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2 years ago

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A retailer needs to purchase 11 printers. The first printer costs $58, and each additional printer costs 5% less than the price

of the previous printer, up to 15 printers. What is the total cost of 11 printers?
$381.99
$500.19
$606.10
$609.00

2 answers:

valkas [14]2 years ago

8 0

The total cost of 11 printers is $500.19.

Given that,

A retailer needs to purchase 11 printers.

The first printer costs $58, and each additional printer costs 5% less than the price of the previous printer, up to 15 printers.

We have to determine,

What is the total cost of 11 printers?

According to the question,

The first printer costs $58.

Each additional printer costs 5% less than the price of the previous printer, up to 15 printers.

The cost of a second printer is,

= 0.95 × 58 = $55.1

The cost of a third printer is,

= 0.95 × 55.1 = $52.345

The cost of a fourth printer is,

= 0.95 × 52.345 = $49.727

The cost of a fifth printer is,

= 0.95 × 49.727 = $47.241

The cost of a sixth printer is,

= 0.95 × 47.241 = $44.879

The cost of a seventh printer is,

= 0.95 × 44.879 = $42.635

The cost of an eight printer is,

= 0.95 × 42.635 = $40.50

The cost of a nineth printer is,

= 0.95 × 40.50 = $38.478

The cost of tenth printer is,

= 0.95 × 38.478 = $36.554

The cost of the eleventh printer is,

= 0.95 × 36.554 = $34.726

Therefore,

The total cost of 11 printers is,

= $58 + $55.1 + $52.345 + $49.727 + $47.241 + $44.879 + $42.635 + $40.50 + $38.478 + $36.554 + $34.726 = $500.19.

Hence, The total cost of 11 printers is $500.19.

For more details refer to the link given below.

brainly.com/question/16048559

svp [43]2 years ago

7 0

Answer:

$500.19 is the answer

Step-by-step explanation:

pls vote branliest

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Answer:

$1 \to 22 \to 0.176$

$2 \to 13 \to 0.104$

$3 \to 18 \to 0.144$

$4 \to 29 \to 0.232$

$5 \to 37 \to 0.296$

$6 \to 6 \to 0.048$

Step-by-step explanation:

Given

$n = 125$

See attachment for proper table

Required

Complete the table

Experimental probability is calculated as:

$Pr = \frac{Frequency}{n}$

We use the above formula when the frequency is known.

For result of roll 2, 4 and 6

The frequencies are 13, 29 and 6, respectively

So, we have:

$Pr(2) = \frac{13}{125} = 0.104$

$Pr(4) = \frac{29}{125} = 0.232$

$Pr(6) = \frac{6}{125} = 0.048$

When the frequency is to be calculated, we use:

$Pr = \frac{Frequency}{n}$

$Frequency = n * Pr$

For result of roll 3 and 5

The probabilities are 0.144 and 0.296, respectively

So, we have:

$Frequency(3) = 125 * 0.144 = 18$

$Frequency(5) = 125 * 0.296 = 37$

For roll of 1 where the frequency and the probability are not known, we use:

$Total \ Frequency = 125$

So:

Frequency(1) added to others must equal 125

This gives:

$Frequency(1) + 13 + 18 + 29 + 37 + 6 = 125$

$Frequency(1) + 103 = 125$

Collect like terms

$Frequency(1) =- 103 + 125$

$Frequency(1) =22$

The probability is then calculated as:

$Pr(1) = \frac{22}{125}$

$Pr(1) = 0.176$

So, the complete table is:

$1 \to 22 \to 0.176$

$2 \to 13 \to 0.104$

$3 \to 18 \to 0.144$

$4 \to 29 \to 0.232$

$5 \to 37 \to 0.296$

$6 \to 6 \to 0.048$

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