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Zepler [3.9K]
4 years ago
15

What is 1 divided by 1/5

Mathematics
2 answers:
77julia77 [94]4 years ago
7 0
1/ 1/5
1/1÷1/5
1/1×5/1
5/1=
5
LenaWriter [7]4 years ago
5 0
First there's 1 divided by 1 which happens to be 1 , Second there's 1 divided by 5 which is 5 so 1 divided by 1/5 is equal to

= 5
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grigory [225]

please refer to this image

please don't report this answer if it is wrong

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5 0
3 years ago
Freezer A measures 1 ft times 1 ft times 5 ft and sells for ​$300. Freezer B measures 1.5 ft times 1.5 ft times 4 ft and sells f
Sladkaya [172]

Answer:

Freezer A is better option as price per cubic foot is less for Freezer A

Step-by-step explanation:

Data provided in the question:

Measure of Freezer A = 1 ft × 1 ft × 5 ft

Measure of Freezer B = 1.5 ft × 1.5 ft × 4 ft

Price of Freezer  A = $300

Price of Freezer  B = $600

Now,

Volume of Freezer A = 1 ft × 1 ft × 5 ft

= 5 ft³

Volume of Freezer B = 1.5 ft × 1.5 ft × 4 ft

= 9 ft³

Now,

Price per cubic foot for Freezer A = $300 ÷ 5 ft³

= $60/ft³

Price per cubic foot for Freezer B = $600 ÷ 9 ft³

= $66.67/ft³

Hence,

Freezer A is better option as price per cubic foot is less for Freezer A

7 0
3 years ago
A manufacturing company regularly conducts quality control checks at specified periods on the products it manufactures. Historic
Rina8888 [55]

Answer:

The probability that none of the LED light bulbs are​ defective is 0.7374.

Step-by-step explanation:

The complete question is:

What is the probability that none of the LED light bulbs are​ defective?

Solution:

Let the random variable <em>X</em> represent the number of defective LED light bulbs.

The probability of a LED light bulb being defective is, P (X) = <em>p</em> = 0.03.

A random sample of <em>n</em> = 10 LED light bulbs is selected.

The event of a specific LED light bulb being defective is independent of the other bulbs.

The random variable <em>X</em> thus follows a Binomial distribution with parameters <em>n</em> = 10 and <em>p</em> = 0.03.

The probability mass function of <em>X</em> is:

P(X=x)={10\choose x}(0.03)^{x}(1-0.03)^{10-x};\ x=0,1,2,3...

Compute the probability that none of the LED light bulbs are​ defective as follows:

P(X=0)={10\choose 0}(0.03)^{0}(1-0.03)^{10-0}

                =1\times 1\times 0.737424\\=0.737424\\\approx 0.7374

Thus, the probability that none of the LED light bulbs are​ defective is 0.7374.

6 0
3 years ago
Are the lines y=-2x and 2x+y=3. Are they paralell, perpendicular, or neither?
ELEN [110]

Answer:

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7 0
3 years ago
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Which is better to buy 12 bars of soap for $10.00 or 5 bars of soap for $4.00 is better which one
Makovka662 [10]

Answer:

5 bars for $4.00

Step-by-step explanation:

10 / 12 = 0.833 repeating

4 / 5 = 0.8

3 0
3 years ago
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