How much of the sphere do you have to paint red so that no matter ow you put a cube in it, some corner touches red?

melamori03 [73]

Look it up online because you don't have a question

3 0

3 years ago

A cold drink is poured out at 52°F. After 2 minutes of sitting in a 72°F room, its temperature has risen to 55°F. Find an equati

I am Lyosha [343]

Answer:

The model for the temperature of the drink can be written as

$T=72-20e^{-0.08t}$

Step-by-step explanation:

For a cold drink in a hotter room, we can say that the rate of change of temperature of the drink is proportional to the difference of temperature between the drink and the room.

We can model that in this way

$\frac{dT}{dt}=k*(T_r-T)$

If we rearrange and integrate

$\int\frac{dT}{(T-Tr)} =-k*\int dt\\\\ln(T-T_r)=-kt+C1\\\\T-T_r=Ce^{-kt}\\\\T=T_r+Ce^{-kt}$

We know that at time 0, the temperature of the drink was 52°F. Then we have:

$T=T_r+Ce^{-kt}\\\\52=72+Ce^0=72+C\\\\C=-20$

We also know that at t=2, T=55°F

$T=T_r+Ce^{-kt}\\\\55=72-20e^{-k*2}\\\\e^{-k*2}=(72-55)/20=0.85\\\\-2k=ln(0.85)=-0.1625\\\\k=0.08$

The model for the temperature of the drink can be written as

$T=72-20e^{-0.08t}$

7 0

4 years ago

If parallel lines a and b are cut by transversal t, then which statement about angles 1 and 2 must be true?

alex41 [277]

Did you not include D or finish writing D since the first three are not right. If one of the possible answers is that they are alternate exterior angles or the same degrees, that would be the answer.

8 0

3 years ago

A recipe states that 5 servings of peanuts have 1,250 calories. How many servings is 375 calories of peanuts? *

inysia [295]

1,250/5 = 250 calories per serving
375/250 = 1.5 servings

6 0

3 years ago

Read 2 more answers

A Packaging Company produces boxes out of cardboard and has a specified weight of 16 oz. A random sample of 36 boxes yielded a s

user100 [1]

Answer:

The margin of error is of 0.73 oz.

The 99% confidence interval for the true mean weight of the boxes is between 14.57 oz and 16.03 oz. This means that we are 99% sure that the true mean weight of all boxed produced by the Packaging Company is between these two values, and that the specified weight is in this interval.

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:

$\alpha = \frac{1 - 0.99}{2} = 0.005$

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

That is z with a pvalue of $1 - 0.005 = 0.995$ , so Z = 2.575.

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.

$M = 2.575\frac{1.7}{\sqrt{36}} = 0.73$

The margin of error is of 0.73 oz.

The lower end of the interval is the sample mean subtracted by M. So it is 15.3 - 0.73 = 14.57 oz.

The upper end of the interval is the sample mean added to M. So it is 15.3 + 0.73 = 16.03 oz.

The 99% confidence interval for the true mean weight of the boxes is between 14.57 oz and 16.03 oz. This means that we are 99% sure that the true mean weight of all boxed produced by the Packaging Company is between these two values, and that the specified weight is in this interval.

4 0

3 years ago