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zavuch27 [327]
3 years ago
13

According to the general equation for conditional probability, if (image attached)

Mathematics
2 answers:
lutik1710 [3]3 years ago
8 0

Answer:

Correct choice is A. P(A|B)=\frac{40}{49}.

Step-by-step explanation:

Given that P(A\cap B)=\frac{5}{7}, P(B)=\frac{7}{8}.

Now using those values , we need to find the value of P(A|B).

So apply the formula of conditional probability:

P(A\cap B)=P(B) \times P(A|B)

Plug the given values into above formula,  we get:

\frac{5}{7}=\frac{7}{8} \times P(A|B)

\frac{7}{8} \times P(A|B)=\frac{5}{7}

P(A|B)=\frac{\frac{5}{7}}{\frac{7}{8}}

P(A|B)=\frac{5}{7}\cdot\frac{8}{7}

P(A|B)=\frac{40}{49}

Hence correct choice is A. P(A|B)=\frac{40}{49}.

katrin2010 [14]3 years ago
5 0

Answer: Option A

P(A|B) = \frac{40}{49}

Step-by-step explanation:

In a probabilistic experiment, when two events A and B are dependent on each other, then the probability of occurrence A since B occurs is:

P(A|B) = \frac{P(A\ and\ B)}{P(B)}

Then if P(A\ and\ B) = \frac{5}{7} and P(B) = \frac{7}{8} then:

P(A|B) = \frac{\frac{5}{7}}{\frac{7}{8}}\\\\P(A|B) = \frac{40}{49}

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Given:

The scale factor between two circles is \dfrac{2x}{5y}.

To find:

The ratio of their areas.

Solution:

We know that, all circles are similar.

The ratio of the areas of similar figures is equal to the ratio of squares of their corresponding sides or equal to the square of ratio of their corresponding sides.

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Ratio of areas of circles = Square of scale factor between two circles

\text{Ratio of areas of circles}=\left(\dfrac{2x}{5y}\right)^2

\text{Ratio of areas of circles}=\dfrac{(2x)^2}{(5y)^2}

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3 years ago
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maria [59]
Answer: 116.93ft^2




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A=15.5•8=124

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What is n x 2 squared= 14
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Step-by-step explanation:

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y'=(t+y)^2-1

Substitute u=t+y, so that u'=y', and

u'=u^2-1

which is separable as

\dfrac{u'}{u^2-1}=1

Integrate both sides with respect to t. For the integral on the left, first split into partial fractions:

\dfrac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)=1

\displaystyle\int\frac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)\,\mathrm dt=\int\mathrm dt

\dfrac12(\ln|u-1|-\ln|u+1|)=t+C

Solve for u:

\dfrac12\ln\left|\dfrac{u-1}{u+1}\right|=t+C

\ln\left|1-\dfrac2{u+1}\right|=2t+C

1-\dfrac2{u+1}=e^{2t+C}=Ce^{2t}

\dfrac2{u+1}=1-Ce^{2t}

\dfrac{u+1}2=\dfrac1{1-Ce^{2t}}

u=\dfrac2{1-Ce^{2t}}-1

Replace u and solve for y:

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y=\dfrac2{1-Ce^{2t}}-1-t

Now use the given initial condition to solve for C:

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y=\dfrac2{1-\frac34e^{2t-6}}-1-t=\boxed{\dfrac8{4-3e^{2t-6}}-1-t}

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