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3 years ago

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The circumference is 111.44 cm the shaded region is three times the whole circle find the area of the shaded region in square ce

ntimeters round your answer to the nearest hundreds

1 answer:

Umnica [9.8K]3 years ago

4 0

Yeah when we are the Jews and this is so weird what’s going on so we’re only seven so please stop

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How to solve -13=5(1 4m)-2m?

Paladinen [302]

First, you would multiply: -13 = 70m - 2m. Then, you subtract the like terms, which would be 70m - 2m: -13 = 68m. You are looking for what m is, so you would divide 68 on both sides: m = - 13 / 68. In math, people would usually like fractions more than decimals, but if you are looking for the decimal number, it would be m ≈ - 0.19.

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3 years ago

I need help can someone help

zaharov [31]

Answer:

yes i can but will have to get back to you in about half a hour just doing maths atm with the teacher

Step-by-step explanation:

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3 years ago

COMBINE LIKE TERMS TO CREATE AN EQUIVALENT EXPRESSION!

d1i1m1o1n [39]

Answer:

0.25k + 1.5 - k - 3.5 = -0.75k - 2.

Step-by-step explanation:

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3 years ago

Determine whether the random variable is discrete or continuous. a. The number of statistics students now reading a book. b. The

andrey2020 [161]

Answer:

Continuous random variables: c and e

Discrete random variables: a, b, d

Step-by-step explanation:

We have to identify whether the random variable is discrete or continuous.

A discrete variable is a variable whose value is obtained by counting.
A continuous random variable X takes all values in a given interval of numbers.
Thus, a continuous variable can have values in decimals but a discrete random variable cannot take values in decimals.

a. The number of statistics students now reading a book.

Discrete random variable since number of students cannot take decimal values.

b. The number of textbook authors now sitting at a computer.

Discrete random variable since number of textbooks cannot be expressed in decimals but counted.

c. The exact time it takes to evaluate 27 plus 72.

It is a continuous random variable as it may take all values within an interval of time.

d. The number of free dash throw attempts before the first shot is made.

It is a discrete random variable since the number of throws can always be whole number.

e. The time it takes to fly from City Upper A to City Upper B.

Time is a continuous random variable.

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3 years ago

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

3 years ago