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3 years ago

13

Scott works as a real-estate agent and earns 8% commission. What is his straight commission on $198,000 worth of total sales?

1 answer:

bogdanovich [222]3 years ago

5 0

Answer:

15,840

Step-by-step explanation:

8% of 198,000 is 15,840

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Which values of a, b, and c represent the answer in simplest form?<br> 7/9 / 4/9

satela [25.4K]

1 3/4 the answer is

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3 years ago

Read 2 more answers

I need help with this :( will mark brainliest

JulsSmile [24]

a. 20 because y is the amount of $$ in account now

b. 12 a $$

c. 260, bc 12*20= 240 240+20=260

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3 years ago

The math club needs to raise more than $552. $.50 for a trip to state competition. The club has raise 12% of the funds which is

ratelena [41]

Each of the members must raise at least $69.46.

To start they need to raise $552.50. However, they have already raised 12% of it. They only need to raise 88% of it.

552.50 x 0.88 = 486.2

Divide the remaining amount by 7 to determine how much each individual must raise.

486.2 / 7 = 69.46

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3 years ago

Alice searches for her term paper in her filing cabinet, which has several drawers. She knows thatshe left her term paper in dra

katen-ka-za [31]

You made a mistake with the probability $p_{j}$ , which should be $p_{i}$ in the last expression, so to be clear I will state the expression again.

So we want to solve the following:

Conditioned on this event, show that the probability that her paper is in drawer $j$ , is given by:

(1) $\frac{p_{j} }{1-d_{i}p_{i} } , if j \neq i,$ and

(2) $\frac{p_{i} (1-d_{i} )}{1-d_{i}p_{i} } , if j = i.$

so we can say:

$A$ is the event that you search drawer $i$ and find nothing,

$B$ is the event that you search drawer $i$ and find the paper,

$C_{k}$ is the event that the paper is in drawer $k, k = 1, ..., n.$

this gives us:

$P(B) = P(B \cap C_{i} ) = P(C_{i})P(B | C_{i} ) = d_{i} p_{i}$

$P(A) = 1 - P(B) = 1 - d_{i} p_{i}$

Solution to Part (1):

if $j \neq i$ , then $P(A \cap C_{j} ) = P(C_{j} )$ ,

this means that

$P(C_{j} |A) = \frac{P(A \cap C_{j})}{P(A)} = \frac{P(C_{j} )}{P(A)} = \frac{p_{j} }{1-d_{i}p_{i} }$

as needed so part one is solved.

Solution to Part(2):

so we have now that if $j$ = $i$ , we get that:

$P(C_{j}|A ) = \frac{P(A \cap C_{j})}{P(A)}$

remember that:

$P(A|C_{j} ) = \frac{P(A \cap C_{j})}{P(C_{j})}$

this implies that:

$P(A \cap C_{j}) = P(C_{j}) \cdot P(A|C_{j}) = p_{i} (1-d_{i} )$

so we just need to combine the above relations to get:

$P(C_{j}|A) = \frac{p_{i} (1-d_{i} )}{1-d_{i}p_{i} }$

as needed so part two is solved.

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4 years ago

Larry has 4-cent stamps and 9-cent stamps, which he can combine to produce various amounts of postage. For example, he can make

Tju [1.3M]

Answer:

the largest number that cannot be made is 23
see below for an explanation

Step-by-step explanation:

<em>Why numbers greater than 23 can be made</em>

From any number that Larry can make that includes an appropriate number of stamps, Larry can ...

add 1 by adding 1×9 and taking away 2×4
add 2 by adding 2×9 and taking away 4×4
add 3 by adding 3×9 and taking away 6×4
add 4 by adding 1×4
add 5 by adding 4 and 1 (add 9, take away 4)
add 6 by adding 4 and 2 (add 2×9, take away 3×4)
add 7 by adding 4 and 3 (add 3×9, take away 5×4)
add 8 by adding 2×4
add 9 by adding 1×9

Clearly, for Larry to be able to continue adding any number, he must start with a number that contains at least 6×4 cents = 24 cents.

Larry can make 24 cents from 6 4-cent stamps. Using the above table, he can add any amount to that.

___

<em>Why 23 cannot be made</em>

For 23 cents, the number of stamps must include 0, 1, or 2 9-cent stamps, so there are 3 cases to try for making 23:

23 cannot be made from 0 9-cent stamps because 23 is not divisible by 4
23 cannot be made using 1 9-cent stamp, because 14 is not divisible by 4
23 cannot be made using 2 9-cent stamps, because 5 is not divisible by 4.

Therefore, 23 cannot be made.

7 0

4 years ago