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1 year ago

What is the difference between scaling up and scaling down a ration?

Mathematics

1 answer:

Anton [14]1 year ago

5 0

When you go up it’s positive and when you do down it’s negative. Hope this helps. Mark as brainliest please

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What is the name of the lake located at letter B on the map above? Answer is C :)

astraxan [27]

Answer:

the answer is C. lake Texoma

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3 years ago

Five plus what equals 2?

N76 [4]

Answer:

-3

Step-by-step explanation:

5+-3 equals 2 because it is technically 5-3.

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3 years ago

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Help me please ^^ I'm getting my grades to all A's for a honerole

UNO [17]

Answer:

Step-by-step explanation:

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2 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

2 years ago

Choose the number pair that has digits of 4 that differ in place by a factor of 1000. 4 A. 49,823 and 81,540 B. 83,419 and 46,85

Ne4ueva [31]

Answer:

The correct option is D) 47,063 and 54,722.

Step-by-step explanation:

Consider the provided information.

Here we need to find the number pair that has digits of 4 that differ in place by a factor of 1000.

Now consider the option A) 49,823 and 81,540

The place value of 4 in 49,823 and 81,540 is 40,000 and 40 respectively.

So the difference=40,000-40=39960 which is not a multiple of 1000.

Consider the option B) 83,419 and 46,859

The place value of 4 in 83,419 and 46,859 is 400 and 40,000 respectively.

So the difference=40,000-400=39600 which is not a multiple of 1000.

Consider the option C) 54,890 and 81,745

The place value of 4 in 54,890 and 81,745 is 4000 and 40 respectively.

So the difference=4000-40=3960 which is not a multiple of 1000.

Consider the option D) 47,063 and 54,722

The place value of 4 in 47,063 and 54,722 is 40,000 and 4000 respectively.

So the difference=40000-4000=36000 which is a multiple of 1000.

Hence, the correct option is D) 47,063 and 54,722.

3 0

2 years ago