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3 years ago

5

WILL MARK BRAINLIETS! Claire has 300 fruits to sell.She sells 75 fruits each day for 3 days.How many fruits does she have left t

o sell after 3 days?

2 answers:

oee [108]3 years ago

8 0

Answer:

75

Step-by-step explanation:

75*3=225

300-225=75 remaining

KengaRu [80]3 years ago

7 0

Answer:

75

Step-by-step explanation:

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Brenda+said+that+5/2+divided+by+1/4+equals+10.+how+do+you+know+without+dividing+whether+brenda's+statement+is+reasonable?+justif

Lapatulllka [165]

<span>Brenda would be correct, another way to find the answer would be to use the keep change flip method. Where you keep 5/2 change division to multiplication then flip 4/1. The equation would then be 5/2 * 4/1 then you just multiply straight across 5*4 = 20 and 2*1 equals 2. You end up with 20/2 which reduces to 10</span>

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3 years ago

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

Translate this phrase into an algebraic expression.

Alekssandra [29.7K]

4/x+10

(4 divided by x + 10)

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3 years ago

Write the following equation in general form Ax+By+C=0. 1/2y-1/3x-1=0

Oksanka [162]

Answer:

1/2 y - 1/3 x -1 = 0

Multiply through by the LCD -6 to clear the fractions and the negative coefficient with the 'x'.

-3y + 2x + 6 = 0

Then reorder

2x - 3y + 6 = 0

Step-by-step explanation:

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3 years ago

50 US states for the namethat start with the letter W what percent of US states have the names that start with the letter W

Vanyuwa [196]

Washington, West Virginia, Wisconsin, Wyoming
Four states out of 50
4 / 50 * 100 = 8 per cent.

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3 years ago