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3 years ago

10

Use the distributive property to remove the parentheses. -6(4u-2y-3)

1 answer:

PilotLPTM [1.2K]3 years ago

4 0

Using the distributive property it would be -24u12y18

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spencer took 21$ with him to the mall. At the music store, he spent 2/3 of his money on a DVD, he spent 1/4 of what he had left

Sunny_sXe [5.5K]

Answer:

$5.25 Dollars

Step-by-step explanation:

I know that if you divide 21 by 3 you get 7. So, 2/3 of 21 is 14. He spent 14 dollars at the music store. This means that he has 7 dollars left. From here he spend 1/4 of 7 dollars on a drink. 1/4 of 7 is $1.75. This means that he has 5.25 dollars left after he bought the drink! :)

<em>Hope this helped! :)</em>

6 0

3 years ago

Find two consecutive integers whose sum is the square of 7

Maksim231197 [3]

Answer:

Step-by-step explanation:

Two consecutive integers be x and (x+1)

x + x +1 = 7²

2x + 1 = 49

2x = 49 - 1

2x = 48

x = 48/2

x = 24

The consecutive integers are: 24,25

3 0

3 years ago

Answer all please? tysm

8_murik_8 [283]

1.A 2.D 3.B These are the answers I think is the choice to your question.

6 0

2 years ago

Read 2 more answers

how far would the cheetah have traveled in the first eight minutes? show how you arrived at your answer

Sav [38]

How far would the cheetah have traveled in the first eight minutes? show how you arrived at your answer

8 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

2 years ago