5. A rectangle ABCD has vertices A(-2, 3), B(3, 3), C(3, -5). Find the coordinates of D.

The average time taken to complete a test follows normal probability distribution with a mean of 70 minutes and a standard devia

Dimas [21]

Under 45 mins is roughly 16%. This is because 68% of the curve exists within 1 SD of the mean. So 16% must be outside and smaller and 16% outside and larger (on average).

It is impossible to determine how likely you are to find someone with exactly the second amount. However, if you are looking for that or less, you would get 84%

3 0

2 years ago

An electric current, I, in amps, is given by I=cos(wt)+√8sin(wt), where w≠0 is a constant. What are the maximum and minimum valu

exis [7]

Take the derivative with respect to t
$- w \sin(wt) + \sqrt{8} w cos(wt)$
the maximum and minimum values occur when the tangent line is zero so we set the derivative to zero
$0 = -w \sin(wt) + \sqrt{8} w cos(wt)$
divide by w
$0 =- \sin(wt) + \sqrt{8} cos(wt)$
we add sin(wt) to both sides

$\sin(wt)= \sqrt{8} cos(wt)$
divide both sides by cos(wt)
$\frac{sin(wt)}{cos(wt)}= \sqrt{8} \\ \\ arctan(tan(wt))=arctan( \sqrt{8} ) \\ \\ wt=arctan(2 \sqrt{2)}$ OR $\\$ $wt=arctan( { \frac{1}{ \sqrt{2} } )$
(wt)=2(n*pi-arctan(2^0.5))
(wt)=2(n*pi+arctan(2^-0.5))
where n is an integer
the absolute max and min will be

$I=cos(2n \pi -2arctan( \sqrt{2} ))$
since 2npi is just the period of cos
$cos(2arctan( \sqrt{2} ))= \frac{-1}{3} 
$
substituting our second soultion we get
$I=cos(2n \pi +2arctan( \frac{1}{ \sqrt{2} } ))$
since 2npi is the period
$I=cos(2arctan( \frac{1}{ \sqrt{2}} ))= \frac{1}{3}$
so the maximum value = $\frac{1}{3}$
minimum value = $- \frac{1}{3}$

4 0

3 years ago

Someone please help me with this question

mrs_skeptik [129]

Answer:

10x

Step-by-step explanation:

By definition

$(f\codt g)(x)=f(x)\cdot g(x)$

So we just need to multiply the two functions

$(f\cdot g)(x)= f(x)\cdot g(x)\\=\sqrt{2x}\cdot \sqrt{50x}\\=\sqrt{(2x)\cdot (50x)}=\sqrt{100x^2}\\=10\sqrt{x^2}\\=10x \text{ since x is non negative}$

8 0

2 years ago

A math professor notices that scores from a recent exam are normally distributed with a mean of 61 and a standard deviation of 8

Alexeev081 [22]

Answer:

a) 25% of the students exam scores fall below 55.6.

b) The minimum score for an A is 84.68.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean of 61 and a standard deviation of 8.

This means that $\mu = 61, \sigma = 8$

(a) What score do 25% of the students exam scores fall below?

Below the 25th percentile, which is X when Z has a p-value of 0.25, that is, X when Z = -0.675.

$Z = \frac{X - \mu}{\sigma}$

$-0.675 = \frac{X - 61}{8}$

$X - 61 = -0.675*8$

$X = 55.6$

25% of the students exam scores fall below 55.6.

(b) Suppose the professor decides to grade on a curve. If the professor wants 0.15% of the students to get an A, what is the minimum score for an A?

This is the 100 - 0.15 = 99.85th percentile, which is X when Z has a p-value of 0.9985. So X when Z = 2.96.

$Z = \frac{X - \mu}{\sigma}$

$2.96 = \frac{X - 61}{8}$