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2 years ago

Mr. Smith saved $15 by buying a tool at a 10% discount. What was the original price for this tool? Please show work.

Mathematics

2 answers:

Mnenie [13.5K]2 years ago

8 0

Well 10%= 1/10 of 100% so you would Multiply 15$ by 10 to get the original price. E.G 15*10=150$

Olegator [25]2 years ago

7 0

Well we know that 10% is going to equal 1/10 of 100%.
So now just multipy 15% times 10 to get the original price.
Your answer is, $150.

Hope this help's!
:D:D:D:D:D:D:D

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2 years ago

2 Rita bought three and forty-eight hundredths pounds of bananas at the store. How is this number written in expanded notation?

boyakko [2]

The cost of bananas = $ 3.48

This can be written as :

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2 years ago

John is planning to purchase a new car that costs $15,500. On average, a new car loses 11% of its value the moment that it is dr

Levart [38]

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4 0

1 year ago

Read 2 more answers

A disadvantage of the contention approach for LANs, such as CSMA/CD, is the capacity wasted due to multiple stations attempting

sammy [17]

Answer:

The overview of the given problem is outlined in the following segment on the explanation.

Step-by-step explanation:

The proportion of slots or positions that have been missed due to numerous concurrent transmission incidents can be estimated as follows:

Checking a probability of transmitting becomes "p".

After considering two or even more attempts, we get

Slot fraction wasted,

= $[1-no \ attempt \ probability-first \ attempt \ probability-second \ attempt \ probability+...]$

On putting the values, we get

= $1-no \ attempt \ probability-[N\times P\times probability \ of \ attempts]$

= $1-(1-P)^{N}-N[P(1-P)^{N}]$

So that the above seems to be the right answer.

8 0

2 years ago

Out of six computer chips, two are defective. If two chips are randomly chosen for testing (without replacement), compute the pr

Ratling [72]

Answer:

The probability that of the two chips selected both are defective is 0.1089.

Step-by-step explanation:

Let X = number of defective chips.

It is provided that there are 2 defective chips among 6 chips.

The probability of selecting a defective chip is:

$P(X)=p=\frac{2}{6}=0.33$

A sample of n = 2 chips are selected.

The random variable X follows a Binomial distribution with parameter n = 2 and p = 0.33.

The probability function of a Binomial distribution is:

$P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0, 1, 2, ...$

Compute the probability that of the two chips selected both are defective as follows:

$P(X=2)={2\choose 2}(0.33)^{2}(1-0.33)^{2-2}=1\times 0.1089\times 1=0.1089$

Thus, the probability that of the two chips selected both are defective is 0.1089.

The sample space of selecting two chips is:

S = (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)

(2, 1), (2, 3), (2, 4), (2, 5), (2, 6)

(3, 1), (3, 2), (3, 4), (3, 5), (3, 6)

(4, 1), (4, 2), (4, 3), (4, 5), (4, 6)

(5, 1), (5, 2), (5, 3), (5, 4), (5, 6)

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5)

3 0

1 year ago