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Mercedes rides her bike 10 miles in the first week. She increases the distance she rides by 2 miles each week.

How many miles will she ride over the course of 8 weeks?

Sum them all up as an Arithmetic Series:

Sum =N/2 [2F + (N - 1)*D]

Sum =8/2 [2*10 + (8 - 1)*2]

Sum =4 [20 + (7*2)]

Sum =4 [20 + 14]

Sum =4 x 34

Sum = 136 miles - that Mercedes will ride over 8 weeks.

10 + 12 + 14 + 16 + 18 + 20 +22 + 24 =136 miles

There are 4 possible pairs Gretchen can pick, and (6C2)−3=12 ways for her to pick socks of two other colors. There are (8C4)=70 total ways Gretchan can pick socks, and thus there is a 4⋅12/70 =24/35 probability Gretchen picks exactly 1 pair.

This number might seen large, but considering there is a 2^4/ 70 =8/35 chance of picking no pairs, and a (4C2)/70 =3/35 chance of two pairs, it adds up (to 1).

https://math.stackexchange.com/questions/2660495/probability-of-choosing-exactly-1-pair-out-of-4

Factor the following:

35 k^3 + 42 k^2 + 25 k + 30

Factor terms by grouping. 35 k^3 + 42 k^2 + 25 k + 30 = (35 k^3 + 42 k^2) + (25 k + 30) = 7 k^2 (5 k + 6) + 5 (5 k + 6):

7 k^2 (5 k + 6) + 5 (5 k + 6)

Factor 5 k + 6 from 7 k^2 (5 k + 6) + 5 (5 k + 6):

=(5k + 6) (7k^2 + 5)

24,690,000,000,000 == {Twenty-four trillion, six hundred ninety billion}

Here is a bell curve with % and standard deviations shown

It does not have 1/2 standard deviations listed, but the graph is near linear in this region so we will interprolate the values for 1/2 SD

For an 'A' , 1 1/2 SD would encompass

  .15  + 2.35 + 13.5/2 % of the students = 9 %  (rounded to 1 decimal point)

B would encompass  the first 1/2 SD  34/2  plus 1/2 of the next SD   13.5/2  = 24% (rounded)

(4 +  11i)  (4 - 11i)

thanks

We need to find  "b"

So  we have

200  =  300 (b)^5        divide both sides by 300

2/3   =  b^5        take the 5th root of both sides

b  ≈ .9221

So....the model is

f(t)  =  300(.9221)^t

Thanks so much!

t²v³/ [ac ]   =

t²v³a^-1c^-1

Here's a "mock"  pic

           C

    10                    20            

A              15                B   

Side  AB  = 15......side  BC  =  20   

If B is bisected....we have this ratio

BC / AB   =  longer segment  DC / shorter segment AD 

20/15   =  DC / AD

4/3  =  DC / AD

Thus.......AC  is split into 7 equal parts...and  DC  is 4 of these....so its length is

(4/7)  * 10   =   40 / 7   units

If he fails because of the practical only...he must have passed the theory part...so....the probability that he failed because of the practical only  is

Pass Theory * Fail Practical =

   0.8              *            0.5     =      0.40

Thanks CPhill!

ab^2/  [ x^3y  ]  =

ab²x^-3y^-1

The angle bisector of B  sets up the following ratio :

BC / AB   =  CE / AE

BC / 8  =  4 / 6       multiply both sides by 8

BC  =   8 * 4  / 6     =    32 / 6   =   16 / 3

hi how are you doing today.

Thanks

Good job  !!!!

I was actually right ouo

Tysm for your help!

HI how are you doing.

We have a right triangle......the hypotense is 8  and we are looking for the adjacent side next to the angle

The cosine relates the adjacent and hypotenuse...so we have

cos (10°)  =  adjacent side /  8       multiply both sides by 8

8*cos(10°)  =  adjacent side   ≈  7.9 ft

( x + y)^4

We can use the 4th row of Pascal's Triangle for the coefficients.....the first row is "row 0"

1   4      6      4     1

So  we have

1x^4  +   4x^3y   +  6x^2y^2  + 4xy^3  + 1y^4

Note, Manuel.....the sum of the exponents in each term  = 4  ......

Thank you ^-^

Unfortunately...it's not true that half the radius would mean half the volume

Let's find the original radius

904.32   =  (4/3) pi  * r^3     divide both sides by   (4/3)pi

904.32

______    =  r^3

  (4/3)pi

215.89  =  r^3

Take the cube root  

6   =  r^3

Half of this   =  3

So...the volume of a ball with half of the original radius is

(4/3) pi  (3)^3   = 

36 pi  =  113.09 in^3     ...the answer that is closest is

113.04  in^3

[ The  slight difference may be due to using 3.14 for  pi  ]