Determine whether each question is biased. Explain your answer. please help!!

The surface area of the similar cone can be determined by multiplying the surface area of the cone by 3. Because it is given that the similar cone is three times large.

Thus, we have;

$SA = 3 \times 54$

$SA=162$

Thus, the surface area of the similar cone is 162 square inches.

6 0

3 years ago

Here is the histogram of a data distribution. What is the shape of this distribution.

tankabanditka [31]

Answer:

Hey there!

I think your answer would be unimodal skewed. This graph only has one maxima, thus it can't be bimodal. However, it's not symmetric, meaning that it is skewed.

Hope this helps :)

6 0

3 years ago

The circumference of a sphere was measured to be 80 cm with a possible error of 0.5 cm. A) Use differentials to estimate the max

siniylev [52]

Answer:

A) The maximum error in the calculated surface area: $25cm^2$

Relative error: $0.013$

B) The maximum error in the calculated volume: $162cm^2$

Relative error: $0.019$

Step-by-step explanation:

A) The formula for the surface area is:

$A=4\pi r^2$

The measured value is the circumference which is equal to:

$C=2\pi r$

then the radius is:

$r=\frac{C}{2\pi}$

Substituting in the formula of the surface:

$A=4\pi(\frac{C}{2\pi})^2\\A=4\pi(\frac{C^2}{4\pi^2})\\A=\frac{C^2}{\pi}$

Using the formula to calculate the error:

$dy=f'(x)dx$

Where $x$ is the variable measured and $y$ is a function of $x$ ( $y=f(x)$ ).

$dA=f'(C)dC\\dA=\frac{2C^{(2-1)}}{\pi}dC\\dA=\frac{2C}{\pi}dC$

We have C=80cm and dC=0.5cm

$dA=\frac{2C}{\pi}dC\\dA=\frac{2(80)}{\pi}(0.5)\\dA=\frac{160}{\pi}(0.5)\\dA=50.9296(0.5)\\dA=25.4648\approx25cm^2$

The relative error is the maximum error divide by the total area. The total area is: $A=\frac{C^2}{\pi}=\frac{(80)^2}{\pi}=\frac{6400}{\pi}=2037.1833cm^2$

$\frac{dA}{A}=\frac{25.4648}{2037.1833} =0.0125\approx0.013$

B) The formula for the volume is:

$V=\frac{4}{3} \pi r^3$

Using $r=\frac{C}{2\pi}$

$V=\frac{4}{3} \pi r^3\\V=\frac{4}{3} \pi (\frac{C}{2\pi})^3\\V=\frac{4}{3} \pi (\frac{C^3}{8\pi^3})\\V=\frac{1}{3}(\frac{C^3}{2\pi^2})\\V=\frac{C^3}{6\pi^2}$

The maximum error is:

$dV=\frac{3C^{3-1}}{6\pi^2}dC\\dV=\frac{C^{2}}{2\pi^2}dC\\dV=\frac{(80)^{2}}{2\pi^2}(0.5)\\dV=\frac{6400}{2\pi^2}(0.5)\\dV=\frac{6400}{2\pi^2}(0.5)\\dV=(324.2278)(0.5)\\dV=162.1139\approx162cm^2$

The calculated volume is:

$V=\frac{C^3}{6\pi^2}\\V=\frac{(80)^3}{6\pi^2}\\V=\frac{512000}{6\pi^2}\\V=8646.0743$

The relative error is:

$\frac{dV}{V}=\frac{162.1139}{8646.0743}=0.0188\approx0.019$

3 0

2 years ago