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kenny6666 [7]
4 years ago
5

For two events and , the probability that occurs is 0.8, the probability that occurs is 0.4, and the probability that both occur

s is 0.2. Given that occurred, what is the probability that also occurred?
Mathematics
1 answer:
sergey [27]4 years ago
7 0

Answer:

P(B|A)=0.25  , P(A|B) =0.5

Step-by-step explanation:

The question provides the following data:

P(A)= 0.8

P(B)= 0.4

P(A∩B) = 0.2

Since the question does not mention which of the conditional probabilities need to be found out, I will show the working to calculate both of them.

To calculate the probability that event B will occur given that A has already occurred (P(B|A) is read as the probability of event B given A) can be calculated as:

P(B|A) = P(A∩B)/P(A)

      = (0.2) / (0.8)  

P(B|A)=0.25

To calculate the probability that event A will occur given that B has already occurred (P(A|B) is read as the probability of event A given B) can be calculated as:

P(A|B) = P(A∩B)/P(B)

          = (0.2)/(0.4)

P(A|B) =0.5

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Answer:

A. Product quality is not affected by size.

Step-by-step explanation:

This statement is true because the quality of a product is not a measure of its size, which is a quantifiable factor. This means that we can have a very small-sized product which has very good quality. Size also depends on the product being manufactured. Quality is rather measured by how good the materials used in the production are.

If inferior materials are used in the production, then the product, no matter how small or big they are, would lack quality. To obtain a good and accurate measurement, a lot of time and energy should be invested.

5 0
3 years ago
Find an equation of the tangent to the curve x =5+lnt, y=t2+5 at the point (5,6) by both eliminating the parameter and without e
svet-max [94.6K]

ANSWER

y = 2x -4

EXPLANATION

Part a)

Eliminating the parameter:

The parametric equation is

x = 5 +  ln(t)

y =  {t}^{2}  + 5

From the first equation we make t the subject to get;

x - 5 =  ln(t)

t =  {e}^{x - 5}

We put it into the second equation.

y =  { ({e}^{x - 5}) }^{2}  + 5

y =  { ({e}^{2(x - 5)}) }  + 5

We differentiate to get;

\frac{dy}{dx}  = 2 {e}^{2(x - 5)}

At x=5,

\frac{dy}{dx}  = 2 {e}^{2(5 - 5)}

\frac{dy}{dx}  = 2 {e}^{0}  = 2

The slope of the tangent is 2.

The equation of the tangent through

(5,6) is given by

y-y_1=m(x-x_1)

y - 6 = 2(x - 5)

y = 2x - 10 + 6

y = 2x -4

Without eliminating the parameter,

\frac{dy}{dx}  =  \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} }

\frac{dy}{dx}  =  \frac{ 2t}{  \frac{1}{t} }

\frac{dy}{dx}  =  2 {t}^{2}

At x=5,

5 = 5 +  ln(t)

ln(t)  = 0

t =  {e}^{0}  = 1

This implies that,

\frac{dy}{dx}  =  2 {(1)}^{2}  = 2

The slope of the tangent is 2.

The equation of the tangent through

(5,6) is given by

y-y_1=m(x-x_1)

y - 6 = 2(x - 5) =

y = 2x -4

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3 years ago
1) x 20+ 5 = 4
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