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svlad2 [7]
3 years ago
9

15 Points!! Please help and explain how to solve them!! I really want to understand it

Mathematics
1 answer:
babymother [125]3 years ago
8 0
1) idk I got none of your options
2)p=42,000÷(1+0.0225÷12)^(12×5)
P=37,535.04
3)PVAO=4900*12[(1/0.029)-(1/0.029*(1+0.029)^10]=504145.41..compare with 500000
So the answer is D
4)lucas
90×87.92=7,912.8
((8,476.20−7,912.8)
÷7,912.8)×100=7.1%
Peton
55×72.03=3,961.65
((4,192.10−3,961.65)
÷3,961.65)×100=5.8%
So the answer is D
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3 years ago
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Johnson Chemicals is considering two options for its supplier portfolio. Option 1 uses two local suppliers. Each has a "unique-e
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a) P=0.0175

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Step-by-step explanation:

For both options we have to take into account that not only the chance of a "superevent" will disable both suppliers.

The other situation that will disable both is that both suppliers have their "unique-event" at the same time.

As they are, by definition, two independent events, we can calculate the probability of having both events at the same time as the product of both individual probabilities.

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b) The probability that both suppliers will be disrupted using option 2:

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Domain and range and function
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. The sandwich shop offers 8 different sandwiches. Jamey likes them all equally. He picks one randomly each day for lunch. Durin
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Answer:

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From the given information:

A sandwich shop offers eight types of sandwiches, and Jamey likes all of them equally.

The probability that Jamey picks any one of them is 1/8

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Y represents the number of times he chooses falafel

Z represents the number of times he chooses veggie

Then  X+Y+Z ≤ 5 and;

5-X-Y-Z represents the no. of time he chooses the remaining

8 - 3 = 5 sandwiches

However, the objective is to determine the P[X=x,Y=y,Z=z] such that 0≤x,y,z≤5

So, since he chooses x no. of salami sandwiches with probability (1/8)x

and y number of falafel with probability (1/8)y

and for z (1/8)z

Therefore, the remaining sandwiches are chosen with probability \dfrac{5}{8} (5-x-y-z)

So. these x days, y days and z days can be arranged within five days in

= \dfrac{5!}{x!y!z!(5-x-y-z)!}

Thus;

P[X=x,Y=y,Z=z]=  \dfrac{5!}{x!y!z!(5-x-y-z)}  \times \dfrac{1}{8}x*\dfrac{1}{8}y* \dfrac{1}{8}z* \dfrac{5}{8}(5-x-y-z)

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