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2 years ago

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The volume of a cube is given by V = x3, where x is the length of an edge of the cube. The area of a square is given by A = x2,

where x is the length of a side of the square. A given cube has a volume of 1,728 cubic inches.

1 answer:

ki77a [65]2 years ago

8 0

X = 12 and A = 144 (If V = 1,728)

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Answer:

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given - 2x < 10 ( divide both sides by - 2 )

Remembering that the symbol is reversed when multiplying/ dividing by a negative quantity.

x > - 5 ← note reversal of symbol

solution is {x | x > - 5 } → C

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2 years ago

What is the result of a dilation of scale factor 3 centered at the origin of the line 2y + 3x=10?? PLEASE HELP PLEASEEEEEEEEE

maks197457 [2]

Given:

The equation of a line is:

$2y+3x=10$

The line is dilated by factor 3.

To find:

The result of dilation.

Solution:

The equation of a line is:

$2y+3x=10$

For $x=0$ ,

$2y+3(0)=10$

$2y+0=10$

$y=\dfrac{10}{2}$

$y=5$

For $x=2$ ,

$2y+3(2)=10$

$2y+6=10$

$2y=10-6$

$2y=4$

Divide both sides by 2.

$y=\dfrac{4}{2}$

$y=2$

The given line passes through the two points A(0,5) and B(2,2).

If the line dilated by factor 3 with origin as center of dilation, then

$(x,y)\to (3x,3y)$

Using this rule, we get

$A(0,5)\to A'(3(0),3(5))$

$A(0,5)\to A'(0,15)$

Similarly,

$B(2,2)\to B'(3(2),3(2))$

$B(2,2)\to B'(6,6)$

The dilated line passes through the points A'(0,15) and B'(6,6). So, the equation of dilated line is:

$y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)$

$y-15=\dfrac{6-15}{6-0}(x-0)$

$y-15=\dfrac{-9}{6}(x)$

$y-15=\dfrac{-3}{2}x$

Multiply both sides by 2.

$2(y-15)=-3x$

$2y-30=-3x$

$2y+3x=30$

Therefore, the equation of the line after the dilation is $2y+3x=30$ .

3 0

2 years ago

Which expression will help find the volume of the figure?

Alex Ar [27]

Answer:

9*4*1*2

Step-by-step explanation:

9×4×1×2

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3 years ago

To evaluate the effect of a treatment, a sample is obtained from a population, with a mean of u = 30, and the treatment is admin

Murrr4er [49]

Answer:

a) $t=\frac{31.3-30}{\frac{3}{\sqrt{16}}}=1.733$

$p_v =2*P(t_{15}>1.733)=0.104$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the we don't have a significant effect for the new treatment at 5% of significance.

b) $t=\frac{31.3-30}{\frac{3}{\sqrt{36}}}=2.6$

$p_v =2*P(t_{15}>2.6)=0.0201$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the we have a significant effect for the new treatment at 5% of significance.

c) When we increase the sample size we increse the probability of rejection of the null hypothesis since the z score tend to increase when the sample size increase.

Step-by-step explanation:

Data given and notation

Part a: If the sample consists of n=16 individuals, are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with alpha = 0.05.

$\bar X=31.3$ represent the sample mean

$s=3$ represent the sample standard deviation

$n=16$ sample size

$\mu_o =30$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is different from 30, the system of hypothesis are :

Null hypothesis: $\mu = 30$

Alternative hypothesis: $\mu \neq 30$

Since we don't know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{31.3-30}{\frac{3}{\sqrt{16}}}=1.733$

P-value

First w eneed to find the degrees of freedom given by:

$df=n-1=16-1 =15$

Since is a two-sided test the p value would given by:

$p_v =2*P(t_{15}>1.733)=0.104$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the we don't have a significant effect for the new treatment at 5% of significance.

Part b: If the sample consists of n=36 individuals, are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with alpha = 0.05.

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{31.3-30}{\frac{3}{\sqrt{36}}}=2.6$

P-value

First w eneed to find the degrees of freedom given by:

$df=n-1=16-1 =15$

Since is a two-sided test the p value would given by:

$p_v =2*P(t_{15}>2.6)=0.0201$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the we have a significant effect for the new treatment at 5% of significance.

Part c; Comparing your answer for parts a and b, how does the size of the sample influence the outcome of a hypothesis test

When we increase the sample size we increse the probability of rejection of the null hypothesis since the z score tend to increase when the sample size increase.

5 0

3 years ago