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Pie
2 years ago
15

I need help solving this question. A large company is considering installing charging stations for electric cars in their compan

y parking lots. Suppose that 1.5% of employees at the main office have electric cars and 2% of employees at the branch office have electric cars.

Mathematics
1 answer:
Oduvanchick [21]2 years ago
6 0

Answer:

C; right answer on Khan Academy

Step-by-step explanation:

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Ricky buys a remote car. He applies a markup of 10%. Write two expressions that represent the price of the car
Softa [21]

Answer:

we have to learn this  in middle school

Step-by-step explanation:

3 0
3 years ago
Which answer describe the type of sequence? 3,9,15,21
Alenkinab [10]

Answer:

The sequence is arithmetic

we know that

In an arithmetic sequence, the difference between consecutive terms is always the same and is called common difference

In this problem we have

3,9,15,21,...

Let

a1=3, a2=9,a3=15,a4=21

a4-a3=21-15=6

a3-a2=15-9=6

a2-a1=9-3=6

The sequence is arithmetic

The common difference is equal to

5 0
3 years ago
Simplify the expression<br><br> x³<br> ÷ x^4×x^5
miss Akunina [59]
Answer is = X^6

Please give me the brainliest

4 0
3 years ago
Hi! This is a fairly simple question for some of you, but I wanted to ask, how do you solve this? Thank you so much! :)
givi [52]
Hey I got to it c c c c c c c c c c c c c c c c 8 8
5 0
3 years ago
The prior probabilities for events A1 and A2 are P(A1) = 0.35 and P(A2) = 0.50. It is also known that P(A1 ∩ A2) = 0. Suppose P(
forsale [732]

Answer:

Step-by-step explanation:

Hello!

Given the probabilities:

P(A₁)= 0.35

P(A₂)= 0.50

P(A₁∩A₂)= 0

P(BIA₁)= 0.20

P(BIA₂)= 0.05

a)

Two events are mutually exclusive when the occurrence of one of them prevents the occurrence of the other in one repetition of the trial, this means that both events cannot occur at the same time and therefore they'll intersection is void (and its probability zero)

Considering that P(A₁∩A₂)= 0, we can assume that both events are mutually exclusive.

b)

Considering that P(BIA)= \frac{P(AnB)}{P(A)} you can clear the intersection from the formula P(AnB)= P(B/A)*P(A) and apply it for the given events:

P(A_1nB)= P(B/A_1) * P(A_1)= 0.20*0.35= 0.07

P(A_2nB)= P(B/A_2)*P(A_2)= 0.05*0.50= 0.025

c)

The probability of "B" is marginal, to calculate it you have to add all intersections where it occurs:

P(B)= (A₁∩B) + P(A₂∩B)=  0.07 + 0.025= 0.095

d)

The Bayes' theorem states that:

P(Ai/B)= \frac{P(B/Ai)*P(A)}{P(B)}

Then:

P(A_1/B)= \frac{P(B/A_1)*P(A_1)}{P(B)}= \frac{0.20*0.35}{0.095}= 0.737 = 0.74

P(A_2/B)= \frac{P(B/A_2)*P(A_2)}{P(B)} = \frac{0.05*0.50}{0.095} = 0.26

I hope it helps!

5 0
3 years ago
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