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Thanks! How can we determine the intervals where the function is positive and negative?

Using a positive z -score table:

Jake scored 20/60 of one S.D. above the mean = .3333 S.D above the mean  

      looking at the z score table , S.D. +.33   he scored better than ~63% of the peole who took this test

We  go from left to  right

Increasing  -  intervals  on x where the function "rises"  from left to right  = (-inf, -2) and (3, inf )

Decreasing  - intervals on x  where the function "falls" from left to right  = (1, 3)

Constant  -  intervals where the function neither rises or falls  =  (-2 , 1 )

Thank you!! I see where I went wrong but im still honestly confused. I really do appreciate your help!! I will not be taking anymore online math courses after this!! In class Math ONLY for me :)

Could you help me with this please..  Its practicing with binomial distribution

A company has determined that 1% of the parts it makes are defective. If the company packs a shipment of 5 parts to a customer, what is the probability that : 

a) none are defective 

b) more than 2 are defective

I have been searching for help online and found a similar question where i came up with these answers

P(defective) = 1-0.05 

a) = 0.95

2- 0.95 

b) = 1.05

I feel like im atleast in the right direction but honestly still dont understand.. :/ 

\(-3x-1 = x^2 - 3x + 4\\ x^2 +5=0\\ \text{This should pretty clearly be seen to have no real solutions}\)

y = floor ( x)  - 3

Let's plot a few values

x = 1  ....floor  = 1  so     y = -2

x = 1.5 ....floor = 1   so    y =  -2 

x = 2   ....floor = 2  so         y = -1

So....the points   (1, - 2)   (1.5, -2)   and ( 2, - 1)    are on the graph

Notice that these points only exist on graph  "D"

Wait, but you already defined b to be 2. Then how can it be 0 or 2?

Thanks!!!

Yeah! Guest, that's what I got too (see my post above)

It's floor.

Are the brackets around the x a floor or ceiling function? I can't tell, so I am unable to tell what it is.

THX, CS !!!

1 /( x - 3)   +  1

1 / ( x - 3 + 7}   + 1 + 2       =    1 / (x + 4)  + 3

So...the function is translated 7 units to the left and   2 units up

Given f(x)=3x^2 +9x and g(x) = x^2 - x -12, find (f/g) (x). 

(f / g) (x)   means  f(x) / g(x)......so

3x^2 + 9x                   3x  ( x + 3)                   3x

____________=      ____________  =      ______          [ x not equal to -3, or 4 ]

x^2 - x - 12                (x - 4)( x + 3)              x -  4

Second one

f(x)  = g(x)

x^2 - 3x + 7  =  x + 3             subtract x + 3 from both sides

x^2 - 4x + 4  =  0                factor

(x - 2)^2 = 0                      take the square root

x - 2 = 0                              add 2 to both sides

 x = 2

Hi Rom

Can you explain to me why the following reasoning is wrong ?

The probability of snow on the first day is 1/3, so the probability that it doesn't snow is 2/3.

Same for the next two days, so the probability of no snow during the first three days is (2/3)^3.

Similarly the probability of no snow during the next four days is (3/4)^4.

So the probablity of no snow during the week is (2/3)^3.(3/4)^4 = 3/32.

If  that's the case then the probability that it snows at least once will be 29/32 ?

Second one:

x^2-3x+7=x+3

x^2-4x+4=0

(x-2)^2=0

x=2

You are very welcome!

:P

The red is equation 1 and blue is equation 2.

Can you find the transformation?

You are very welcome!

:P

Thanks!!!

Thanks!!!

Thanks to all of you!!!

Let's find the chance that it *doesn't* snow.

There's a 2/3 chance for the first three days and 3/4 chance for the last four.

So... \((\frac{2}{3})^3\times (\frac{3}{4})^4=\frac{3}{32}\)

Now, \(1-\frac{3}{32}=\frac{29}{32}\) so there is a 29/32 chance it will snow at least once.

You are very welcome!

:P

Thanks!!!

Thanks!!!