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2 years ago

Evaluate: 4vw/-8 -|v| when v=-2 and w=3

Mathematics

2 answers:

Kaylis [27]2 years ago

6 0

Answer: 1

Step-by-step explanation:

{[4x(-2)x(3)]/(-8)}-2=1

Marta_Voda [28]2 years ago

3 0

Answer:

Step-by-step explanation:

4*-2*3=-24 -24/-8

reduced=3

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Help with #9<br> ASAP PLEASE.

yawa3891 [41]

you add the amount of people in each section that fits into the category. So, 11 people from 6-6:29 and 15 people from 6:30-6:59 and then 8 people from 7:30-7:59 so you get 34 people total

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3 years ago

14. Find the value of the function:<br> If f(x) = 4x + 4x - 4, find f(0).<br> f(0) =

n200080 [17]

Answer:

Step-by-step explanation:

$f(x)=4x^2+4x-4\ or\ f^*(x)=4x+4x-4=8x-4\\\\f(0)-\text{put}\ x=0\ \text{to}\ f(x):\\\\f(0)=4(0)^2+4(0)-4=4(0)+0-4=0-4=-4\\\\f^*(0)=8(0)-4=0-4=-4$

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3 years ago

Suppose GDP is $12 trillion, taxes are $3.6 trillion, private saving is $1.5 trillion, and public saving is $0.8 trillion. Assum

Marianna [84]

Assuming this economy is closed: Consumption $6.9 trillion; Government Purchases $11.2 trillion; National Saving $2.3 trillion; Investment $2.3 trillion.

<h3>Gross domestic product</h3>

a. Consumption

Private S = ( Y – T – C )

C = Y - T - Private S

C = $12 - $3.6 - $1.5 =$6.9 trillion

b. Government purchases

Public S = ( T - G )

G = T - Public S

G = $12 - $0.8

G = $11.2 trillion

c and d. National saving and investment

National savings = Public S + Private S

National savings = $0.8 + $1.5

National savings =$2.3 trillion

Investment=Savings=$2.3 trillion

Therefore, Consumption $6.9 trillion; Government Purchases $11.2 trillion; National Saving $2.3 trillion; Investment $2.3 trillion.

Learn more about GDP here:brainly.com/question/1383956

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2 years ago

Which value is the reciprocal of −238?

ch4aika [34]

The answer is -1/238

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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

3 years ago