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4 years ago

8

Sam buys tickets to a basketball game for $92.00 total. He must also pay 11% sales tax on the tickets. Write and simply a numeri

cal expression that represents the total cost.

1 answer:

Flauer [41]4 years ago

7 0

Sales tax 11%
11/100 x $92.00
=$10.12

Total cost = x
x= $92.00 + $ 10.12
= $102.12

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A store has a 28% off sale on hammers. Before the discount, the price of a hammer is $49. What is the sale price?

zheka24 [161]

28% of 49 is 13.72. So subtract that from 49 and you get 35.28. 35.28 is the price after the sale

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2 years ago

If i started at 11:40 A.M. And my elapsed time was 5 hours and 20 minutes. What is my end time?

kiruha [24]

Answer:

5:00 pm

Step-by-step explanation:

11:00 + 5 is 4.00 so you'd be a 4:40 + 20 minutes, that ranks up another hour so its at 5:00 pm

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3 years ago

Joe concluded that the expression 3x + 15 + 2x is equivalent to the expression 5(x + 3). Which of these best explains how Joe re

marta [7]

Answer:

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Step-by-step explanation:

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3 years ago

Use estimation to help you choose the correct answer.<br><br> 0.945 x 7.6 =

hoa [83]

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The thing you must not do is write 7.182 as your answer. That is the exact answer found using a calculator.

What to do.

What you should do is multiply two rounded numbers together.

7.6 rounded is 8 roughly

0.945 is almost 1.

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7 0

3 years ago

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