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3 years ago

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PLEASE HELP Consider eight, eight-cubic centimeter (8 cm3) sugar cubes stacked so that they form a single 2 x 2 x 2 cube. How do

es the surface area of the single, large cube compare to the total surface area of the individual eight cubes? Report your answer as a ratio. Be sure to show all calculations leading to an answer.

1 answer:

victus00 [196]3 years ago

5 0

$8=2*2*2\\8=2*2*2\\2*2*2*2*2*2=4*4*4\\s_{large}=4\\s_{small}=2\\\\SA_{large}=6s^2\\SA_{large}=6(4)^2\\SA_{large}=6(16)\\SA_{large}=96\\\\SA_{small}=8s^3\\SA_{small}=8(2)^3\\SA_{small}=8(8)\\SA_{small}=64\\\\96:64\\\frac{96}{32}:\frac{64}{32}\\3:2$

3 : 2

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Find the other endpoint of the line segment with the given endpoint and midpoint.

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Answer:

(-2,12)

Step-by-step explanation:

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3 years ago

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Tensile-strength tests were carried out on two different grades of wire rod. Grade 1 has 10 observations yielding a sample mean

LuckyWell [14K]

Answer:

$t = \frac{1085-1034}{\sqrt{\frac{52^2}{10} +\frac{61^2}{15}}} = 2.240$

$df = n_1 +n_2 -2 = 10+15-2= 23$

$p_v = 2*P(t_{23} >2.240) = 0.035$

Since the p value is lower than the significance level we have enough evidence to conclude that the true means are different at 5% of significance

Step-by-step explanation:

Data given

$\bar X_1 = 1085$ sample mean for group 1

$\bar X_2 = 1034$ sample mean for group 2

$n_1 = 10$ sample size for group 1

$n_2 = 15$ sample size for group 2

$s_1 = 52$ sample deviation for group 1

$s_2 = 61$ sample deviation for group 2

Solution

We want to check if the two means are equal so then the system of hypothesis are:

Null hypothesis: $\mu_1= \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

And the statistic is given by:

$t = \frac{\bar X_1 -\bar X_2}{\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}$

And replacing we got:

$t = \frac{1085-1034}{\sqrt{\frac{52^2}{10} +\frac{61^2}{15}}} = 2.240$

The degrees of freedom are given by:

$df = n_1 +n_2 -2 = 10+15-2= 23$

And the p value would be:

$p_v = 2*P(t_{23} >2.240) = 0.035$

Since the p value is lower than the significance level we have enough evidence to conclude that the true means are different at 5% of significance

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4 years ago

Please Help! 30 POINTS! Given the following three points, find by hand the quadratic function they represent. (−1,−8), (0,−1),(1

ycow [4]

Answer:

(1) B

(2) D

Step-by-step explanation:

(1)

Let the quadratic function be:

$y = ax^{2} + bx + c$

For the point, (0,-1),

$y = ax^{2} + bx + c$

$-1=(a\times0)+(b\times0}+c\\-1=c\\c=-1$

Then the equation is:

$y = ax^{2} + bx -1$

For the point (-1, -8) ,

$y = ax^{2} + bx -1$

$-8=(a\times (-1)^{2})+(b\times -1)-1\\-8=a-b-1\\a-b=-7...(i)$

For the point (1, 2) ,

$y = ax^{2} + bx -1$

$2=(a\times (1)^{2})+(b\times 1)-1\\2=a+b-1\\a+b=3...(ii)$

Add the two equations and solve for <em>a</em> as follows:

$a-b=-7\\a+b=3\\\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\2a = -4\\a = -2$

Substitute <em>a</em> = -2 in (i) and solve for <em>b</em> as follows:

$a-b=-7\\-2-b=-7\\b=5$

Thus, the quadratic function is:

$f(x)=-2x^{2}+5x-1$

The correct option is (b).

(2)

The ordered pairs are:

(5, 7), (7, 11), (9, 14), (11, 18)

Represent them in an XY table as follows:

X : 5 | 7 | 9 | 11

Y : 7 | 11 | 14 | 18

Compute the difference between the <em>Y</em> values as follows:

Diff = 11 - 7 = 4

Diff = 14 - 11 = 3

Diff = 18 - 14 = 4

Now compute the difference between the Diff values:

d = 3 - 4 = -1

d = 4 - 3 = 1

Since the differences between the differences of the y-values is not consistent, the ordered pairs do not represent a quadratic equation.

The correct option is D.

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3 years ago

Prove that the cube root of 2 is irrational.

harina [27]

The cube root of 2 is irrational. The proof that the square root of 2 is irrational may be used, with only slight modification. Assume the cube root of 2 is rational. Then, it may be written as a/b, where a and b are integers with no common factors. (This is possible for all nonzero rational numbers). Since a/b is the cube root of 2, its cube must equal 2. That is, (a/b)3 = 2 a3/b3 = 2 a3 = 2b3. The right side is even, so the left side must be even also, thatis, a3 is even. Since a3 is even, a is also even (because the cube of an odd number is always odd). Since a is even, there exists an integer c such that a = 2c. Now, (2c)3 = 2b3 8c3 = 2b3 4c3 = b3. The left side is now even, so the right side must be even too. The product of two odd numbers is always odd, so b3 cannot be odd; it must be even. Therefore b is even as well. Since a and b are both even, the fraction a/b is not in lowest terms, thus contradicting our initial assumption. Since the initial assumption cannot have been true, it must <span>be false, and the cube root of 2 is irrational.
</span>

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4 years ago

o-na [289]

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3 years ago