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3 years ago

If you know the answer to these plz answer.

Mathematics

1 answer:

kondaur [170]3 years ago

7 0

Answer:

Step-by-step explanation:

You can readily see from the diagram, above, that the side length of the middle cube will be between 3 and 4. You want to determine to the nearest hundredth what value between 3 and 4 represents the side length of the cube whose volume is 45 units^3.

Please note: the middle cube has been mislabeled. Instead of volume = 30 units^3, the volume should be 45 units^3.

Here's the procedure:

Guess an appropriate s value. Let's try s = side length = 3.5

Cube this: (3.5 units)^3 = 42.875. Too small. Choose a larger possible side length, such as 3.7: 3.7^3 = 50.653. Too large.

Try s = 3.6: 3.6^3 = 46.66. Too large.

Choose a smaller s, one between 3.5 and 3.6: 3.55^3 = 44.73. This is the best estimate yet for s. Continue this work just a little further. Try s = 3.57. Cube it. How close is the result to 45 cubic units?

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I've been having struggles with math recently and would greatly appreciate help

BARSIC [14]

ANSWER:

$-4x^3+4x^2-4x+2$

STEP-BY-STEP EXPLANATION:

We have the following expression:

$(4x^3+6x^2-9x+1)-(8x^3+2x^2-5x-1)$

In order to operate, we must first break the parentheses and then group similar terms, like this:

$\begin{gathered} 4x^3+6x^2-9x+1-8x^3-2x^2+5x+1 \\ (4x^3-8x^3)+(6x^2-2x^2)+(-9x+5x)+(1+1) \\ \text{ therefore:} \\ -4x^3+4x^2-4x+2 \\ (4x^3+6x^2-9x+1)-(8x^3+2x^2-5x-1)=-4x^3+4x^2-4x+2 \end{gathered}$

6 0

1 year ago

Calculate the sample standard deviation and sample variance for the following frequency distribution of hourly wages for a sampl

lidiya [134]

Answer:

(a) The sample variance is 16.51

(a) The sample standard deviation is 4.06

Step-by-step explanation:

Given

$\begin{array}{cc}{Class} & {Frequency} & 8.26 - 10.00 & 20 &10.01-11.75 & 38 &11.76 - 13.50& 36 & 13.51-15.25 &25&15.26-17.00 &27 &\ \end{array}$

Solving (a); The sample variance.

First, calculate the class midpoints.

This is the mean of the intervals.

i.e.

$x_1 = \frac{8.26+10.00}{2} = \frac{18.26}{2} = 9.13$

$x_2 = \frac{10.01+11.75}{2} = \frac{21.76}{2} = 10.88$

$x_3 = \frac{11.76+13.50}{2} = \frac{25.26}{2} = 12.63$

$x_4 = \frac{13.51+15.25}{2} = \frac{28.76}{2} = 14.38$

$x_5 = \frac{15.26+17.00}{2} = \frac{32.26}{2} = 16.13$

So, the table becomes:

$\begin{array}{ccc}{Class} & {Frequency} & {x} & 8.26 - 10.00 & 20&9.13 &10.01-11.75 & 38 &10.88&11.76 - 13.50& 36 &12.63& 13.51-15.25 &25&14.38&15.26-17.00 &27 &16.13\ \end{array}$

Next, calculate the mean

$\bar x = \frac{\sum fx}{\sum f}$

$\bar x = \frac{20*9.13 + 38 * 10.88+36*12.63+25*14.38+27*16.13}{20+38+36+25+27}$

$\bar x = \frac{1845.73}{146}$

$\bar x = 12.64$

Next, the sample variance is:

$\sigma^2 = \frac{\sum f(x - \bar x)^2}{\sum f - 1}$

So, we have:

$\sigma^2 = \frac{20*(9.13-12.63)^2 + 38 * (10.88-12.63)^2 +...........+27 * (16.13 -12.63)^2}{20+38+36+25+27-1}$

$\sigma^2 = \frac{2393.6875}{145}$

$\sigma^2 = 16.51$

The sample standard deviation is:

$\sigma = \sqrt{\sigma^2}$

$\sigma = \sqrt{16.51}$

$\sigma = 4.06$

6 0

3 years ago

You roll 2 dice what is the probability that the sum of the dice is greater than 8 and that one die shows a six

MAXImum [283]

It'll be rare to do it

7 0

3 years ago

Simplify the following expression by combining like terms.<br> 3x + 4x² - 7x+ 9x2

EleoNora [17]

Answer:

13x² - 4x

Step-by-step explanation:

3x + 4x² - 7x + 9x² ← collect like terms

= (4x² + 9x² ) + (3x - 7x)

= 13x² + (- 4x)

= 13x² - 4x

7 0

2 years ago

I have no idea what I'm supposed to do for this question please explain

maks197457 [2]

For part B
the fixed cost is 500 dollars because it is a set inital amount, the varibale cost is .15 because it is the cost dependent on something else (per indicated variable part)

y=.15x+500
.15 cents per flyer, add 500 because that is the set amount it costs to even create the flyer

Plug in 500 for the equation
y=.15(500)+500
y=75+500
y=575

for part a b=ka for direct variations so the direct one is C - two items directly multiplied y=mx+b for partial variation so its A,D - basically the equation of a line A is neither - shows neither of these relationships

5 0

3 years ago