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thanks cphill see you in the afternoon

the answer is 87 if you turn your head upside down the numbers are going by step

This car's parking spot is  89.....I assigned the parking spots and this goofball parked in the wrong one.

thank cphill

Good job, QS   !!!

I know this one...so.....I'll let someone else answer  !!!

f(x)  = 2x - 3        g(x)  = x^2  + 1

a)  f(g(x)  =    2 [ x^2 + 1] - 3    =   2x^2 + 2 - 3   =   2x^2   - 1

b)  g(f(x) )  =   [ 2x - 3 ] ^2 + 1  =  4x^2 - 12x + 9 + 1  =  4x^2 - 12x + 10

Ans:87

Thanks, lynx......!!!!

1)

Note, NSS....characteristics of inverse variation :

x will increase and y will decrease   or   x will decrease and y will increase

And   each  x, y association will multiply to the same thing

Note that the second choice is correct

x increases  and y decreases      ...and ....

each  x, y association multiplies to the same thing  ⇒  60

Tom is incorrect
Lower bound=55000 
Upper bound=65000 
Lower bound=5350 
Upper bound=5450 
10%×55000=5500 

answer:

the second table shows an inverse variation

explanation:

Only the second table has pairs whose product is always the same value (60).

2)

y  = k / x

8 = k / 2   multiply both sides by 2

16  = k

So we have that

y  =  16  / x

3)   Just look  at the table, NSS

Multiplying each  n  by its associated c    gives  us   100

So 

 n * c  = 100

answer:84.3%

median score=116

lower quartile=102

explanation:

The Median is simply the number in the exact middle, or the 7th number in the list =140

Find the 25th percentile of the list:
(80, 90, 100, 110, 120, 130, 140, 150, 160, 170, 180, 190, 200)

The 25th percentile of the ordered list (80, 90, 100, 110, 120, 130, 140, 150, 160, 170, 180, 190, 200) should be the median of the lower half of the list.
Since the list has an odd number of elements, we take the average of the medians of the two lists that are closest to being the lower half of the list:
(80, 90, 100, 110, 120, 130) and (80, 90, 100, 110, 120, 130, 140)

To find the median of the ordered list (80, 90, 100, 110, 120, 130), we must find the two elements in the middle of the list.
The two elements in the middle of the list (80, 90, 100, 110, 120, 130) are 100 and 110:
(80, 90, 100, 110, 120, 130)

The median is the average of the two elements in the middle of the sorted list.
The median is the average of the two middle elements, 100 and 110. This average is:
(100 + 110)/2 = 105

The median of the ordered list (80, 90, 100, 110, 120, 130, 140) is the element in the middle of the list.
The median of the list (80, 90, 100, 110, 120, 130, 140) is the element in the middle, which is:110
 

Now that we know that the median of {80, 90, 100, 110, 120, 130} is 105 and the median of {80, 90, 100, 110, 120, 130, 140} is 110, we have only to find the average of the two medians.
The 25th percentile is the average of 105 and 110. This average is:
(105 + 110)/2 = 215/2 =107.50

Tom is incorrect
Lower bound=55000 
Upper bound=65000 
Lower bound=5350 
Upper bound=5450 
10%×55000=5500 
 

5)   Here's what we need

N =  k * A / P

The second sentence translates to :

1800  = k * 34000 / 25         multiply both sides by 25 / 34000

25 * 1800 / 34000  =  k  =  45/34

So we have

N  =  (45/34) * 42000 / 25     =    2224  dolls

4)

Here's the equation we need :

r = ks / t

And when r = 2, s = 3

So .......  we need to find k

2  = k* 3 / 12        

2 = (1/4) k      multiply both sides by 4

8  = k

So

r  =  8 (5) / 4

r  =  40 / 4

r = 10

Answered better by lynx above  !!!!

Triangle AHI is equilateral. We know BC, DE, and FG are all parallel to HI and AB = BD = DF = FH.
What is the ratio of the area of the trapezoid FGIH to the area of triangle AHI?
Express your answer as a common fraction.

\(\text{Let $FG = s $} \\ \text{Let $HI = c $} \\ \text{Area of the triangle $AHI = A_{AHI}$ } \\ \text{Area of the triangle $AGF = A_{AGF}$ } \\ \text{Area of the trapezoid $FGIH = A_{FGIH}$ } \\ \text{Let $AK = h$ (height of the triangle$_{AGF}$) } \\ \text{Let $AL = H$ (height of the triangle$_{AHI}$) } \\ \text{Let $AF = \frac34 c $}\)

1.

\(\begin{array}{|rcll|} \hline A_{AHI} &=& A_{AGF} + A_{FGIH} \\\\ \dfrac{cH}{2} &=& \dfrac{sh}{2} + \left( \dfrac{s+c}{2}\right)(H-h) \quad & | \quad \times 2 \\\\ cH &=& sh + (s+c)(H-h) \\\\ \not{cH} &=& \not{sh} + sH-\not{sh}+\not{cH}-ch \\\\ \mathbf{ch} &\mathbf{=}& \mathbf{ sH } \qquad (1) \\\\ &\text{or}& \\\\ \mathbf{\dfrac{s}{c}} &\mathbf{=}& \mathbf{\dfrac{h}{H}} \qquad (2) \\ \hline \end{array}\)

2.

\(\begin{array}{|rcll|} \hline \text{ratio} &=& \dfrac{ A_{FGIH} } {A_{AHI}} \\\\ &=& \dfrac{ \left( \dfrac{s+c}{2}\right)(H-h) } {\dfrac{cH}{2}} \\\\ &=& \dfrac{(s+c)(H-h)}{cH} \\\\ &=& \dfrac{sH-sh+cH-ch}{cH} \quad & | \quad ch=sH \qquad (1) \\\\ &=& \dfrac{cH-sh}{cH} \\\\ &=& 1-\dfrac{sh}{cH} \quad & | \quad \dfrac{s}{c} = \dfrac{h}{H} \qquad (2) \\\\ &=& 1-\dfrac{h^2}{H^2} \\\\ \mathbf{\text{ratio}} & \mathbf{=} & \mathbf{ 1-\left(\dfrac{h}{H}\right)^2} \\\\ && \boxed{\mathbf{3.}\\ \dfrac{h}{\frac34 c} = \dfrac{H}{c} \\ h = \frac34 c \cdot \dfrac{H}{c} \\ h = \frac34 \cdot H \\ \mathbf{\dfrac{h}{H} = \frac34} } \\\\ \mathbf{\text{ratio}} & \mathbf{=} & \mathbf{ 1-\left(\frac34 \right)^2} \\\\ &=& 1- \frac{9}{16} \\\\ &=& \frac{16-9}{16} \\\\ &\mathbf{=}& \mathbf{\dfrac{7}{16}} \\ \hline \end{array}\)

So  the ratio  of areas is \(\mathbf{\dfrac{7}{16}}\)

The probability that the last digit is NOT a 1 is 9/10. The probability that the penultimate digit (i.e. first digit of the minutes portion) is NOT a 2 is 5/6. The probability that the hour portion does not contain a 2 is 8/12, for 12-hour displays, or 12/24, for 24-hour displays. (The teaser needs to state whether it is a 12- or 24-hour display.) Thus, for 12-hour displays, the probability of at least one 2 is: 1 - (9/10 * 5/6 * 8/12) = 1/2 and for 24-hour displays, is: 1 - (9/10 * 5/6 * 1/2) = 3/8

565 kg to the nearest 5 kg  means that the total weight of the people on

board, W,   could be represented as

562.5 < W < 567.5

But Fred's weight, F, could be represented as

87.5 < F < 88.5

Let's assume the lower bound for Fred....this makes the elevator as heavy as possible 

So....when he gets out of the elevator......the greatest possible total weight will now be  less  than

567.5 - 87.5  =    480 kg

So...it is safe to ride

I have always had better luck when the actual property name had no spaces. So I would suggest you copy the Description down to the property name and remove the spaces. I'm guessing that's 2014?

Actually....I think an infinite number of things could work

Note

-3 + 6  = 3

But

(-3)^5  + 6^5  =  7533

yes  nice try cphill