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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 \(b_n = b_0 + 7n\).