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2 years ago

Is 60% greater than 5/9

Mathematics

1 answer:

Aleks04 [339]2 years ago

8 0

60% = 60/100 = 6/10 = 54/90

5/9 = 50/90

so its false because 60% smaller than 5/9

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Sergio039 [100]

Answer:

Number 3) 8x + 11 = 315

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3 years ago

Read 2 more answers

Choose all the numbers that are common multiples of 3 and 12.

spayn [35]

<h3>Answers: 48 and 72</h3>

=========================================================

Explanation:

The number 12 is a multiple of 3 because 3*4 = 12.

So when looking for common multiples of 3 and 12, we simply need to look at multiples of 12.

The multiples of 12 are:

12, 24, 36, 48, 60, 72, 84, 96, 120, ...

We see that 48 and 72 are on the list. The values 21, 27, 63, 81 are not on the list, so cross them out.

Now we could keep that list of multiples going to see if 844 is on there or not. A better method is to divide 844 over 12. If we get a whole number, then it's a multiple of 12.

844/12 = 70.333 approximately.

This shows that 844 is <u>not</u> a multiple of 12. So we cross 844 from the list.

Only 48 and 72 are multiples of 12 (and also multiples of 3).

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2 years ago

What is log b^b^6x equivalent to (The b is under log and ^b)? How do you know?

Arturiano [62]

Answer: 6x

Work Shown:

For each step, the logs are all base b. This is to save time and hassle of writing tricky notation of having to write the smaller subscript 'b' multiple times. The first rule to use is that log(x^y) = y*log(x) for any base of a logarithm. The second rule is that $\log_b(b) = 1$ meaning that the log base of itself is 1

log(b^(6x)) = 6x*log(b) .... pull down exponent using the first rule above

log(b^(6x)) = 6x*1 .... use the second rule mentioned

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3 years ago

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2 years ago

The daily revenues of a cafe near the university are approximately normally distributed. The owner recently collected a random s

lbvjy [14]

Answer:

The sample size to obtain the desired margin of error is 160.

Step-by-step explanation:

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Rearranging this equation in terms of n gives

$n=\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2$

Now the Margin of Error is reduced by 2 so the new M_2 is given as M/2 so the value of n_2 is calculated as

$n_2=\left[z_{crit}\times \dfrac{\sigma}{M_2}\right]^2\\n_2=\left[z_{crit}\times \dfrac{\sigma}{M/2}\right]^2\\n_2=\left[z_{crit}\times \dfrac{2\sigma}{M}\right]^2\\n_2=2^2\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2\\n_2=4\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2\\n_2=4n$

As n is given as 40 so the new sample size is given as

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So the sample size to obtain the desired margin of error is 160.

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3 years ago