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8090 [49]
4 years ago
5

Which statement is true about the geometric figure that can contain these points?

Mathematics
1 answer:
snow_lady [41]4 years ago
5 0

I believe the answer is o<span>ne plane can be drawn so it contains all three points for this question.</span>

<span />

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The sum of two numbers is 126 and their difference is 42. what are two numbers
Snowcat [4.5K]

Answer:

<h2>84, 42</h2>

Step-by-step explanation:

x,\ y-numbers\\\\\text{The system of equations:}\\\\\underline{+\left\{\begin{array}{ccc}x+y=126\\x-y=42\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad2x=168\qquad\text{divide both sides by 2}\\.\qquad x=84\\\\\\\text{Put the value of}\ x\ \text{to the first equation:}\\\\84+y=126\qquad\text{subtract 84 from both sides}\\y=42

8 0
3 years ago
Which is the logarithmic form of 9^2 = 81?
pochemuha

log_{9}(81)  = 2
8 0
3 years ago
what impact did Ida b. wells have on the civil rights movement in the late 1800's?
const2013 [10]
<span>D. She wrote articles and gave speeches about hate crimes against blacks in the south.</span>
4 0
3 years ago
Read 2 more answers
The following scenarious represent relations that can be graphed. for which of the graphs should the data values be continous ?
Blizzard [7]

Answer:

I would say c

Step-by-step explanation:

5 0
4 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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