All Answers 354303 Answers

"B" is correct.....

See answer No. 4 here: https://web2.0calc.com/questions/help-plz_44

On a rainy day, every kid at Margo's school brings a raincoat, an umbrella, or both. 8% of the kids bring both. The total number of kids with umbrellas is twice the total number of kids with raincoats.

What percent of the kids bring only an umbrella?

For simplicity, let's suppose that there are 100 kids at the school

If 8% bring both, then 8 bring both

Let the number who bring raincoats = x

Let the number who bring umbrellas =  2x

So....the ones that bring only raincoats = x - 8

And the ones that bring only umbrellas = 2x - 8

So......we have this equation

Kids who bring only umbrellas + kids who bring both + kids who only bring raincoats = 100

(2x -8)  +  8  + (x - 8)  =  100      simplify

3x - 8   = 100      add 8 to both sides

3x = 108      divide both sides by 3

x = 36

So....the ones who bring only a rain coat =  36 - 8  =  28

And the ones who bring only an umbrella  = 2(36) - 8  =  72 - 8  = 64 = 64%

Check

Only umbrellas +  Both + Only Raincoats = 100  ???

64  +  8  +  28   =  100   (Check)

See the previoous picture of bell curve....you should be able to tell which statement is true..... 

you're gonna have to translate this to complete sentences

\(\text{let }N \text{ be the total number of kids}\\ R \text{ is the set of kids that bring raincoats}\\ U \text{ is the set of kids that bring umbrellas}\\ |U|=2|R|\\ |U\cap R|=0.08N\)

\(N = |U \cup R| = |U| + |R| - |U \cap R|\\ N = |U|+\dfrac{|U|}{2} - 0.08 N\\ 0.92N = \dfrac 3 2 |U|\\ |U| = \dfrac 2 3 0.92 N = \dfrac{184}{300}N\)

\(|\text{kids that only bring umbrellas}| = |U| - |U\cap R| = \\ \dfrac{184}{300}N - \dfrac{8}{100}N = \dfrac{160}{300}N = \dfrac{8}{15}N\\ \dfrac{8}{15} \approx 53.3\%\)

\(554_b = (24_b)^2\\ 5b^2+5b+4 = (2b+4)^2 = 4b^2+16b+16\\ b^2- 11b-12=0\\ (b-12)(b+1) = 0\\ b=12,~-1\\ \text{-1 is not a base}\\ b=12\)

\(554 _{12} = 4+60+720 = 784_d\\ 24_{12}=24+4=28_d\\ 748=28^2\)

Answer D

thanks!

yea this is what i struggled with, because r can be in any directions

thank you for answering

From the graph below of a standard bell curve      B is true  

13! = 6,227,020,800 = 2^10 * 3^5 * 5^2 * 7 * 11 * 13

N =5^2 x 3^4 x 2^10 =2,073,600

a(44) mod 45 = 9

2000 ( 1 + .062 / 4 ) ^( 13 * 4)  ≈ $ 4450.27

AC = BD , CB = DA 

Given

AB = AB

Reflexive property of equality

So.....by SSS

Triangle ABC is congruent to Triangle BAD

Have you drawn this Manuel?

You must draw it.  

What are the tests that you can use to show that two triangles are congruent? (congruent means 'the same') ?

Point-Slope

(y - y1) = m(x - x1)

Where m is the slope  and (x1, y1) is the point

Slope-Intercept

y = mx +  b

Where m is the slope and "b" is the y intercept

We can answer this with a Venn diagram :

Kinda Pink only =  60 - 27 =  33

Both = 27

Purply Only = 23

Neither = 17

I count 17 "oddly powerful integers":
(1, 8, 27, 32, 64, 125, 128, 216, 243, 343, 512, 512, 729, 1000, 1024, 1331, 1728)= 17 -"oddly powerful" integers!!.

1^3=1, 2^3 =8, 3^3=27, 2^5 =32, 4^3=64, 5^3 =125, 2^7=128, 6^3 =216, 3^5 =243, 7^3 =343, 2^9 =512, 8^3 =512, 9^3 =729, 10^3=1,000, 4^5 =1,024, 11^3 =1,331, 12^3 =1,728.

3.

The scale factor from EFGH to JKLM is   JK  / EF  =   11.2  /  28  =  2 / 5 =   0.4

The first pair of polygons at the top left are not similar because

18  / 14   =

9 / 7   does not equal to 10 / 7

The pair of polygons at the bottom left are similar  because

18 / 10    =    27 / 15  =  7.2 / 4 = 10.8 / 6  =   9 / 5

The pair of polygons at the top right are similar because they are both equilateral triangles....and all equilateral  are similar

The pair of polygons at the bottom right are similar......each side of the smaller polygon is 3/4 that of its corresponding side in the larger polygon....

\(P[\text{winning}]=0.25\\ P[\text{playing at home}]=0.8\\ P[\text{winning AND playing at home}] = 0.2\\ P[\text{winning}]P[\text{playing at home}] = (0.25)(0.8) = 0.2\\P[\text{winning AND playing at home}]= P[\text{winning}]P[\text{playing at home}] \\ \text{so the two events are independent}\)

now that you've been shown the method apply it to (b)

Maybe. Now I do feel like an idiot if it is

I think Chris MAY have answered your question.....but you did  not ask for #people who think 'purply ONLY'      , you asked for 'purply'

   'purply' would include the 'purply-only'  23     AND the  'kinda-pink and Purply'  27      23 +27 = 50 think purply.     This is consistent with your statement that 60 people think it is kinda pink which includes the 27 who thought BOTH.  

(purply ONLY = 23 , but that was not what was asked in your question)     Thanx, Chris!