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2 years ago

9

each month for 6 month, kelsey completes 5 paintings. how many more paintinga does she need to complete before she completed 38

painting?

1 answer:

eimsori [14]2 years ago

5 0

She needs to complete 8 more paintings. If you multiply 6 by 5, you get 30. Subtract 30 from 38, you would get 8.

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Christine Wong has asked Dave and Mike to help her move into a new apartment on Sunday morning. She has asked them both in case

olga nikolaevna [1]

Answer:

(a) The probability that both Dave and Mike will show up is 0.25.

(b) The probability that at least one of them will show up is 0.75.

(c) The probability that neither Dave nor Mike will show up is 0.25.

Step-by-step explanation:

Denote the events as follows:

<em>D</em> = Dave will show up.

<em>M</em> = Mike will show up.

Given:

$P(D^{c})=0.55\\P(M^{c})=0.45$

It is provided that the events of Dave of Mike showing up are independent of each other.

(a)

Compute the probability that both Dave and Mike will show up as follows:

$P(D\cap M)=P(D)\times P (M)\\=[1-P(D^{c})]\times [1-P(M^{c})]\\=[1-0.55]\times[1-0.45]\\=0.2475\\\approx0.25$

Thus, the probability that both Dave and Mike will show up is 0.25.

(b)

Compute the probability that at least one of them will show up as follows:

P (At least one of them will show up) = 1 - P (Neither will show up)

$=1-P(D^{c}\cup M^{c})\\=P(D\cup M)\\=P(D)+P(M)-P(D\cap M)\\=[1-P(D^{c})]+[1-P(M^{c})]-P(D\cap M)\\=[1-0.55]+[1-0.45]-0.25\\=0.75$

Thus, the probability that at least one of them will show up is 0.75.

(c)

Compute the probability that neither Dave nor Mike will show up as follows:

$P(D^{c}\cup M^{c})=1-P(D\cup M)\\=1-P(D)-P(M)+P(D\cap M)\\=1-[1-P(D^{c})]-[1-P(M^{c})]+P(D\cap M)\\=1-[1-0.55]-[1-0.45]+0.25\\=0.25$

Thus, the probability that neither Dave nor Mike will show up is 0.25.

6 0

3 years ago

Read 2 more answers

Please help me due tomorrow

Nostrana [21]

Answer:

y=3

Step-by-step explanation:

(i used point-slope form which is (y1-y2)=m(x1-x2))

(10-y)=7(4-3)

10-y=7

-y=-3

y=3

7 0

2 years ago

Steven conjectures that for |x|>5, it is true that x^3>125. Is his conjecture correct? Why or why not?

vredina [299]

Conjectures are conclusions formed from evidences. Steven's conjecture that: $x^3 > 125$ is true because: $x > 5$ or $x > -5$ for $|x| > 5$

Given that:

$|x| > 5$

Split the inequality as follows:

$x > 5$ or $x > -5$

Take the cube of both sides

$x^3 > 5^3$ or $x^3 > -5^3$

Evaluate all exponents

$x^3 > 125$ or $x^3 > -125$

Steven's conjecture that: $x^3 > 125$ is true because:

$x > 5$ or $x > -5$ for $|x| > 5$

Read more about absolute inequalities at:

brainly.com/question/4688732

5 0

3 years ago

20-5+15(25)=b+3+65(23)

vlada-n [284]

Answer:

-1108

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Jasper has 7 more checkers left than Karen does in a game of checkers. Jasper has 9 left. Write and solve and addition equation

hammer [34]

Answer:

Karen has only 2 checkers left

Step-by-step explanation:

Jasper has 7 checkers than Karen

Jasper has 9 checkers

The equation can be written as

7 + k = 9

This means that 7 plus Karen checkers equal 9 which is jasper checkers

It can also be written as

9 - 7 = k

k = 2

Therefore Karen has only 2 checkers left

3 0

3 years ago