17/64 + 47/64 + 7/8 = B. 1/16 134 151/64 1% D.

It takes 45 minutes or forty-five over sixty of an hour to cook a chicken. How many fourths of an hour is that?

lawyer [7]

Answer:3/4

Step-by-step explanation:60 minutes divided by 4 is 15. 3 15 minute increments is 45. 45 minutes is 3 /4 of an hour.

3 0

3 years ago

A bin contains seven red chips, nine green chips, three yellow chips, and six blue chips. Find the probability of drawing a yell

igomit [66]

Answer:

a) 3/100

Step-by-step explanation:

There are 25 total chips

P(Yellow) = 3/25

With no replacement P(Blue) = 6/24 = 1/4

Multiply 3/25* 1/4= 3/100

3 0

3 years ago

Please help plssssssss

sergey [27]

Answer:

.38

Step-by-step explanation:

Given that they have green eyes, we only look at the green row

Total is 3+5+5 = 13

Red hair is 5

P( red hair ) red / total = 5/13 =.384615385

8 0

3 years ago

Read 2 more answers

Hello! Any help is appreciated (:

lara [203]

Answer:

yes

Step-by-step explanation:

This relation is a function

Each input value goes to only one output value so the relation is a function

6 0

3 years ago

Read 2 more answers

The probability that a professor arrives on time is 0.8 and the probability that a student arrives on time is 0.6. Assuming thes

saul85 [17]

Answer:

a)0.08 , b)0.4 , C) i)0.84 , ii)0.56

Step-by-step explanation:

Given data

P(A) = professor arrives on time

P(A) = 0.8

P(B) = Student aarive on time

P(B) = 0.6

According to the question A & B are Independent

P(A∩B) = P(A) . P(B)

Therefore

${A}'$ & ${B}'$ is also independent

${A}'$ = 1-0.8 = 0.2

${B}'$ = 1-0.6 = 0.4

part a)

Probability of both student and the professor are late

P(A'∩B') = P(A') . P(B') (only for independent cases)

= 0.2 x 0.4

= 0.08

Part b)

The probability that the student is late given that the professor is on time

$P(\frac{B'}{A})$ = $\frac{P(B'\cap A)}{P(A)}$ = $\frac{0.4\times 0.8}{0.8}$ = 0.4

Part c)

Assume the events are not independent

Given Data

P $(\frac{{A}'}{{B}'})$ = 0.4

= $\frac{P({A}'\cap {B}')}{P({B}')}$ = 0.4

$P$ $({A}'\cap {B}')$ = 0.4 x P $({B}')$

= 0.4 x 0.4 = 0.16

$P({A}'\cap {B}')$ = 0.16

i)

The probability that at least one of them is on time

$P(A\cup B)$ = 1- $P({A}'\cap {B}')$

= 1 - 0.16 = 0.84

ii)The probability that they are both on time

P $(A\cap B)$ = 1 - $P({A}'\cup {B}')$ = 1 - $[P({A}')+P({B}') - P({A}'\cap {B}')]$

= 1 - [0.2+0.4-0.16] = 1-0.44 = 0.56

6 0

3 years ago

17/64 + 47/64 + 7/8 =B.1/16134151/641%D.​

17/64 + 47/64 + 7/8 =B.1/16134151/641%D.