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likoan [24]
3 years ago
14

Solve this math problem

Mathematics
1 answer:
laila [671]3 years ago
6 0
1-.8=.2
za/2=.2/2=.1
invNorm(.1)=1.28
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Please help me with this question
Amanda [17]
The graph shows the maximum residual is ...
  ◉  Data point (10, 10); Residual = 4.50

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Apparently, this question makes no use of the line of best fit.

6 0
3 years ago
WILL GIVE BRAINLIEST!
Veseljchak [2.6K]

Answer:

it's A please mark mine good .............

5 0
3 years ago
When two 6-sided dice are rolled, there are 36 possible outcomes. Find the probability that the sum is 8
katrin2010 [14]

Answer:

5/36

Step-by-step explanation:

Probability is the outcome that a  event will occur

Probability = Expected outcome/Total outcome

Since two sided dice are rolled, the total outcome will be expressed as;

Total outcome = 36

Since we are to find the probability that the sum is 8, the values that gives a sum of 8a re;

(2,6)(6,2), (5, 3), (3,5), (4, 4)

The expected outcome = 5

Probability = 5/36

Hence the probability that the sum is 8 is 5/36

4 0
3 years ago
I definitely need help, big time. What's the step by step solution to this math problem -(80/-10)^2?
mojhsa [17]
The final answer is -64
6 0
3 years ago
Read 2 more answers
The route used by a certain motorist in commuting to work contains two intersections with traffic signals. The probability that
tekilochka [14]

Answer:

a) P(X∩Y) = 0.2

b) P_1 = 0.16

c) P = 0.47

Step-by-step explanation:

Let's call X the event that the motorist must stop at the first signal and Y the event that the motorist must stop at the second signal.

So, P(X) = 0.36, P(Y) = 0.51 and P(X∪Y) = 0.67

Then, the probability P(X∩Y) that the motorist must stop at both signal can be calculated as:

P(X∩Y) = P(X) + P(Y) - P(X∪Y)

P(X∩Y) = 0.36 + 0.51 - 0.67

P(X∩Y) = 0.2

On the other hand, the probability P_1 that he must stop at the first signal but not at the second one can be calculated as:

P_1 = P(X) - P(X∩Y)

P_1 = 0.36 - 0.2 = 0.16

At the same way, the probability P_2 that he must stop at the second signal but not at the first one can be calculated as:

P_2 = P(Y) - P(X∩Y)

P_2 = 0.51 - 0.2 = 0.31

So, the probability that he must stop at exactly one signal is:

P = P_1+P_2\\P=0.16+0.31\\P=0.47

7 0
3 years ago
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