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$\text{Let }\Phi(x) \text{ be the CDF of the standard normal, (that thing you have the table for)}\\ P[\text{response between 400 and 500 seconds}] = \\ \Phi\left(\dfrac{500-450}{50}\right) - \Phi\left(\dfrac{400-450}{50}\right)=\\ \Phi(1)-\Phi(-1) \approx 0.683\\ 0.683 \times 160 \approx 109.23$

Thank you!

Yeah i probably will later

We have the form

y = ax^3  + bx^2   + cx  +  d

Because f(0)  = 0.....then d  =  0

And we have this system

a(-1)^3 + b(-1)^2 + c(-1) = 15

a(1)^3 + b(1)^2 + c(1)   = 5

a(2)^3 + b(2)^2 + c(2)  = 12

-a + b - c  =15  (1)

a  + b + c   = 5   (2)  

8a   + 4b + 2c  = 12     (3)

Add the first two equations  and we get that   2b = 20  ⇒  b = 10

Multiply (2)  by -2  ⇒  -2a -2b -2c = -10

Add this to (3)

6a + 2b = 2

6a + 2(10) = 2

6a = -18

a = -3

And using   (2)

-3 + 10 + c = 5

c = -2

So....the polynomial is

f(x)  =  -3x^3 + 10x^2  - 2x  

This graph shows the x intercepts :  https://www.desmos.com/calculator/0qrtw2nob8

x^2 + 2xy

________  =   -1

y^2  - x^2

x^2  + 2xy  =  -1 ( y^2 - x^2)

x^2 + 2xy   =  x^2 - y^2

2xy   =  -y^2

y^2 + 2xy = 0

y ( y + 2x)  =  0        set both factors to  0

y = 0      and subbing this into the original equation  produces  

x^3  =  3

x=  (3)^(1/3)

So...another  point  is  ( 3^(1/3) , 0 )

P(B|A)=  P(A and B)                0.2

             _________   =         _____   =  0.8  =  P(B)

                   P(A)                    0.25 

P(A|B)  =    P ( A and B)              0.2

                  __________ =        ______    = 0.25 =  P(A)

                      P(B)                       0.8 

So

(D) Having soup and salad for lunch are independent events because P(A|B)=P(A) and P(B|A)=P(B) .

Thank you so much for your answer.

I have a further question, though. How would I determine this other coordinate if I didn't have access to a graph?

Is it possible though algebra alone? If so, could you please show me how? 

Thanks once again in advance,

bqrs01

Perhaps this graph will help:

integrating w substitution

$\large{\int \limits_{0}^{2} \dfrac{x}{ \sqrt{1+2x^2} }dx}$

$\begin{array}{|rcll|} \hline && \mathbf{\large{\int \limits_{x=0}^{x=2} \dfrac{x}{ \sqrt{1+2x^2} }dx}} \\ && \quad \boxed{ \text{substitute: } \\ u = 1+2x^2,\ du=4x\ dx,\\ x=0 \Rightarrow u=1+2\cdot 0^2=1,\\ x=2\Rightarrow u=1+2\cdot 2^2=9 } \\ &=& \int \limits_{u=1}^{u=9} \dfrac{x}{ \sqrt{u} }\dfrac{du}{4x} \\\\ &=& \dfrac{1}{4} \int \limits_{u=1}^{u=9} \dfrac{1}{ \sqrt{u} }du \\\\ &=& \dfrac{1}{4} \int \limits_{u=1}^{u=9} u^{-\frac{1}{2}} du \\\\ &=& \dfrac{1}{4} \left[ \dfrac{u^{\frac{1}{2}}}{\frac{1}{2}} \right]_{u=1}^{u=9} \\\\ &=& \dfrac{1}{4}\cdot \dfrac{2}{1} \left[ u^{\frac{1}{2}} \right]_{u=1}^{u=9} \\\\ &=& \dfrac{1}{2} \left( 9^{\frac{1}{2}}-1^{\frac{1}{2}} \right) \\\\ &=& \dfrac{1}{2} ( \sqrt{9} -1 ) \\\\ &=& \dfrac{1}{2} ( 3 -1 ) \\\\ &=& \dfrac{2}{2} \\\\ &\mathbf{=}& \mathbf{1} \\ \hline \end{array}$

Yes, 1 is the correct answer:

Are you sure you have the limits written correctly in your second integral?

Thankyou for you help

1. The normal monthly rainfalll at the Seattle-Tacoma Airport can be approximated by the model R = 3.121 +2.399sin(0.524t+1.377), where R is measured in inches and t is the time in months. Use the model to approximate the normal yearly rainfall. 

Here is the approximate annual rainfall graph

the horizontal axis is Time in months and the vertical axis is Rainfall in inches.

The approximate annual rainfall is the area between the horizonal axis and the curve from t=0 to t=12

i.e.

$yearly\;\;rainfall\\\approx \displaystyle\int_0^{12}\; 3.121+2.399sin(0.524t+1.377)\,dt\\ =\left[\;\; 3.121t-2.399*0.524cos(0.524t+1.377) \;\; \right]_0^{12}\\ = [3.121*12-2.399*0.524cos(0.524*12+1.377) ] - [3.121*0-2.399*0.524cos(0.524*0+1.377) ] \\ = [37.452-1.257076cos(4.425) ] + [1.257076cos(1.377) ] \\$

(37.452-1.257076cos(4.425) ) + (1.257076cos(1.377) ) = 38.050411914390589152

So annual rainfall is approximately  38.05 inches.     (note that this answer has been edited, there was a small error before)

You STILL need to check my working though   

www.wolframalpha.com

is a good site for you to check these (and many other questions) for yourself.

I already added the first one for you :)

What is the sum of this sequence?
1000^2 + 999^2 - 998^2 - 997^2 +996^2 + 995^2 - .........+ 4^2 + 3^2 - 2^2 - 1^2.

$\begin{array}{|rcll|} \hline && \mathbf{1000^2 + 999^2 - 998^2 - 997^2 +996^2 + 995^2 - \ldots + 4^2 + 3^2 - 2^2 - 1^2} \\ &=& \sum \limits_{k = 1}^{250} \Big( (4k)^2+ (4k-1)^2- (4k-2)^2 -(4k-3)^2 \Big) \\ &=& \sum \limits_{k = 1}^{250} \Big( 16k^2 +16k^2-8k+1-16k^2+16k-4 -16k^2+24k-9 \Big) \\ &=& \sum \limits_{k = 1}^{250} \Big( -8k+1 +16k-4 +24k-9 \Big) \\ &=& \sum \limits_{k = 1}^{250} \Big( 32k-12 \Big) \\ &=& \sum \limits_{k = 1}^{250}32k-\sum \limits_{k = 1}^{250}12 \\ &=& 32\sum \limits_{k = 1}^{250}k- 12\cdot 250 \\ &=& 32\left(\dfrac{1+250}{2}\right)\cdot 250- 12\cdot 250 \\ &=& 16\cdot 251 \cdot 250- 12\cdot 250 \\ &=& 250( 16\cdot 251 - 12) \\ &=& 250\cdot 4004 \\ &\mathbf{=}& \mathbf{1001000} \\ \hline \end{array}$

Given triangle ABC,

sinC=(sinA+sinB)/(cosA+cosB),

What is the shape of triangle ABC? 

$\begin{array}{|rcll|} \hline \sin(C) &=& \dfrac{\sin(A)+\sin(B)}{\cos(A)+\cos(B)} \\ && \boxed{\sin(A)+\sin(B) =2\cdot \sin\left(\dfrac{A+B}{2}\right)\cdot \cos\left(\dfrac{A-B}{2}\right)} \\ && \boxed{\cos(A)+\cos(B) =2\cdot \cos\left(\dfrac{A+B}{2}\right)\cdot \cos\left(\dfrac{A-B}{2}\right)} \\ \sin(C) &=& \dfrac{2\cdot \sin\left(\dfrac{A+B}{2}\right)\cdot \cos\left(\dfrac{A-B}{2}\right)} {2\cdot \cos\left(\dfrac{A+B}{2}\right)\cdot \cos\left(\dfrac{A-B}{2}\right)} \\ \sin(C) &=& \dfrac{ \sin\left(\dfrac{A+B}{2}\right) } { \cos\left(\dfrac{A+B}{2}\right) } \\ && \boxed{\sin(C)=\sin\Big(180^\circ-(A+B)\Big)=\sin(A+B)} \\ \sin(A+B) &=& \dfrac{ \sin\left(\dfrac{A+B}{2}\right) } { \cos\left(\dfrac{A+B}{2}\right) } \\ && \boxed{\sin(A+B) =2\cdot \sin\left(\dfrac{A+B}{2}\right)\cdot \cos\left(\dfrac{A+B}{2}\right) } \\ 2\cdot \sin\left(\dfrac{A+B}{2}\right)\cdot \cos\left(\dfrac{A+B}{2}\right) &=& \dfrac{ \sin\left(\dfrac{A+B}{2}\right) } { \cos\left(\dfrac{A+B}{2}\right) } \\ 2\cdot \cos\left(\dfrac{A+B}{2}\right) &=& \dfrac{ 1 } { \cos\left(\dfrac{A+B}{2}\right) } \\ \cos^2\left(\dfrac{A+B}{2}\right) &=& \dfrac{ 1 }{2} \\ \cos\left(\dfrac{A+B}{2}\right) &=& \dfrac{ 1 }{\sqrt{2}} \\ \cos\left(\dfrac{A+B}{2}\right) &=& \dfrac{ \sqrt{2} }{2} \\ \dfrac{A+B}{2} &=& \arccos\left( \dfrac{ \sqrt{2} }{2} \right) \\ \dfrac{A+B}{2} &=& 45^\circ \\ \mathbf{ A+B } & \mathbf{=} & \mathbf{ 90^\circ} \quad \text{ so } C=90^\circ \\ \hline \end{array}$

The shape is a right-angled triangle

I have just been counting. 

I got

(6+5+4+3+2+1)+(10+8+6+4+2)+(12+9+6+3)+(12+8+4)+(10+5)+6 = 126

thank you so much! would there be any way i could start problem 2 out? :)

asdf just told you that he thinks both answers are wrong.  :/

It happens   

$\dfrac{d}{dx} \displaystyle \int_{a(x)}^{b(x)}f(t)~dt = \\ f(b(x))\dfrac{db}{dx} - f(a(x))\dfrac{da}{dx}$

$\dfrac{d}{dx} \displaystyle \int_3^{x^2}\dfrac{\sqrt{1+2t}}{t}~dt = \\ \dfrac{\sqrt{1+2x^2}}{x^2}\cdot 2x - \dfrac{\sqrt{1+2(3)}}{3}\cdot 0 = \\ \dfrac{2\sqrt{1+2x^2}}{x}$

Triangle ABC has a right angle at C. sinA sinB and sinC forms a geometric sequence.

Please determine the value of sinA. 

$\text{Let $r=$ ratio }$

$\begin{array}{lcll} C=90^\circ \\ \sin(C) = \sin(90^\circ) = 1 \\ \end{array}$

$\begin{array}{lcll} B=90^\circ-A \\ \sin(B) = \sin(90^\circ-A) = \cos(A) \\ \end{array}$

Geometric sequence:

$\begin{array}{|rcll|} \hline a_1 &=& a \\ &=& \sin(A) \\\\ a_2 &=&a\cdot r \quad &| \quad a=\sin(A) \\ &=& \sin(A)\cdot r \quad &| \quad a_2 = \sin(B)=\cos(A)\\ \mathbf{\cos(A)} & \mathbf{=}& \mathbf{\sin(A)\cdot r} \qquad (1) \\\\ a_3 &=& a\cdot r^2 \quad &| \quad a=\sin(A) \\ &=& \sin(A)\cdot r^2 \quad &| \quad a_3 =\sin(C)=1 \\ \mathbf{1} &\mathbf{=}& \mathbf{\sin(A)\cdot r^2} \qquad (2) \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \cos(A) &=&\sin(A)\cdot r & (1) \\ r &=& \dfrac{\cos(A)}{\sin(A)} \\ \hline 1 &=& \sin(A)\cdot r^2 & (2) \quad | \quad r = \dfrac{\cos(A)}{\sin(A)} \\ 1 &=& \sin(A)\cdot \left(\dfrac{\cos(A)}{\sin(A)}\right)^2 \\ 1 &=& \sin(A)\cdot \dfrac{\cos^2(A)}{\sin^2(A)} \\ 1 &=& \dfrac{\cos^2(A)}{\sin(A)} \\ \sin(A) &=& \cos^2(A) & \quad | \quad \cos^2(A)=1-\sin^2(A) \\ \sin(A) &=& 1-\sin^2(A) \\ \sin^2(A)+\sin(A) -1 &=& 0 \\\\ \sin(A) &=& \dfrac{-1\pm \sqrt{1-4\cdot(-1)} } {2} \\ \sin(A) &=& \dfrac{-1\pm \sqrt{5} } {2} \\ \sin(A) &=& \dfrac{-1{\color{red}+}\sqrt{5} } {2} \\ \mathbf{\sin(A) } &\mathbf{=}& \mathbf{\dfrac{-1 + \sqrt{5} } {2} } \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline r^2 &=& \dfrac{1}{\sin(A)} \\ r &=& \dfrac{1}{\sqrt{\sin(A)}} \\ \mathbf{r} &\mathbf{=}& \mathbf{ \dfrac{1}{\sqrt{\dfrac{-1 + \sqrt{5} } {2}}} }\\ \hline \end{array}$

$\begin{array}{|rcll|} \hline a_1 = \sin(A) &=& \dfrac{-1 + \sqrt{5} } {2} \\ a_2 = \sin(B) &=& \sqrt{\dfrac{-1 + \sqrt{5} } {2}} \\ a_3 = \sin(C) &=& 1 \\ \hline \end{array}$

The value of $\sin(A)$ is  $\mathbf{\dfrac{-1 + \sqrt{5} } {2} }$ 

$\text{a linear factor is of the form }(x-a)$

$f(x) = x^3 + 24x^2 - 2x + 6 \text{ (I'm assuming what was written was a typo)}\\ \text{This doesn't seem to have any linear factors}$

$f(x) = x^{16}+72x^6 + x = x(x^{15}+72x^5+1)\\ \text{This has one linear factor, }x$

$f(x)=(x+2)\text{ clearly has a single linear factor which is itself}$

I can't really agree with the answer given.  We have no idea what the form of f(x) is so trying to cast it as linear or quadratic is confusing.

3a) All we have to do here is take the graph of f(x) and shift it 4 units up the y axis.

3b) Here we shift f(x) 2 units to the right along the x-axis, and then 2 units up the y axis