.7)^{15}.(0.3)^{0} = 0.0047$

Then:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.2061 + 0.2186 + 0.1700 + 0.0916 + 0.0305 + 0.0047 = 0.7215$

0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

A similar problem is given at brainly.com/question/24863377

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2 years ago