Identify the class width for the given frequency distribution

A computer consulting firm presently has bids out on three projects. Let Ai = {awarded project i}, for i = 1, 2, 3, and suppose

Karo-lina-s [1.5K]

Step-by-step explanation:

The data below is what was provided in the question and it is what I solved the question with

P(A1) = 0.23

P(A2) = 0.25

P(A3) = 0.29

P(A1 n A2 ) = 0.09

P(A1 n A3) = 0.11

P(A2 n A3) = 0.07

P(A1 n A2 n A3) = 0.02

a

P(A2|A1) = P(A1 n A2)/P(A1)

= 0.09/0.23

= 0.3913

We have 39.13% confidence that event A2 will occur given that event A1 already occured

b.)

P(A3 n A3|A1) = P(A2 n A3)n A1)/P(A1)

= 0.02/0.23

= 0.08695

We have about 8.7% chance of events A2 and A3 occuring given that A1 already occured.

C.

P(A2 u A3|A1)

= P(A1 n A2)u(A1 n A3)/P(A1)

= P( A1 n A2) + P(A1 n A3) - P(A1 n A2 n A3) / P(A1)

= (0.09+0.11-0.02)/0.23

= 0.18/0.23

= 0.7826

We have 78.26% chance of A2 or A3 happening given that A1 has already occured.

6 0

3 years ago

What is the expression written using each base only once? 48 x 43 O A 411 O B. 1211 O C. 424 O D. 6411

Gemiola [76]

Answer:

D i think

Step-by-step explanation:

I think 4 is the same thing, but you multiply by the powered numbers.

4 0

3 years ago

What is the minimum of y=1/3 x^2 + 2x + 5

lawyer [7]

Answer:

min at x = -3

Step-by-step explanation:

steps are in the pic above.

5 0

3 years ago

(c) what is the probability that at least one adult prefers milk chocolate to dark chocolate?

Schach [20]

Part A:

Given that 47% of adults prefer milk chocolate to dark chocolate, for a random sample of n = 4 adults, the probability that all four adults say that they prefer milk chocolate to dark chocolate is given by

$P(4)=(0.47)^4=0.049$

Part B:

The probability that exactly two of the four adults say they prefer milk chocolate to dark chocolate is given by

$P(2)={ ^4C_2(0.47)^2(1-0.47)^2} \\ \\ =6(0.2209)(0.2809)=0.372$

Part C:

The probability that at least one adult prefers milk chocolate to dark chocolate is given by

$P(at \ least \ 1)=P(X\geq1) \\ \\ =P(1)+P(2)+P(3)+P(4) \\ \\ =1-P(0)=1-{ ^4C_0(0.47)^0(1-0.47)^4} \\ \\ =1-(0.53)^4=1-0.0789=0.921$

6 0

3 years ago

I have no clue what the answer is. Please help ASAP

ivanzaharov [21]

The answer should be B, but I might be wrong.

4 0

3 years ago

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