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brilliants [131]

3 years ago

9

HELP PLS -- A square is rotated around the y-axis of coordinate plane. Which of the following three-dimensional figures is forme

d? a. cilinder b.circle c.sphere d. cube

2 answers:

Doss [256]3 years ago

7 0

<h3>Answer: A. Cylinder</h3>

The square must be on the y axis in some way, and the square must be oriented such that its sides are parallel to the x and y axis. Rotating said square around the y axis will form a cylinder. Think of a revolving door. The door is the square while it sweeps out a 3D cylinder shape/space as it rotates around. Another example would be a fan or turbine that rotates in a similar fashion.

aksik [14]3 years ago

3 0

Answer:

The Answer is A. a cylinder

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You want to buy an item that costs $100. Which of these is the most cost-effective choice for buying the item?

Evgen [1.6K]

The answer is C. This would make the total cost 90 dollars without paying for a membership card or shipping.

3 0

2 years ago

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Two different radioactive isotopes decay to 10% of their respective original amounts. Isotope A does this in 33 days, while isot

Andrews [41]

Answer:

The approximate difference in the half-lives of the isotopes is 66 days.

Step-by-step explanation:

The decay of an isotope is represented by the following differential equation:

$\frac{dm}{dt} = -\frac{t}{\tau}$

Where:

$m$ - Current mass of the isotope, measured in kilograms.

$t$ - Time, measured in days.

$\tau$ - Time constant, measured in days.

The solution of the differential equation is:

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$

Where $m_{o}$ is the initial mass of the isotope, measure in kilograms.

Now, the time constant is cleared:

$\ln \frac{m(t)}{m_{o}} = -\frac{t}{\tau}$

$\tau = -\frac{t}{\ln \frac{m(t)}{m_{o}} }$

The half-life of a isotope ( $t_{1/2}$ ) as a function of time constant is:

$t_{1/2} = \tau \cdot \ln2$

$t_{1/2} = -\left(\frac{t}{\ln\frac{m(t)}{m_{o}} }\right) \cdot \ln 2$

The half-life difference between isotope B and isotope A is:

$\Delta t_{1/2} = \left| -\left(\frac{t_{A}}{\ln \frac{m_{A}(t)}{m_{o,A}} } \right)\cdot \ln 2+\left(\frac{t_{B}}{\ln \frac{m_{B}(t)}{m_{o,B}} } \right)\cdot \ln 2\right|$

If $\frac{m_{A}(t)}{m_{o,A}} = \frac{m_{B}(t)}{m_{o,B}} = 0.9$ , $t_{A} = 33\,days$ and $t_{B} = 43\,days$ , the difference in the half-lives of the isotopes is:

$\Delta t_{1/2} = \left|-\left(\frac{33\,days}{\ln 0.90} \right)\cdot \ln 2 + \left(\frac{43\,days}{\ln 0.90} \right)\cdot \ln 2\right|$

$\Delta t_{1/2} \approx 65.788\,days$

The approximate difference in the half-lives of the isotopes is 66 days.

4 0

3 years ago

Read 2 more answers

Can some one explain how to answer this

Softa [21]

You would first have to find the area of the other part.

6 0

3 years ago

The median of the values:29,24,30,23,28,18,

joja [24]

Answer:

26

Step-by-step explanation:

18,23,24,28,29,30

(6 in total + 1)/2=3.5.

- the number which is 3.5 is between 24 and 28.

- middle number between these two is 26.

6 0

1 year ago

Read 2 more answers

2. Suppose over several years of offering AP Statistics, a high school finds that final exam scores are normally distributed wit

nirvana33 [79]

Answer:

By the Central Limit Theorem, the mean is 78, the standard deviation is $s = \frac{6}{\sqrt{n}}$ and the shape is approximately normal.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 78 and a standard deviation of 6

This means that $\mu = 78, \sigma = 6$

Samples of n:

This means that the standard deviation is:

$s = \frac{\sigma}{\sqrt{n}} = \frac{6}{\sqrt{n}}$

What are the mean, standard deviation, and shape of the distribution of x-bar for n?

By the Central Limit Theorem, the mean is 78, the standard deviation is $s = \frac{6}{\sqrt{n}}$ and the shape is approximately normal.

6 0

2 years ago