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First one :   The top left table

Second one : 

The total number of students =  95 + 105 + 145 + 220 = 565

% 7 - 9  =   95/565 =  .168

The top right table must be correct

Third one :

P(-1/3) < 0

Sum of probabilities = 2/3

Last one :

P( x = 2  or x = 0)  =  P(x = 2) + P( x = 0) - P(x=2 and x = 0)  =

0.1 + 0.3  - (0.1 * 0.3)  =

0.4 - .03 =

0.37

I am so happy! I am glad I am starting to understand things I see the improvement.

Could you check this one as well?

V=250(1.28)^d 

Since one year  =  d / 10.........we have

V = 250 (1.28)^(d/10)  =    

250 [(1.28)^(1/10)]^d  

250  [≈ 1.025]^d

250 [ 1 + .025 ]^d

So....the annual growth rate is   .025  ≈   2.5%

Need to be careful, here......assuming that we have

 3/4 +(1/3) /( 2/12) - 1/2 

We always perform any divisions from left to right before  any additions

So.....we really have

 3/4 +( [1/3] / [2/12] ) - 1/2 =

So.....we  are doing this, first :     (1/3 ) /  ( 2/12)   =    (1/3 ) / (1/6)  =  (1/3) * (6/1)  = 2

So....now we have

3/4 + 2 - 1/2  =

3/4 + 8/4 - 2/4  = 

11/4 - 2/4  =

9 / 4

That question didn't give the answer i needed 

to make this a tad clearer

Max Wong has noted that

$1,005,010,010,005,001 = \\ 1\times 10^{15} + 5 \times 10^{12} + 10\times 10^9 + 10\times 10^6 + 5 \times 10^3 + 1\times 10^0\\ \text{This is }\\ \sum \limits_{k=0}^5~\dbinom{5}{k} \left(10^{3}\right)^k \cdot 1^{5-k} = (10^3+1)^5 = 1001^5$

$\text{Now the problem becomes find the largest prime factor of }1001\\ \text{I don't know any way of doing this other than brute force which is what Max Wong used}$

First one : Correct  !!!

If we let x = 1.5, then   floor [ 1.5 - 1] = floor [0.5]  =  0

So...(1.5, 0 )  is on the graph

Second one : Correct !!!

Third one : Correct !!!

You're getting good at this stuff!!!

When you're adding or subtracting fractions, you always want to start by giving them all a common denominator. So, we have $\frac {3/4 + 1/3}{2/12} - 1/2 = \frac {9/12 + 4/12}{2/12} - 6/12$. This helps us calculate the value without messing anything up. So adding the top part, we get $\frac {13/12}{2/12} - 6/12$. The fraction on the bottom is pesky, so we flip 2/12 to 12/2, and now it's multiplying. $13/12 \cdot 12/2 -6/12 = (13/2) \cdot 6 - 1/2$. 6 times 13 is 78, but we can't forget the denominator of two, so it turns out to be 39 - 1/2. The answer is 38 1/2. 

Hope this helps!

Yes. Good job!

Bascilly divide 60 by 1/4 and then 60 by 2/5 and add the answers together

I found only 1 "triple" that seems to balance the equation:

p =3, q =7, r =733

3 x 7 x 733 - 18*(3 +7 + 733) = 2019

15,393 - 18*(743)                    =2019

15,393 - 13,374                        =2019

2019                                         =2019

$\text{Remarks: In my opinion, the ultimate objective of Math is to express the most complicated things by using simplest methods.}$

Define $f(n) = n+n-n\times n \div n$.

Then your expression becomes $f(14) + f(3)+f(13)+f(24)+15+15\times15\div 15$.

Simplifying f(n): $f(n) = n+n-n\times n\div n = 2n - n = n$.

Therefore, the answer of the expression is $14 + 3 + 13 + 24 + 15 + 15 \times 15 \div 15$, which is 84.

Obviously, $m\angle \text{BOA} = m\angle\text{DOC}= 64^{\circ}$

Considering the sum of interior angles of $\triangle \text{BOA}$,

$m\angle \text{BAO} + m\angle \text{BOA} + m\angle \text{ABO} = 180^{\circ}\\ x + 62^{\circ} + 64^{\circ} = 180^{\circ}\\ x = 54^{\circ}$

Then, considering the sum of interior angles of $\triangle \text{DOC}$,

$m\angle \text{DCO} + m\angle \text{DOC} + m\angle \text{CDO} = 180^{\circ}\\ y+64^{\circ}+54^{\circ}= 180^{\circ}\\ y=62^{\circ}$

Finally, $x + y = 54^{\circ} + 62^{\circ} = 116^{\circ}$.

Therefore, option B is correct.

1,005,010,010,005,001

As you see the pattern 1,5,10,10,5,1 in the number, the number is possibly $(10^n + 1)^5$, where n is a natural number.

As the number has 16 digits and $10^{15} = \left(10^3\right)^5$ also contains 16 digits, the number is possibly $(10^3 + 1)^5$ which is 1001⁵.

What we need to do now is to find the prime factorisation of the number, and we can find the prime factorisation of that huge number.

After some trial divisions, $1001 = 7\times 11\times 13$.

Therefore, it immediately follows that $1\;005\;010\;010\;005\;001 =(7\times 11\times 13)^5 = 7^5 \times 11^5 \times 13^5$.

As a result, the largest prime factor of this number is 13 :)

Multiplying by 1/4 means dividing by 4.

Melody, thank you for clarifying the question.

$7 + (10-4)^2\div 4 \times \left(\dfrac{1}{2}\right)^2\\ =7+36\div 4 \div 4\\ =7+\dfrac{9}{4}\\ =\dfrac{37}{4}$

$\quad\text{Surface area}\\ =2\times 2 + 4\times\left(\dfrac{2\times 5}{2}\right)\\ = 4 + 20\\ = 24 \;\text{inch}^2$

If I work and get $7/hour, then for every working hours, my balance forms an arithmetic sequence.

Let's say my original balance is $$b_0$, and my balance after n hours of working is $$b_n$, then the recursive formula is $b_n = b_{n-1} + 7$ and the explicit formula is $b_n = b_0 + 7n$.