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3 years ago

6

Determine the effect on the balance sheet after the following transaction. You sell a product purchased for $500 for $1500. This

means

2 answers:

andrey2020 [161]3 years ago

6 0

You gained $1000 dollars from over pricing an item. Therefore you made profit which the sum was $1000.

Vlad1618 [11]3 years ago

5 0

You have made a profit of 300% (500=100% therefore 500*3=1500 =300%)
You have also made £1000 profit

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A class of 20 boys and 15 girls is divided into n groups so that each group has x boys and y girls. Find x, y and n. what values

melamori03 [73]

20 = 2 x 2 x 5
15 = 3 x 5

common divisor (CD) of 20 and 15 is 5 so n should be 5 and each group has 4 boys and 3 girls.

If there are more than one CD, then there are more than one answer.

If they ask for the largest number of groups, greatest common divisor (GCD) will be the answer.

4 0

3 years ago

If 1/2 equals .5 what does it equal as a percentage

Dmitriy789 [7]

To convert decimal over to percent, we move the decimal place two places to the right. So 0.5 becomes 50%.

6 0

3 years ago

Let A and B be events with =PA0.7, =PB0.3, and =PA or B0.9. (a) Compute PA and B. (b) Are A and B mutually exclusive? Explain. (

kvasek [131]

Answer:

a) $P(A \cup B) = P(A) +P(B) - P(A\cap B)$

And if we solve for $P(A \cap B)$ we got:

$P(A \cap B) = P(A) + P(B) -P(A\cup B)= 0.7+0.3-0.9 = 0.1$

b) False

The reason is because we don't satisfy the following relationship:

$P(A\cup B) = P(A) + P(B)$

We have that:

$0.9 \neq 0.3+0.7 =1$

c) False

In order to satisfy independence we need to have the following condition:

$P(A \cap B) = P(A) *P(B)$

And for this case we don't satisfy this relation since:

$0.1 \neq 0.7*0.3 = 0.21$

Step-by-step explanation:

For this case we have the following probabilities given:

$P(A) = 0.7, P(B) =0.7, P(A \cup B) =0.9$

Part a

We want to calculate the following probability: $P(A \cap B)$

And we can use the total probability rule given by:

$P(A \cup B) = P(A) +P(B) - P(A\cap B)$

And if we solve for $P(A \cap B)$ we got:

$P(A \cap B) = P(A) + P(B) -P(A\cup B)= 0.7+0.3-0.9 = 0.1$

Part b

False

The reason is because we don't satisfy the following relationship:

$P(A\cup B) = P(A) + P(B)$

We have that:

$0.9 \neq 0.3+0.7 =1$

Part c

False

In order to satisfy independence we need to have the following condition:

$P(A \cap B) = P(A) *P(B)$

And for this case we don't satisfy this relation since:

$0.1 \neq 0.7*0.3 = 0.21$

4 0

3 years ago

Abby earns $48 for babysitting for 6 hours . At that rate , how many hours will it take her to make $72

never [62]

Answer:

nine hours

Step-by-step explanation:

48 divided by 6 is 8 and 8 times 9 is 72

8 0

3 years ago

Read 2 more answers

Which table represents a quadratic relationship?

svp [43]

In each case, the x-values are equally-spaced. Thus looking at second differences will tell you if the relation is quadratic. If the second differences are non-zero and constant, then the values have a quadratic relationship.

A. First differences are 2-4 = -2, 1-2 = -1, 0.5-1 = -0.5. Second differences are -1-(-2) = 1, -0.5-(-1) = 0.5. Since 1 ≠ 0.5, this relation is not quadratic. (It is exponential with a base of 1/2.)

B. First differences are 128-135 = -7, 105-128 = -23, 72-105 = -33. Second differences are -23-(-7) = -16, -33-(-23)=-10. Since -16 ≠ -10, this relation is not quadratic. (It is cubic, since 3rd differences are constant at +4.)

C. First differences are -23.2-(-23.4) = 0.2, -23.0-(-23.2) = 0.2, -22.8-(-23.0) = 0.2. Second differences are zero, so this is not a quadratic relation. (It is linear, with a slope of 0.2.)

D. First differences are 56-90 = -34, 26-56 = -30, 0-26 = -26. Second differences are -30-(-34) = 4, -26-(-30) = 4. These are constant (=4), so the relation is quadratic.

The appropriate choice is ...

... D. x -1 0 1 2 3 4

... f(x) 90 56 26 0 -22 -40

4 0

4 years ago

Read 2 more answers