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Erik often runs his laptop computer on battery power. He knows that thelonger he runs on battery power, the less power the battery will have.
A. The number of hours the battery will run is a function of the
number of computers Erik has.
B. Erik's choice of a laptop computer is a function of the amount of
power the battery will hold.
C. The amount of power in the battery is a function of the time Erik
runs the computer on the battery.

Answers

Answer 1

Answer:

Answer:

Step-by-step explanation:

Because the more power the more battery. Therefore it is a function of time.

Answer 2

Answer: C!
I’ve had this question before! Brainlist me!! :) xoxo

Related Questions

3/10 times what equals 1
A home security system is designed to have a 99% reliability rate. Suppose that nine homes equipped with this system experience an attempted burglary. Find the probabilities of these events:_________.A. At least one alarm is triggered.B. More than seven alarms are triggered.C. Eight or fewer alarms are triggered.
Suppose you are climbing a hill whose shape is given by the equation z = 1300 − 0.005x2 − 0.01y2, where x, y, and z are measured in meters, and you are standing at a point with coordinates (100, 120, 1106). The positive x-axis points east and the positive y-axis points north. (a) If you walk due south, will you start to ascend or descend? ascend descend
Two guy wires from the top of a 200-foot radio antenna are anchored to the ground by two large concrete posts. The antenna is perpendicular to the ground and between the two posts. If the angles of depression from the top of the antenna to each of the posts is 46o and 48o respectively, then the distance between the two posts is:_______
A fair coin is tossed 25 times. what is the probability that at most 24 heads occur?

What is the length of the hypotenuse of the right angle defined by the points (-3,-1),(1,-1),and (1,2)?A)square root of 3
B)square root of 3
C) 2 timessquare root of 3
D)5

Answers

To find the length of the triangle first find the two coordinates with a common x coordinate, (1,-1) and (1,2). Now find the difference between their y coordinates, -1 and 2. This means that the length of the triangle is 3.
To find the width of the triangle first find the two coordinates with a common y coordinate, (-3,-1) and (1,-1). Now find the difference between their x coordinates, -3 and 1. This means that the width of the triangle is 4.
By Pythagoras' theorem, we know that $hypotenuse = \sqrt{ width^(2) + length^(2) }$ . Inserting our newfound values into this equation, we an find thelength of the hypotenuse:
$hypotenuse = \sqrt{ 4^(2) +3^(2)} hypotenuse = √(16+9) hypotenuse = √(x) 25hypotenuse = 5$
Therefore, your answer is D0 5. To visualise this question better, you can plot the points on a graph.

Find two unit vectors orthogonal to both given vectors. i j k, 4i k

Answers

The cross product of two vectors gives a third vector $\mathbf v$ that is orthogonal to the first two.

$\mathbf v=(\vec i+\vec j+\vec k)*(4\,\vec i+\vec k)=\begin{vmatrix}\vec i&\vec j&\vec k\n1&1&1\n4&0&1\end{vmatrix}=\vec i+3\,\vec j-4\,\vec k$

Normalize this vector by dividing it by its norm:

$(\mathbf v)/(\|\mathbf v\|)=(\vec i+3\,\vec j-4\,\vec k)/(√(1^2+3^2+(-4)^2))=\frac1{√(26)}(\vec i+3\,\vec j-4\vec k)$

To get another vector orthogonal to the first two, you can just change the sign and use $-\mathbf v$ .

30 POINTS FOR JUST ONE QUESTION! :)

Answers

Answer:

I believe the third one

Step-by-step explanation:

Thanks.

Sand is poured onto a surface at 13 cm3/sec, forming a conical pile whose base diameter is always equal to its altitude. How fast is the altitude of the pile increasing when the pile is 1 cm high? Note that the volume of a cone is 13πr2h where r is the radius of the base and h is the height of the cone.

Answers

Answer:

Altitude of the pile will increase by 16.56 cm per second.

Step-by-step explanation:

Sand is poured onto a surface at the rate = 13 cm³ per second

Or $(dV)/(dt)=13$

It forms a conical pile with a diameter d cm and height of the pile = h cm

Here d = h

Volume of the pile $V=(1)/(3)* \pi r^(2)h$ cm³per sec.

Since h = d = 2r [r is the radius of the circular base]

r = $(h)/(2)$

$V=(1)/(3)\pi ((h)/(2))^(2)h$

$V=(1)/(3)\pi ((h^(2)))/(4)(h)$

$V=(1)/(12)\pi h^(3)$

$(dV)/(dt)=(1)/(12)\pi * 3(h)^(2)(dh)/(dt)$

$(dV)/(dt)=(1)/(4)\pi * h^(2)* (dh)/(dt)$

Since $(dV)/(dt)=13$ cm³per sec.

13 = $(1)/(4)\pi (1)^(2)(dh)/(dt)$ [For h = 1 cm]

$(dh)/(dt)=(13*4)/(\pi )$

$(dh)/(dt)=(52)/(3.14)$

$(dh)/(dt)=16.56$ cm per second.

Therefore, altitude of the pile will increase by 16.56 cm per second.

Final answer:

To solve this problem, we first find the expression for the volume of the cone in terms of the height. We then differentiate this expression to get the relation between the rates of change of the volume and the height. By substituting the given values, we can find the rate of change of the height when the cone is 1 cm high.

Explanation:

The question is related to the application of calculus in Physics, specifically rates of change in the context of real-world problem involving a three dimensional geometric shape - a cone. The student asks how fast the altitude of a pile of sand is increasing at a given time if sand is being poured onto a surface at a constant rate and the pile forms a cone whose base diameter is always equal to its altitude.

We know that the volume of a cone is given by V = (1/3)πr²h, where r is the radius of the base and h is the altitude. Since in this problem the base diameter is always equal to its altitude, we have d = 2r = h, or r = h/2.

Replace r in the volume formula, yielding V = (1/3)π(h/2)²h = (1/12)πh³. Differentiate this expression with respect to time (t) to find the rate of change of V with respect to t, dV/dt = (1/4)πh² * dh/dt.

Given that sand is poured at a constant rate of 13 cm³/sec (that is, dV/dt = 13), we can solve for dh/dt when h = 1cm. Substituting the given values into the equation, 13 = (1/4)π(1)² * dh/dt, we find dh/dt = 13/(π/4) = 52/π cm/sec. Therefore, when the conical pile is 1 cm high, the altitude is increasing at a rate of 52/π cm/sec.

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"The municipal transportation authority determined that 58% of all drivers were speeding along a busy street. In an attempt to reduce this percentage, the city put up electronic speed monitors so that drivers would be warned if they were driving over the speed limit. A follow-up study is now planned to see if the speed monitors were effective. The null and alternative hypotheses of the test are H0 : π = 0.58 versus Ha : π < 0.58. It is planned to use a sample of 150 drivers taken at random times of the week and the test will be conducted at the 5% significance level. (a) What is the most number of drivers in the sample that can be speeding and still have them conclude that the alternative hypothesis is true? (b) Suppose the true value of π is 0.52. What is the power of the test? (c) How could the researchers modify the test in order to increase its power without increasing the probability of a Type I Error?"

Answers

Answer:

a) X=77 drivers

b) Power of the test = 0.404

c) Increasing the sample size.

Step-by-step explanation:

This is a hypothesis test of proportions. As the claim is that the speed monitors were effective in reducing the speeding, this is a left-tail test.

For a left-tail test at a 5% significance level, we have a critical value of z that is zc=-1.645. This value is the limit of the rejection region. That means that if the test statistic z is smaller than zc=-1.645, the null hypothesis is rejected.

The proportion that would have a test statistic equal to this critical value can be expressed as:

$p_c=\pi+z_c\cdot\sigma_p$

The standard error of the proportion is:

$\sigma_p=\sqrt{(\pi(1-\pi))/(n)}=\sqrt{(0.58*0.42)/(150)}\n\n\n \sigma_p=√(0.001624)=0.04$

Then, the proportion is:

$p_c=\pi+z_c\cdot\sigma_p=0.58-1.645*0.04=0.58-0.0658=0.5142$

This proportion, with a sample size of n=150, correspond to

$x=n\cdot p=150\cdot0.5142=77.13\approx 77$

The power of the test is the probability of correctly rejecting the null hypothesis.

The true proportion is 0.52, but we don't know at the time of the test, so the critical value to make a decision about rejecting the null hypothesis is still zc=-1.645 corresponding to a critical proportion of 0.51.

Then, we can say that the probability of rejecting the null hypothesis is still the probability of getting a sample of size n=150 with a proportion of 0.51 or smaller, but within a population with a proportion of 0.52.

The standard error has to be re-calculated for the new true proportion:

$\sigma_p=\sqrt{(\pi(1-\pi))/(n)}=\sqrt{(0.52*0.48)/(150)}\n\n\n \sigma_p=√(0.001664)=0.041$

Then, we calculate the z-value for this proportion with the true proportion:

$z=(p-\pi')/(\sigma_p)=(0.51-0.52)/(0.041)=(-0.01)/(0.041)=-0.244$

The probability of getting a sample of size n=150 with a proportion of 0.51 or lower is:

$P(p<0.51)=P(z<-0.244)=0.404$

Then, the power of the test is β=0.404.

The only variable left to change in the test in order to increase the power of the test is the sample size, as the significance level can not be changed (it is related to the probability of a Type I error).

It the sample size is increased, the standard error of the proprotion decreases. As the standard error tends to zero, the critical proportion tend to 0.58, as we can see in its equation:

$\lim_(\sigma_p \to 0) p_c=\pi+ \lim_(\sigma_p \to 0)(z_c\cdot\sigma_p)=\pi=0.58$

Then, if the critical proportion increases, the z-score increases, and also the probability of rejecting the null hypothesis.

As sample size increases, which of the following is true for a t-distribution-Distribution will get taller and SD will increase
-distribution sill get taller and SD will decrease
-distribution will get shorter and SD will decrease
Distribution will get shorter and SD will increase

Answers

Answer:

Distribution will get taller and SD will decrease.

Step-by-step explanation:

Sample Size and Standard Deviation:

In a t-distribution, sample size and standard deviation are inversely related.

A larger sample size results in decreased standard deviation and a smaller sample size will result in increased standard deviation.

Sample Size and Shape of t-distribution:

As we increase the sample size, the corresponding degree of freedom increases which causes the t-distribution to like normal distribution. With a considerably larger sample size, the t-distribution and normal distribution are almost identical.

Degree of freedom = n - 1

Where n is the sample size.

The shape of the t-distribution becomes more taller and less spread out as the sample size is increased

Refer to the attached graphs, where the shape of a t-distribution is shown with respect to degrees of freedom and also t-distribution is compared with normal distribution.

We can clearly notice that as the degree of freedom increases, the shape of the t-distribution becomes taller and narrower which means more data at the center rather than at the tails.

Also notice that as the degree of freedom increases, the shape of the t-distribution approaches normal distribution.

Final answer:

In a t-distribution, as the sample size increases, the distribution becomes 'shorter', and the standard deviation decreases following the law of large numbers. The increased sample size reduces variability and introduces less deviation from the mean.

Explanation:

As the sample size increases for a t-distribution, the distribution tends to approach a normal distribution shape, which means the distribution will get 'shorter'. Additionally, the standard deviation (SD) would generally decrease as the sample size increases. This is due to the fact that when sample size increases, a smaller variability is introduced, hence less deviation from the mean.

To illustrate, imagine rolling a dice. If you roll it a few times, you may end up with quite a bit of variation. If you roll it a hundred times, however, the numbers should average out closer to the expected value (3.5 for a six-sided dice), and the standard deviation (a measure of variability) would decrease.

In conclusion, when the sample size increases, a t-distribution will get 'shorter' and SD will decrease. This concept is often referred as the law of large numbers.

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