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3 years ago

7

Consider the circle with a radius of 30 cm. What is the approximate circumference of the circle? (use 3.14 as an approximation o

f pi.)

1 answer:

Goshia [24]3 years ago

8 0

Answer: 188.4 cm

Step-by-step explanation:

Given

the radius of the circle is $r=30\ cm$

Also, circumference is given by $P=2\pi r$

Putting values

$\Rightarrow P=2\times 3.14 \times 30=188.4\ cm$

The circumference of the circle is $188.4\ cm$

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This exercise illustrates that poor quality can affect schedules and costs. A manufacturing process has 90 customer orders to fi

svp [43]

Answer:

a) 0.0645 = 6.45% probability that the 90 orders can be filled without reordering components.

b) 0.4062 = 40.62% probability that the 100 orders can be filled without reordering components.

c) 0.9034 = 90.34% probability that the 100 orders can be filled without reordering components

Step-by-step explanation:

For each component, there are only two possible outcomes. Either it is defective, or it is not. The components can be assumed to be independent. This means that 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.

3% of the components are identified as defective

This means that $p = 0.03$

a. If the manufacturer stocks 90 components, what is the probability that the 90 orders can be filled without reordering components?

0 defective in a set of 90, which is $P(X = 0)$ when $n = 90$ . So

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

$P(X = 0) = C_{90,0}.(0.03)^{0}.(0.97)^{90} = 0.0645$

0.0645 = 6.45% probability that the 90 orders can be filled without reordering components.

b. If the manufacturer stocks 102 components, what is the probability that the 100 orders can be filled without reordering components?

At most 102 - 100 = 2 defective in a set of 102, so $P(X \leq 2)$ when $n = 102$

Then

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

In which

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

$P(X = 0) = C_{102,0}.(0.03)^{0}.(0.97)^{102} = 0.0447$

$P(X = 1) = C_{102,0}.(0.03)^{1}.(0.97)^{101} = 0.1411$

$P(X = 2) = C_{102,2}.(0.03)^{2}.(0.97)^{100} = 0.2204$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0447 + 0.1411 + 0.2204 = 0.4062$

0.4062 = 40.62% probability that the 100 orders can be filled without reordering components.

c. If the manufacturer stocks 105 components, what is the probability that the 100 orders can be filled without reordering components?

At most 105 - 100 = 5 defective in a set of 105, so $P(X \leq 5)$ when $n = 105$

Then

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

In which

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

$P(X = 0) = C_{105,0}.(0.03)^{0}.(0.97)^{105} = 0.0408$

$P(X = 1) = C_{105,0}.(0.03)^{1}.(0.97)^{104} = 0.1326$

$P(X = 2) = C_{105,2}.(0.03)^{2}.(0.97)^{103} = 0.2133$

$P(X = 3) = C_{105,3}.(0.03)^{3}.(0.97)^{102} = 0.2265$

$P(X = 4) = C_{105,4}.(0.03)^{4}.(0.97)^{101} = 0.1786$

$P(X = 5) = C_{105,5}.(0.03)^{5}.(0.97)^{100} = 0.1116$

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0408 + 0.1326 + 0.2133 + 0.2265 + 0.1786 + 0.1116 = 0.9034$

0.9034 = 90.34% probability that the 100 orders can be filled without reordering components

3 0

2 years ago

Is 6 squared plus 9 sqaured <br> rational?

iris [78.8K]

Simplify the expression

6² + 9²

6² = 6 x 6 = 36
9² = 9 x 9 = 81

36 + 81 = 117

117 is a rational number ∴ <span>6</span>²<span> plus 9</span>² is rational

hope this helps

5 0

3 years ago

A six-sided number cube is tossed and a coin is flipped.The sample space is {1H, 2H, 3H, 4H, 5H, 6H, 1T, 2T, 3T, 4T, 5T, 6T}.Wha

aleksandr82 [10.1K]

Since you're already given the sample space, we can simply count how many outcomes satisfy the request, and divide that number by the cardinality of the sample space.

In other words, we're using the formula

$P(E) = \dfrac{\text{Cases in favour of event }E}{\text{Number of possible cases}}$

So, the outcomes with a number higher than two are

3H, 4H, 5H, 6H, 3T, 4T, 5T, 6T

out of these outcomes, we can filter those with heads flipped:

3H, 4H, 5H, 6H

So, there are 4 cases out of 12, and the probability is thus

$\dfrac{4}{12} = \dfrac{1}{3} \approx 0.33$

4 0

3 years ago

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Find a recursive rule for the nth term of the sequence.<br><br> -8, 3, 14, 25, ...

LuckyWell [14K]

Answer:

Step-by-step explanation:

a1=-8

a2=3

D=11

n=-8+11(n-1)

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2 years ago

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Write an equation for each line in slope-intercept form.

Katarina [22]

Answer:

Line 1 : y = -2x - 5

Line 2 : y = -3x + 3

Step-by-step explanation:

note the slope, and then note the y intercept

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2 years ago

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