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2 years ago

What is the volume, in cubic inches, of one cube with an edge length of inch?

Mathematics

1 answer:

Alex_Xolod [135]2 years ago

7 0

Answer:

We know that each cube with a -inch edge length has a volume of cubic inch, because . Since the prism is built using 24 of these cubes, its volume, in cubic inches, would then be , or 3 cubic inches.

Step-by-step explanation:

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4 0

3 years ago

PLEASE HELP ASAP!!

Elza [17]

Answer:

The parabolic equation does not intersect the linear equation.

Explanation:

Given equation's:

y = x² − 5x + 9
y = -2x

To solve , use substitution method by inserting 2nd equation into 1st

$\sf y = x^2 - 5x + 9$

when substituted y = -2x

$\sf -2x = x^2 - 5x + 9$

collect terms

$\sf x^2 - 5x+ 2x + 9=0$

simplify

$\sf x^2 - 3x + 9=0$

Find discriminant:

$\sf b^2 - 4ac$

$\sf (-3)^2 - 4(1)(9)$

$\sf -27$

As the discriminant is negative, it tell us that they both does not intersect each other.

7 0

1 year ago

How do you work these out with explanation please x

bixtya [17]

1. 2 (x + 3) = x - 4
2x + 6 = x- 4
2x - x = -4 - 6
x = -4 + -6 = -10.
x = 10
2. 4 (5x - 2) = 2 (9x + 3)
20x - 8 = 18x + 3
20x - 18x = 3 + 8
2x = 11
x = 11/2 = 5.5
x = 11/2 or 5.5
These are the Answers.
Thank You!
Please mark me the Brainliest.

4 0

3 years ago

Read 2 more answers

Assume that the helium porosity of coal samples taken from any particular seam is Normally distributed with true standard deviat

riadik2000 [5.3K]

Answer:

a) $4.85-2.33\frac{0.75}{\sqrt{20}}=4.46$

$4.85+2.33\frac{0.75}{\sqrt{20}}=5.24$

b) $4.56-2.33\frac{0.75}{\sqrt{16}}=4.12$

$4.56+2.33\frac{0.75}{\sqrt{16}}=4.99$

c) $n=(\frac{1.960(0.75)}{0.2})^2 =54.02 \approx 55$

Step-by-step explanation:

Part a

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The Confidence is 0.98 or 98%, the value of $\alpha=0.02$ and $\alpha/2 =0.01$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.01,0,1)".And we see that $z_{\alpha/2}=2.33$

Now we have everything in order to replace into formula (1):

$4.85-2.33\frac{0.75}{\sqrt{20}}=4.46$

$4.85+2.33\frac{0.75}{\sqrt{20}}=5.24$

Part b

$4.56-2.33\frac{0.75}{\sqrt{16}}=4.12$

$4.56+2.33\frac{0.75}{\sqrt{16}}=4.99$

Part c

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

And on this case we have that ME =0.4/2 =0.2 we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got $z_{\alpha/2}=1.960$ , replacing into formula (b) we got:

$n=(\frac{1.960(0.75)}{0.2})^2 =54.02 \approx 55$

5 0

3 years ago

5/2 as a mixed numer

Lady_Fox [76]

5/2=2.5 So you can write this as a fraction. 2 5/10 and then simplify 2 1/2. I hope this helps. :) Brainliest answer please. :)

5 0

3 years ago

Read 2 more answers