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3 years ago

What's the simplest from for 15/72

Mathematics

1 answer:

pychu [463]3 years ago

4 0

Divide them both by three and you will get your answer, which is 5/24

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Aliun [14]

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3 years ago

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At the gift shop, they sell small greeting cards and large greeting cards. The cost of a small greeting card is $1.60 and the co

erik [133]

<h2>Explanations:</h2>

From the given question, we have the following parameters

Cost of a small greeting card = $1.60

The cost of a large greeting card = $4.05.

Cost of 5 small greeting cards = 5(1.60)

Cost of 5 small greeting cards = $8.00

Cost of 4 large greeting cards = 4(4.05)

Cost of 4 large greeting cards = $16.2

Cost of 5 small and 4 large = $8.00 + $16.2

Cost of 5 small and 4 large $24.2

Hence it would cost $24.2 to get 5 small greeting cards and 4 large greeting cards

Similarly;

Cost of x small greeting cards = 1.60x

Cost of y large greeting cards = 4.05y

Cost of small greeting cards and y large greeting cards = 1.60x + 4.05y

Hence it would cost $(1.60x + 4.05y) to get x small greeting cards and y large greeting cards

5 0

1 year ago

What is the best way to Solve: 25 – 5x = 80

Art [367]

25 - 5x = 80
+ 5x + 5x

25 = 5x + 80
- 80 -80

-55 = 5x

-55/5 = 11

x = 11

Hope I helped!

Let me know if you need anything else!

~ Zoe

5 0

3 years ago

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The Wall Street Journal reported that automobile crashes cost the United States $162 billion annually (2008 data). The average c

ivanzaharov [21]

Answer:

$n=(\frac{2.326(500)}{120})^2 =93.928 \approx 94$

So the answer for this case would be n=94 rounded up to the nearest integer

Step-by-step explanation:

Information given

$\bar X$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma= 500$ represent the population standard deviation

n represent the sample size

Solution to the problem

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{s}{\sqrt{n}}$ (a)

And on this case we have that ME =120 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

The confidence2 level is 98% or 0.98 then the significance level would be $\alpha=1-0.98=0.02$ and $\alpha/2=0.01$ , the critical value for this case would be $z_{\alpha/2}=2.326$ , replacing into formula (b) we got:

$n=(\frac{2.326(500)}{120})^2 =93.928 \approx 94$

So the answer for this case would be n=94 rounded up to the nearest integer

8 0

3 years ago

I need help asap!!!!!!!!!!!!!

Tatiana [17]

Answer:

Step-by-step explanation:

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3 years ago