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3 years ago

6

A regular square pyramid just fits inside a cube (the base of the pyramid is congruent to a face of the cube and

1 answer:

zlopas [31]3 years ago

4 0

Volume of reg sq. pyramid = vp
vp = 1/3×s^2×h, where s = side and h = height
Volume of cone = vc =1/3×h×pi×r^2
Now we know that h is the same for both, and the cones diameter = s of square base, so radius (r) = 1/2 s
so now vc = 1/3×h×pi×(1/2s)^2
let's remove the same items for both vc and vp
so 1/3 and h
now let's plug an arbitrary number into each:
vc = pi (10/2)^2 = 3.14×25 = 78.54
vp = s^2 = 10^2 = 100
So any square pyramid has slightly more volume than the cone

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Ten kids line up for recess. There are five boys and five girls. How many ways are there for the 10 kids to line up so that all

Nana76 [90]

Answer:

14400

Step-by-step explanation:

We have 2 groups of 5 elements each

there is a condition all boys will be ahead always, therefore in the group of boys there are

5! = 5*4*3*2*1 = 120 differents form

And the girls at the same time can change theirs places according to

5! = 120

Then total ways for the line up in the conditions of the statement of the problem are

T = 120*120

14400

3 0

3 years ago

The grid you see below is in the shape of angle What is the area, in square units of the shaded part?

kramer

Answer:

Around 7.5

Since the shaded region is very slightly less than half, we can estimate that the answer is somewhere around or exactly 7.5 since it is half of 15.

3 0

2 years ago

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almond37 [142]

1.) no
2.)no
3.)no
4.)no
5.)no
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5 0

3 years ago

This is the second one that i have no clue how to do lol so pls helppp

LekaFEV [45]

Answer:

1.25

Step-by-step explanation:

5 0

3 years ago

A normal curve with a mean of 500 and a standard deviation of 100 is shown. Shade the region under the curve within one standard

N76 [4]

Answer:

And for this case we want to find how much of the data within the shaded region given in the figure attached.

And we can use the zscore formula given by:

$z = \frac{X -\mu}{\sigma}$

And if we find the z score for the limits we got:

$z = \frac{400-500}{100}=-1$

$z = \frac{600-500}{100}= 1$

And we want to find this probability using the normal standard distribution or excel:

$P(-1$

So we would expect about 68.2 % of the data within one deviation from the mean

Step-by-step explanation:

For this case we know that the variable of interest let's say x follows a normal distribution with the following parameters:

$X \sim N(\mu = 500, \sigma =100)$

And for this case we want to find how much of the data within the shaded region given in the figure attached.

And we can use the zscore formula given by:

$z = \frac{X -\mu}{\sigma}$

And if we find the z score for the limits we got:

$z = \frac{400-500}{100}=-1$

$z = \frac{600-500}{100}= 1$

And we want to find this probability using the normal standard distribution or excel:

$P(-1$

So we would expect about 68.2 % of the data within one deviation from the mean

6 0

4 years ago