Can you help me with 10-13 please? I don’t really understand it

Jim's work evaluating 2 (three-fifths) cubed is shown below. 2 (three-fifths) cubed = 2 (StartFraction 3 cubed Over 5 EndFractio

sammy [17]

Answer:

Jim's error is " He did not multiply Three-fifths by 2 before applying the power "

Step-by-step explanation:

Jim's evaluating expression is $2(\frac{3}{5})^3$

To verify Jim's error :

Jim's steps are

$2(\frac{3}{5})^3$

$=2(\frac{3^3}{5})$

$=2(\frac{3\times 3\times 3}{5})$

$=2(\frac{27}{5})$

$=\frac{54}{5}$

Therefore $2(\frac{3}{5})^3=\frac{54}{5}$

Jim's error is " He did not multiply Three-fifths by 2 before applying the power "

That is the corrected steps are

$2(\frac{3}{5})^3$

$=2(\frac{3^3}{5^3})$ ( using the property $(\frac{a}{b})^m=\frac{a^m}{b^m}$ )

$=2(\frac{3\times 3\times 3}{5\times 5\times 5})$

$=2(\frac{27}{125})$

$=\frac{54}{125}$

$2(\frac{3}{5})^3=\frac{54}{125}$

6 0

3 years ago

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The Ball Corporation's beverage can manufacturing plant in Fort Atkinson, Wisconsin, uses a metal supplier that provides metal w

vampirchik [111]

Answer:

The 95 percent confidence interval for the true mean metal thickness is between 0.2903 mm and 0.2907 mm

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96\frac{0.000586}{\sqrt{59}} = 0.0002$

The lower end of the interval is the sample mean subtracted by M. So it is 0.2905 - 0.0002 = 0.2903 mm

The upper end of the interval is the sample mean added to M. So it is 0.2905 + 0.0002 = 0.2907 mm

The 95 percent confidence interval for the true mean metal thickness is between 0.2903 mm and 0.2907 mm

4 0

3 years ago

What is the equation of the line parallel to the given line with an x-intercept of 4? y = x +

Nady [450]

Answer:

Step-by-step explanation:

The equation of the line parallel to

y = x

has a slope of m = 1

This means that the family of lines with the same slope are of the form

y = x - (a)

where -a represents the y intercept, and (a) the x- intercept

With a = 4, the equation intercepts the x axis at x = 4

Please see attached graph

8 0

3 years ago

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Find the circumference of the circle and show work

zaharov [31]

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$\large\blue\textsf{\textbf{\underline{\underline{Question:-}}}}$

Find the circumference of the circle and show work. [refer to

attachment]

$\large\blue\textsf{\textbf{\underline{\underline{Answer and How to Solve:-}}}}$

Formula for Circumference of a Circle:-

$\dashrightarrow\star\bigstar\boxed{\boxed{C=2\pi r}}}$

Where

C = circumference
π = pi (3.14…)
r = radius, which in this case is equal to 8

Substitute the value of r and solve:-

C=2π(8)

On simplification,

C=16π

On further simplification,

C=50.3

Therefore, we conclude that the circumference of this circle is

$\dashrightarrow\star\bigstar\boxed{\boxed{\underline{\bold{C=50.3\:yd}}}}\bigstar\star\dashleftarrow$

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7 0

2 years ago

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The volume of a rectangular prism is (x^3 – 3x^2 + 5x – 3), and the area of its base is (x^2 – 2). If the volume of a rectangula

bixtya [17]

Answer:

The height of the prism is equal to $h=(x^{3}-3x^{2}+5x-3)/(x^{2}-2)$

Step-by-step explanation:

we know that

The volume of a rectangular prism is equal to

$V=Bh$

where

B is the area of the base of the prism

h is the height of the prism

In this problem we have

$V=x^{3}-3x^{2}+5x-3$

$B=x^{2}-2$

substitute in the formula and solve for h

$x^{3}-3x^{2}+5x-3=(x^{2}-2)h$

$h=(x^{3}-3x^{2}+5x-3)/(x^{2}-2)$

3 0

4 years ago