What is the equivalent of 4 meters?

pictured below are stacks of solid cubes. determine the number of cubes in the stack and the number of faces glued together.

Iteru [2.4K]

Solution

For this case we need to find the number of cubes in the stack given so then we can do this:

The first line present 5 cubes

The second line present 5 cubes

And the last line we got:

10 cubes

Then the total of cubes are:

10 +5+5 = 20 cubes

If we need to find the number of faces we can do the following:

5+ 4+ 4+4 +4 +1 + (5+4+4+4+4+1 ) + (5+4+4+4+4+1 ) + (5+ 4+4+4 +4 )

And solving we got:

22+22+ 22+21= 87

5 0

1 year ago

There are 5 cards with a picture of a rose and 1 card with a picture of a daisy.

Lostsunrise [7]

Answer:

the correct answer is 1/5

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

PLEASE HELP!!

QveST [7]

Just add up all the numbers

7 0

3 years ago

Two equal sized plots of land were planted with corn. The first plot reproduced 3,500 bushels of corn.The science plot produced

madreJ [45]

3500:4200 reduce to 35:42 see if you can reduce anymore. you can to 5:6

7 0

3 years ago

Read 2 more answers

For many years businesses have struggled with the rising cost of health care. But recently, the increases have slowed due to les

kaheart [24]

Answer:

The confidence interval would be given by this formula

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

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

The margin of error for this case is given by:

$ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

And replacing we got:

$ME = 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.0259$

And replacing into the confidence interval formula we got:

$0.52 - 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.4941$

$0.52 + 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.5459$

And the 95% confidence interval would be given (0.4941;0.5459).

Step-by-step explanation:

Data given and notation

n=1000 represent the random sample taken

$\hat p=0.52$ estimated proportion of of U.S. employers were likely to require higher employee contributions for health care coverage

$\alpha=0.05$ represent the significance level (no given, but is assumed)

Solution to the problem

The confidence interval would be given by this formula

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

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

The margin of error for this case is given by:

$ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

And replacing we got:

$ME = 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.0259$

And replacing into the confidence interval formula we got:

$0.52 - 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.4941$

$0.52 + 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.5459$

And the 95% confidence interval would be given (0.4941;0.5459).

5 0

3 years ago