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2 years ago

Help here please. thank you

Mathematics

1 answer:

NISA [10]2 years ago

4 0

Answer: A=C+7

Step-by-step explanation:

Let’s say hypothetically C=6, then B would be 4 since you would have to subtract 2 and A would be 13 since 4+9=13

The difference between 13 and 6 is 7, but that would work with any number and i’m just using c=6 as an example to make the problem easier to understand

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The prior probabilities for events A1 and A2 are P(A1) = 0.30 and P(A2) = 0.55. It is also known that P(A1 ∩ A2) = 0. Suppose P(

k0ka [10]

We're given the following probabilities:

$P(A_1)=0.30$

$P(A_2)=0.55$

$P(A_1\cap A_2)=0$

$P(B\mid A_1)=0.20$

$P(B\mid A_2)=0.05$

(a) Yes, $A_1$ and $A_2$ are mutually exclusive. This is exactly what zero probability of their intersection means. The two events cannot occur simultaneously.

(b) Use the definition of conditional probability to expand:

$P(A_1\cap B)=P(A_1)P(B\mid A_1)=0.30\cdot0.20=0.06$

$P(A_2\cap B)=P(A_2)P(B\mid A_2)=0.55\cdot0.05=0.0275$

$P(B)=P(A_1\cap B)+P(A_2\cap B)=0.06+0.0275=0.0875$

(d) Bayes' theorem says

$P(A_1\mid B)=\dfrac{P(A_1)P(B\mid A_1)}{P(B)}=\dfrac{0.30\cdot0.20}{0.0875}\approx0.686$

$P(A_2\mid B)=\dfrac{P(A_2)P(B\mid A_2)}{P(B)}=\dfrac{0.55\cdot0.05}{0.0875}\approx0.314$

3 0

2 years ago

Help! Give brainliest!

dsp73

Initial population=145,201

Total years=2037-2001=36

$\\ \sf\longmapsto S=P\left(1+\dfrac{r}{100}\right)^t$

$\\ \sf\longmapsto S=145201\left(1+\dfrac{5}{100}\right)^{36}$

$\\ \sf\longmapsto S=145201\left(\dfrac{105}{100}\right)^{36}$

$\\ \sf\longmapsto S=145201(1.05)^{36}$

$\\ \sf\longmapsto S=145201(5.79)$

$\\ \sf\longmapsto S\approx 840713$

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