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Amiraneli [1.4K]

2 years ago

13

Find the area of the shaded

1 answer:

spayn [35]2 years ago

3 0

Answer:

area = 84 in²

Step-by-step explanation:

area = (9x12) - (6x8x0.5) = 84 in²

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The thickness of each type of coin is shown in the table. How much thicker is stack of dollars worth of nickels than a dollars w

Lilit [14]

Let our basis be worth 1 dollar. A nickel's worth is $0.05. In order to come up with $1, the number of nickels should be:

Number of nickels = $1 * 1 nickel/$0.05 = 20 nickels
Thickness of 20 nickels = 20 nickels * 1.95 mm = 39 mm

Let's do the same for the quarters. Each quarter is worth $0.25.

Number of quarters = $1 * 1 quarter/$0.25 = 4 quarters
Thickness of 4 quarters = 4 quarters * 1.75 mm = 7 mm

Find the ratio of the two:
39 mm/7 mm = 5.57

Therefore, a stack of nickels is 5.57 times thicker than a stack of quarters worth one dollar.

7 0

3 years ago

Perform the indicated operation. Express answer as a simplified fraction (You may have

timofeeve [1]

Step-by-step explanation:

$4 + \frac{2}{3} = \frac{3 \times 4 + 2}{3} = \frac{14}{3} = 4 \frac{2}{3}$

$\frac{67}{71} \times \frac{381}{399} = \frac{67 \times 381}{71 \times 399} = \frac{25527}{28329} = \frac{8509}{9443}$

$\frac{684}{795} \div \frac{24}{37} = \frac{684}{795} \times \frac{37}{24} = \frac{25308}{19080} = 1.33$

6 0

3 years ago

A shipment of 11 printers contains 2 that are defective. Find the probability that a sample of size 2, drawn from the 11, will

svet-max [94.6K]

The required probability is $\frac{36}{55}$

<u>Solution:</u>

Given, a shipment of 11 printers contains 2 that are defective.

We have to find the probability that a sample of size 2, drawn from the 11, will not contain a defective printer.

Now, we know that, $\text { probability }=\frac{\text { favourable outcomes }}{\text { total outcomes }}$

Probability for first draw to be non-defective $=\frac{11-2}{11}=\frac{9}{11}$

(total printers = 11; total defective printers = 2)

Probability for second draw to be non defective $=\frac{10-2}{10}=\frac{8}{10}=\frac{4}{5}$

(printers after first slot = 10; total defective printers = 2)

Then, total probability $=\frac{9}{11} \times \frac{4}{5}=\frac{36}{55}$

7 0

3 years ago

A store had 5 packs of paper for $7.80. How much would it cost if you were to buy 3 packs

valentina_108 [34]

To buy 3 packs, the cost would be $4.68

4 0

2 years ago

Please help me with these questions

Jlenok [28]

Posted 5 days ago no need for an answer now.

3 0

3 years ago