When creating equivalent fractions (creating common denominators) you only need to change the denominators.

Which situation is best modeled with a division expression?

Solnce55 [7]

Answer:

Any situation which requires splitting up or separation of terms and items amongst certain groups of people. Division is useful when simplifying fractions and in word problems involving probability, or sharing of items, or proportioning.

4 0

3 years ago

Organisms A and B start out with the same population size. Organism A’s population doubles every day. After five days, the popul

Alinara [238K]

Answer:

1. 252N = ( $2^{2}$ x $3^{2}$ x $7^{1}$ )N

2. Factor = $2^{5}$

Step-by-step explanation:

Let the starting population size for organisms A and B be represented by N.

For organism A;

Day 0: N

Day 1: 2N

Day 2: $2^{2}$ N

Day 3: $2^{3}$ N

Day 4: $2^{4}$ N

Day 5: $2^{5}$ N

Day 6: $2^{4}$ N

Day 7: $2^{3}$ N

Day 8: $2^{2}$ N

For organism B;

Day 0: N

Day 1: 2N

Day 2: $2^{2}$ N

Day 3: $2^{3}$ N

Day 4: $2^{4}$ N

Day 5: $2^{5}$ N

Day 6: $2^{6}$ N

Day 7: $2^{7}$ N

Day 8: $2^{8}$ N

1. After the eight days, organism B's population is larger than organism A's population by;

$2^{8}$ N - $2^{2}$ N = 256N - 4N

= 252N

= ( $2^{2}$ x $3^{2}$ x $7^{1}$ )N

2. Organism A's population grows by $2^{5}$ .

7 0

3 years ago

Which of the following is NOT an example of mixed fraction?

Anika [276]

Answer:

C

Step-by-step explanation:

a mixed fraction consists of a whole number and a proper fraction

a

$\frac{29}{15}$ = 1 $\frac{14}{15}$ ← a mixed number

b

$\frac{35}{12}$ = 2 $\frac{11}{12}$ ← a mixed number

c

$\frac{12}{13}$ ← cannot be expressed as a mixed number

6 0

2 years ago

A 3rd degree binomial with a constant term of 8

rewona [7]

Answer:

-5x^3 + 8

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

A factory received a shipment of 11 hammers, and the vendor who sold the items knows there are 2 hammers in the shipment that ar

zhenek [66]

Question:

If a sample of 2 hammer is selected

(a) find the probability that all in the sample are defective.

(b) find the probability that none in the sample are defective.

Answer:

a $Pr = \frac{2}{110}$

b $Pr = \frac{72}{110}$

Step-by-step explanation:

Given

$n = 11$ --- hammers

$r = 2$ --- selection

This will be treated as selection without replacement. So, 1 will be subtracted from subsequent probabilities

Solving (a): Probability that both selection are defective.

For two selections, the probability that all are defective is:

$Pr = P(D) * P(D)$

$Pr = \frac{2}{11} * \frac{2-1}{11-1}$

$Pr = \frac{2}{11} * \frac{1}{10}$

$Pr = \frac{2}{110}$

Solving (b): Probability that none are defective.

The probability that a selection is not defective is:

$P(D') = \frac{9}{11}$

For two selections, the probability that all are not defective is:

$Pr = P(D') * P(D')$

$Pr = \frac{9}{11} * \frac{9-1}{11-1}$

$Pr = \frac{9}{11} * \frac{8}{10}$

$Pr = \frac{72}{110}$

8 0

3 years ago