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how is everyone in general doing today?

y = −x2 + 10x + 24 for −1 ≤ x ≤ 3. 

concave down parabola

Area under the curve.

\(y=-x^2+10x+24\\ y=-(x^2-10x-24)\\ y=-(x-12)(x+2)\\\)

zeros at x= -2   and   x= 12

So there are no roots between x=-1 and x=3 and the curve is above the x axis for this entire region.

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the above stuff is only relevant if you want the absolute value of the areas, if you do not care if negative areas cancel postiive areas then you do not need to worry about it.

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Since this is a parabola, if you use simpson's Rule you will need only one subunit because Simpson's Rule gives exact results with parabolas.   (You didn't say what numerical method you had to use)

check:

\(area=\frac{3--1}{6}(13+4*33+45)\\ area=\frac{4}{6}(190)\\ area=126.\dot6\;\;units^2\)

check:

\(area=\displaystyle\int_{-1}^3\;-x^2+10x+24\;\;dx\\ area=\left[\;\frac{-x^3}{3}+5x^2+24x\;\right]_{-1}^{3}\\ area=\left[-9+45+72\;\right] - \left[\;\frac{1}{3}+5-24\;\right]\\ area=108 - \left[-19+\frac{1}{3}\;\right]\\ area=108 +19-\frac{1}{3}\\ area=126.\dot6\;units^2\\\)

Hi Guest, certainly I will adjudicate :)    

I'd like you to become a member so next time I will know who you are. I'd like that a lot   

Gino's answer is correct, CPhill just misread the question.

Ben rolls four fair 20-sided dice, and each of the dice has faces numbered from 1 to 20. What is the probability that exactly two of the dice show an even number?

This is the same as rolling a die 4 times, it doesn't matter that there are 20 sides, becasue we only care whether the roll is even or odd.

The probability of any individual roll being even (sucess) is 0.5 and 

The probability of any individual roll being odd (failure) is 0.5 

You want exactly 2 even and therfore 2 odd rolls

P(exactly 2 successes)

=[ number of ways 2 items can be chosen from 4] * [prob of success^2] * [ prob of failure^2]

=                                         4C2                             *        (1/2)^2              *     (1/2)^2

=                                           6                                *          1/4                  *     1/4

=      6/16

=       3/8

Yeah, sure! I will help if you post problems to me.

Sorry Chris but you read the question incorrectly :(

You have found the probability of rolling exactly 2 heads in 20 rolls of the dice but there are only 4 rolls.

Continued here:

Melody: Please adjudicate the right answer between the above by "geno3141" and this by "CPhill":

http://web2.0calc.com/questions/please-help-asap_10

Thank you.

The sum of 18 consecutive odd integers is a perfect fifth power of n.

If p is the smallest first number of the series,

then the product np=?

arithmetic progression:

\(\begin{array}{|lcll|} \hline p + 0 \cdot 2,~ p + 1 \cdot 2,~ p+ 2 \cdot 2,~ p+ 3 \cdot 2,~\ldots,~ p+ 17 \cdot 2 \\ p,~ p +2,~ p+ 4,~ p+ 6,~ \ldots,~ p+ 34 \\ \hline \end{array}\)

sum of a arithmetic progression:

\(\begin{array}{|rcll|} \hline && a_1 = p \qquad a_m = p+34 \qquad d = 2 \\\\ s_m &=& \left( \frac{a_1+a_m}{2} \right) \cdot m \\ \hline \end{array} \)

\(\begin{array}{rcll} s_{18} &=& \left( \frac{p+(p+34)}{2} \right) \cdot 18 \\ s_{18} &=& \left( \frac{2p+34}{2} \right) \cdot 18 \\ s_{18} &=& (p+17) \cdot 18 \\ s_{18} &=& 18p+17\cdot 18 \quad & | \quad s_{18} = n^5 \\ n^5 &=& 18p+17\cdot 18 \\ n^5 - 17\cdot 18 &=& 18p \quad & | \quad : 18 \\ \frac{n^5}{18} - 17 &=& p \\ p&=& \dfrac{n^5}{18} - 17 \quad & | \quad 18 = 2^1 \cdot 3^2 \\\\ p&=& \dfrac{n^5}{2^1 \cdot 3^2} - 17 \quad & | \quad n = 2^i\cdot 3^j \\\\ p&=& \dfrac{(2^i\cdot 3^j)^5}{2^1 \cdot 3^2} - 17 \\\\ p&=& \dfrac{2^{5i}\cdot 3^{5j}}{2^1 \cdot 3^2} - 17 \\\\ p&=& 2^{5i-1}\cdot 3^{5j-2} - 17 \\\\ p_{\text{min}}&=& 2^{5i-1}\cdot 3^{5j-2} - 17 \quad & | \quad p_{\text{min}} \text{ if }i = 1 \text{ and } j = 1 \\\\ p_{\text{min}}&=& 2^{4}\cdot 3^{3} - 17 \\ p_{\text{min}}&=& 16\cdot 27 - 17 \\ \mathbf{p_{\text{min}}}& \mathbf{=}& \mathbf{415} \\\\ n &=& 2^i\cdot 3^j \quad & | \quad i = 1 \text{ and } j = 1 \\ n &=& 2^1\cdot 3^1 \\ \mathbf{n}& \mathbf{=}& \mathbf{6} \\\\ n\cdot p &=& 6\cdot 415 \\ &=& 2490 \end{array}\)

csc^4x-2csc^2x+1=cot^4x

LHS=

\(LHS=cosec^4x-2cosec^2x+1\\ LHS=(cosec^2x-1)^2\\ LHS=\left(\frac{1-sin^2x}{sin^2x}\right)^2\\ LHS=\left(\frac{cos^2x}{sin^2x}\right)^2\\ LHS=\left(cot^2x\right)^2\\ LHS=cot^4x\\~\\ LHS=RHS \qquad QED\)

Your answer is the correct solution to the problems of getting exactly 3 questions correct.

Solve                                           A  =  (1/2)(b₁ + b₂)h

Multiply both sides by 2:            2A  =  (b₁ + b₂)h

Divide both sides by b₁ + b₂:     2A/(b₁ + b₂)  =  h

Write                                      y = (-2/3)x - 7 in standard form

Multiply both sides by 3:      3y  =  -2x - 21

Add 2x to both sides:           2x + 3y  =  -21

Write in slope-intercept form:      5x + 2y  =  -6

Subtract 5x from both sides:               2y  =  -5x - 6

Divide both sides by 2:                          y  =  (-5/2)x - 3

Each 20-sided die has 10 odd-numbered sides and 10 even-numbered sides.

To show exactly two even numbers, you must also show two odd numbers.

The probability of showing an even number is 1/2; the probability of showing an odd number is also 1/2.

There are 6 combinations that show two even numbers and two odd numbers:

     6 x (1/2)² x (1/2)²  =  6/16  =  3/8

a)     Let x = number of hot dogs sold for the month

        Let m = amount of money made from hot dog sales

        m  =  3.25(x - 4)

b)     m  =  3.25(55 - 4)  =  165.75

c)     Let x = number of hamburgers sold for the month

        Let m = amount of money made from hamburger sales

        m  =  4.75(x - 2)

d)     Since the number of hamburgers sold is probably not the same as the number of hot dogs sold, the work shown is           

        incorrect.

        You need to use  different variable for the number of hamburgers sold than for the number of hot dogs sold.

         A possible equation is:  m  =  3.25(x - 4) + 4.75(y - 2)

            where  x = number of hot dogs sold for the month

                        y = number of hamburgers sold for the month

                        m = amount of money made from the sales of both hamburgers and hot dogs.

To find the y-intercept of  7x + 2y  =  -12, replace the x-term with 0:

     7x + 2y  =  -12     --->     7(0) + 2y  =  -12     --->     0 + 2y  =  -12     --->     2y  =  -12      --->     y  =  -6

To find the x-intercept of  -3x - 5y  =  21, replace the y-term with 0:

     -3x - 5y  =  21     --->     -3x - 5(0)  =  21     --->      -3x - 0  =  21    --->      -3x  =  21     --->     x  =  -7

Projected profit  =  projected revenues - projected costs

Projected profit  =  (3x + 22,050) - (2x - 1,560)

                          =  3x + 22,050 - 2x + 1,560

                          =  x + 23,610 

The probability of rolling a 1 is 1/6.

The probability of rolling anything but a 1 is 5/6.

You want one 1 and nine not 1s:  (1/6) x (5/6)⁹.

Since there are ten different places the 1 could be rolled, you need to multiply the answer in the above line by10.

Therefore, the probability of exactly one die, out of ten, shows a 1 is:  10 x (1/6) x (5/6)⁹  =  10 x 1 953 125 / 60 466 176

     =  19 531 250 / 60 466 176

     =  0.323  (approximately)

a      40v + 15d ≤ 175

b     40(2) +  15d  ≤ 175

       80 +  15d ≤  175    subtract 80 from both sides

       15d  ≤   95        divide both sides by 15

          d ≤  6 + 1/3      but   d  must be a whole quantity....so.....d  = 6

He can buy  - at most - 6 DVDs 

c    No DVDs..solve this

40 v ≤  175      divide both sides by 40

v ≤  4 + 3/8...........  ⇒  4  video games

Look  at the graph here....

https://www.desmos.com/calculator/ylop9twasi

 x + y  is maximized at the corner point  (-5/9, 31/9)

So .....  the max sum is just   26/9

Here's 21

You  almost had it correct.....

The probability of A union B is equivalent to the sum of  the two separate probabilities  less the product of their probabilities

This will be mutually exclusive whenever  P(A intersect B  )  =   0

Here's19

Let us suppose that   100000 total bulbs are manufactured by both factories...55000 by A and 45000 by B

So.....the number of defects are   55000 * .007  +  45000  * .003  =    520

So....out of the 520 defects........45000*.003  =  135  come from  B

So.....the probability that a defective bulb comes from  B  given that we have a defective bulb  is just

 135 / 520  = .2596

Just as you found  !!!

(2a−5)x^2−2(a−1)x+3=0

Let b =  -2(a - 1)  =   2 - 2a

Let a  =  (2a - 5)

Let c  =  3

This will  happen when

b^2 -  4ac  = 0    ....so we have.....

(2 - 2a)^2  - 4(2a - 5)(3)   =  0     simplify

4a^2 - 8a + 4   -  24a  + 60  = 0

4a^2 - 32a + 64  =  0

a^2 - 8a + 16  = 0

(a - 4)^2  = 0

And a   =  4 

This will  not be exact  and I'm using degrees 

Because we have 365 days......then each  day will be 360 / 365  =  (72/73)°......thus, the funkiness of the measurements....!!!!

I'm using the sine function, here.....

The midline of this function will be    [15.3 + 8.7] / 2  =  24/2  = 12

The amplitude  is  3.3....and the graph will be shifted up 12 units

I'm letting  x = 0   =   Jan 1st......this is arbitrary......it won't make any difference as long as we shift the day count back by 1 unit......in other words....Jan 2nd  will be x = 1  on our graph instead of x = 2.......even so.....the true day count and days after Jan 1st will be "off" a little

So.......Dec 21st  is 11 days before Jan 1st

Because the  normal minimum for the sine  = -90°  we want to let the minimum occur at x = -11

So....we are shifting the graph  by   90  +  (-11)(72/73)  ≈ 79.1056° to the right

So......the approximate graph is

y = 3.3sin ( (72x)/73 -  79.1056) + 12

March 27th is the 86th day of the year - x ≈ 85 on the graph since Jan 1st  = 0 ......the approximate number of daylight hours on this day will be ≈  12.27 hrs   ....this makes sense....on March 21st there will be 12 hours and the days grow longer after this until June 21st

And October 2nd  is the 275th day of the year - x ≈  274 on the graph........and the approximate daylight hours at that time will be ≈ 11.365 hrs.....this also makes sense......the days grow shorter after Sept 21st (when there are 12 hours of daylight)  until Dec 21st

Here's the graph :https://www.desmos.com/calculator/wmtc8h5fss

Log_7 sqrt(2) * sqrt(1 + 1/2 + 1/4........)

Log_7 sqrt(2) * sqrt(2)........because 1+1/2+1/4 + 1/8..........etc. This series sums up to 2.

Log_7 (2) =Log(2) / Log(7) =0.356207......

1/log_4 sqrt(3) * sqrt[1+∑[(2^n)/(3^n)], n, 1,∞] This series sums up to 3.

1/log_4 sqrt(3) * sqrt(3)

1/log_4 (3) =Log(4)/Log(3) =1.261859507........

Last one......pretty easy, too

Midline = 0

No vertical shift

Amplitude  = .5

Period  =  4pi....so  (1/2)  periods in 2pi

Sine curve shifted to the right by pi units 

The equation is :

y  =  .5sin (1/2)(x - pi)   =   .5sin (  [ x - pi ] / 2 )

Here's the graph : https://www.desmos.com/calculator/uziyarngnz

Third one....this one is easier

Midline  = 0

No vertical shift

Amplitude  = 2

Period  = pi....so....2 periods in 2 pi

Phase shift....sine curve shifted to the left by pi/6 units

So.....the equation is

y = 2sin2(x  + pi/6)

Here's the graph : https://www.desmos.com/calculator/jufztepinz