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2 years ago

A full glass of water cam hold 1/6 of a bottle

2 answers:

5 0

3 1/3 divided by 1/6

3x3=9
9+1=10

So

10/3 divided by 1/6

10/3 x 6/1 = 60/3
Because
6x10=60
3x1=3

Simplify:
60/3 = 20

Taya2010 [7]2 years ago

4 0

Answer: 20

Explanation: 3 1/3 divided by 1/6 is 20.

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geniusboy [140]

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1 year ago

Paolo paid $ 28 $ 28 for a hat that was originally priced at $ 35 $ 35 . By what percent was the hat discounted?

devlian [24]

The original price of the hat is $35. Paolo paid $28 for the hat.

So discount for the hat is $(35-28) = $7

$7 is discounted from $35. We have to find the percentage of the discount.

To find the percentage we have to divide the discount amount by the original price and then have to multiply it by 100.

So the discount percentage =

((7/35)×100) %

We can simplify 7/35 by dividing 7 and 35 both by 7. So we will get 1/5 after simplifying.

((1/5) ×100) %

(100/5)% = 20%

So the hat was discounted by 20%.

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2 years ago

Change a Relation to Make It a Function The relation R is shown below as a list of ordered pairs. R = { (1, 4), (1, 3), (-1, 3),

nexus9112 [7]

Answer: (1,4) and (1,3) because the have the same x-value

Step-by-step explanation:

A function is defined so that for each input (known as the domain), there is no more than one output (known as the range).

8 0

1 year ago

Matt is the catcher for his school's baseball team. The catcher must be able to throw from home plate to second base. What is th

Oduvanchick [21]

Answer:

Step-by-step explanation:

127.3

4 0

2 years ago

Read 2 more answers

How many students must be randomly selected to estimate the mean weekly earnings of students at one college? We want 95% confide

kari74 [83]

Answer:

The sample of students required to estimate the mean weekly earnings of students at one college is of size, 3458.

Step-by-step explanation:

The (1 - α)% confidence interval for population mean (μ) is:

$CI=\bar x\pm z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}$

The margin of error of a (1 - α)% confidence interval for population mean (μ) is:

$MOE=z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}$

The information provided is:

σ = $60

MOE = $2

The critical value of z for 95% confidence level is:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

Compute the sample size as follows:

$MOE=z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}$

$n=[\frac{z_{\alpha/2}\times \sigma }{MOE}]^{2}$

$=[\frac{1.96\times 60}{2}]^{2}$

$=3457.44\\\approx 3458$

Thus, the sample of students required to estimate the mean weekly earnings of students at one college is of size, 3458.

8 0

2 years ago