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Maurinko [17]
3 years ago
12

How can you determine which two whole numbers a mixed will be located on a number on a number line

Mathematics
1 answer:
Bad White [126]3 years ago
5 0
I'm not exactly sure how to answer your question but i'll do my best. It is basically adding two numbers. If you have the problem 2+3 you will get five, so you count 5 spaces from zero on the number line. There is no such thing as adding two whole numbers and getting a mixed number on a number line. You have to add a mixed number+ a fraction to get a mixed number, and in some cases add a mixed number+ a mixed number to get a mixed number.
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How can you make 0.40 into a fraction??
melamori03 [73]
<h3>2/5 is your answer. </h3><h3>▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓</h3>

☞ First, 0.40 is the same as 40/100.

But we are not done.

<u>Now, doing this you need to find the GCF (Greatest common factor) of 40, 100.</u>

2 x 2 x 5 = 20.

☞ So our GCF is 20.

Divide both the numerator and the denominator by the GCF.

40 ÷ 20 / 100 ÷ 20.

<h2>Simplify to get ☞ 2/5.</h2><h2>▬▬ι══════════════ι▬▬</h2>

Best of luck on your assignment!

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- Emacathy | Top Answerer ♫♪.ılılıll|̲̅̅●̲̅̅|̲̅̅=̲̅̅|̲̅̅●̲̅̅|llılılı.♫♪

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8 0
3 years ago
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4 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}

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3 years ago
Estimate the awnser to 17 x 193
8_murik_8 [283]
I think i know this one. it's 3,281.
8 0
3 years ago
Read 2 more answers
Brian rolls a fair dice 12 times. How many times would Brian expect to roll a number greater than 4?
igor_vitrenko [27]

Answer:

Numbers greater than 4 are 5 and 6

So we have 2 numbers out of 6 numbers so the probability is equal to 2/6 = 1/3

And 1/3 of 12 is equal to 4

So Brian expect to roll a number greater than 4,

Four times.

5 0
3 years ago
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