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A teacher has found that the probability that a student studies for a test is 0.610.61, and the probability that a student gets

Ann [662]

Answer with step-by-step explanation:

Let

A=Student studies for a test

B=Student gets good grade on a test

The probability that a student studies for a test=P(A)=0.61

The probability that a student gets a good grade on a test=P(B)=0.79

The probability that both occur= $P(A\cap B)=0.56$

a.We have to find the events are independent

We know that if two events A and B are independent then

$P(A)\cdot P(B)=P(A\cap B)$

$P(A)\cdot P(B)=0.61\times 0.79=0.4819$

$P(A\cap B)\neq P(A)\cdot P(B)$

Hence, given events are not independent.

b.We have to find $P(B/A)$

$P(B/A)=\frac{P(A\cap B)}{P(A)}$

$P(B/A)=\frac{0.56}{0.61}=0.92$

c. We have to find $P(A/B)$

$P(A/B)=\frac{P(A\cap B}{P(B)}$

$P(A/B)=\frac{0.56}{0.79}=0.71$

7 0

3 years ago

Find the general term of sequence defined by these conditions.

disa [49]

Answer:

$\displaystyle a_{n} = (2)^{2n -1} - (3) ^{n-1 }$

Step-by-step explanation:

we want to figure out the general term of the following recurrence relation

$\displaystyle \rm a_{n + 2} - 7a_{n + 1} + 12a_n = 0 \: \: where : \: \:a_1 = 1 \: ,a_2 = 5,$

we are given a linear homogeneous recurrence relation which degree is 2. In order to find the general term ,we need to make it a characteristic equation i.e

${x}^{n} = c_{1} {x}^{n - 1} + c_{2} {x}^{n - 2} + c_{3} {x}^{n -3 } { \dots} + c_{k} {x}^{n - k}$

the steps for solving a linear homogeneous recurrence relation are as follows:

Create the characteristic equation by moving every term to the left-hand side, set equal to zero.
Solve the polynomial by factoring or the quadratic formula.
Determine the form for each solution: distinct roots, repeated roots, or complex roots.
Use initial conditions to find coefficients using systems of equations or matrices.

Step-1:Create the characteristic equation

${x}^{2} - 7x+ 12= 0$