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1 year ago

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What is the reciprocal Consider the expression (5) of the base? Write your answer as a fraction, using the / symbol, like this:

42/53

1 answer:

lora16 [44]1 year ago

3 0

Reciprocal of a fraction<h2>Answer</h2>

8/7

<h2>Explanation</h2>

In this case, the base is 7/8 and the exponent is -4.

In order to find the reciprocal of a fraction we just have to flip it upside down:

That is why the answer is 8/7.

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How do I solve this question?

Elena L [17]

Answer:

47

Step-by-step explanation:

should be the correct answer

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2 years ago

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A volunteer has 12 pounds of birdseed to put in the bird feeders in all of the city parks. There are 13/4 cups of birdseed in a

kozerog [31]

Answer:

$\large \boxed{28}$

Step-by-step explanation:

1. Calculate the number of cups

$\text{Cups} = \text{12 lb}\times\dfrac{1\frac{3}{4} \text{ cups}}{\text{1 lb}} =12\times\dfrac{\text{7 cups}}{4} = 3\times7\text{ cups}= \text{21 cups}$

2. Calculate the number of bird feeders

$\text{ Feeders}= \text{21 cups} \times \dfrac{\text{1 feeder}}{\frac{3}{4}\text{ cup}}=\text{21 cups} \times \dfrac{\text{4 feeders}}{\text{3 cups}} = 7 \times \text{ 4 feeders}\\\\= \textbf{28 feeders}\\\text{The volunteer can fill $\large \boxed{\textbf{28 feeders}}$}$

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3 years ago

Simplify the expression and identify which property is used in each step.

777dan777 [17]

Answer:

1.2

Step-by-step explanation:

(0.85 + 0.50 + 0.15) - 0.30 Given

Simplify by adding in the parentheses first

(0.85 + 0.50 + 0.15) - 0.30

(1.35 + 0.15) - 0.30

(1.5) - 0.30

Subtraction Property

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2 years ago

Find 3/4ths of 40 ......................................................

Mkey [24]

30.

3/4 is equal to 75%, so take 75% of 40 by multiplying .75 by 40 to get 30

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2 years ago

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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