All categories

3 years ago

If 1 16 oz can cost. $1 how much is 8 40oz cans

Mathematics

1 answer:

Studentka2010 [4]3 years ago

3 0

Answer:

8 I think

Step-by-step explanation:

I did 40÷16=2.5 then 2.5×8=20

You might be interested in

Helpppp lol pleaseeeeee

makkiz [27]

Answer: i thinks it's D. none of the above because they are all angles?

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Need alot of help please!!!

kirza4 [7]

A=x
B= 2x
C=3x
If you add them up =
6x =180
÷6 both sides
x=30
A=30
B=60
C=90
one angle is 90 degrees so the triangle is classified as a right triangle

8 0

3 years ago

Read 2 more answers

Find the Volume. Can someone explain how they did it. thank you

hodyreva [135]

6 x 5 x 7 = 210
The volume formula is length x width x height :)

4 0

3 years ago

Read 2 more answers

2) At the store Julie bought a shirt for $23.97, which was 25% off the original price. Which of these should she use to find x,

Scrat [10]

Answer:

0.75x = 23.97

Step-by-step explanation:

The total is 23.97. 0.25% is off so you have to subtracting 25% from 100% and you will get 75% if you convert it to decimal you will get 0.75.

5 0

3 years ago

A researcher wishes to be 95% confident that her estimate of the true proportion of individuals who travel overseas is within 3%

andre [41]

Answer: a) 683 b) 1067

Step-by-step explanation:

The confidence interval for population proportion is given by :-

$p\pm z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}$

a) Given : Significance level : $\alpha=1-0.95=0.05$

Critical value : $z_{\alpha/2}}=\pm1.96$

Margin of error : $E=0.03$

Formula to calculate the sample size needed for interval estimate of population proportion :-

$n=p(1-p)(\dfrac{z_{\alpha/2}}{E})^2\\\\=0.2(0.8)(\dfrac{1.96}{0.03})^2=682.951111111\approx683$

Hence, the required sample size would be 683 .

b) If no estimate of the sample proportion is available then the formula to calculate sample size will be :-

$n=0.25(\dfrac{z_{\alpha/2}}{E})^2\\\\=0.25(\dfrac{1.96}{0.03})^2=1067.11111111\approx1067$

Hence, the required sample size would be 1067 .

3 0

3 years ago