CPhill

We will have a right triangle , here....the hypotenuse is 12   and the side opposite the angle of interest will be 11.8

So....we have the opposite side and the hypotenuse.....

To    find this angle.....use the   sine inverse (arcsin)....so we have

arcsin (11.8 / 12)  =  79.52°

So...not safe  !!!!

Here's a pic: 

\( \frac{5}{24} + \left|x-\frac{11}{48}\right| < \frac{5}{16}\)

subtract 5/24 from both sides

l x - 11/48  l  < 5/16 - 5/24

l x - 11/48 l  <  15/48 - 10/48

l x -  11/48 l  <  5/48        we have two equations, here

x - 11/48 < 5/48            x  - 11/48 >  - 5/48

                          add  11/48 to  both sides

x < 16/48                   x  > 6/48

x < 1/3                x > 1/8

x  = ( 1/8 ,  1/3)

What are the probabilities of coming up  heads and talls???....they aren't showing up for me...

What hectictar said  !!!!

Here's a cake.....!!!

The answer is  given by :

[ First term + Last term] * 10 / 2

The frist term is 14

The last  is  14 + 3(10 -1)  =  14 + 3(9) =  41    ....  so  we have.......

[ 14 + 41 ] * 10 / 2  =

[55 * 10]  / 2  =

$275

P(Both Even )  =   (2/5) (1/4)  =  2/20  =  1/10

P(Both Prime)  = (3/5)(2/4) =  6/20  =  3/10

P (sum of two numbers is odd)  ....we have C(5,2)  = 10 possible sets

Sets where the sum of two integers is odd :

(1,2) (1,4)

(2,3) (2,5)

(3, 4) 

So...P (Sum of two numbers is odd) =  5/10   =  1/2

P (Product of two numbers is odd)

(1,3) (1,5)

(3, 5)

So.....P (Product of two numbers is odd)  = 3 / 10

It seems  that   "John" and "Bob"  have somehow morphed into "Grogg"  and "Winnie"...

But, anyway.....call the number of minutes that  one of tthem could be late  = x

And call the number of minutes that the other could be late  = y

So......the total possible minutes that they could be late is 60 minutes each...so...we have this equation of a line

x + y  ≤  120    (1)

But....we  are looking for this  x + y  ≤ 45    (2)

Look at the graph here : https://www.desmos.com/calculator/90isasjiab

The area  of the triangle formed by (1)  in the first quadrant is  (1/2)base * height =

(1`/ 2) (120) (120)   =  14400 / 2 =  7200 units^2

The area  of (2)  in the first quadrant  is   (1/2)(45) (45)  = 1012.5 units^2

So....the probability  that the sum of the minutes that both are late is ≤ 45  is

1012.5 / 7200   =    9 / 64

Last one

P(Red )  =  25 / ( 25 + 95)  = 25/120

P(Blue)  =  95 / 120

These events are complementary because  25/120 + 95 / 120  =  120 / 120  =  1

P (L l R)  =  P(L and R) / P(R)

Rearranging this, we have

P(R)  =  P(L and R)  / P(L l R )

P(R)  = 0.03  / 0.0625

Second one

P(Single l Has Pet )  =  16 / (16 + 24)  =  16/40   =  2/5  = .40

P (Single l Has No Pet )  = 12 /  (18 + 12)  = 12 / 30  = 2/5  = .40 true

P (Married l Has No Pet )  = 18 / (18 + 12)  = 18 /30 = 3/5  = .60

P(Married l Has Pet )  = 24 / ( 24 + 16)  = 24 / 40  =  3/5  = .60  true

RP....I believe that   C  is correct for the last one....for instance

If P( Q)  = .30     and P (T)  =  .40

Then

P(Q l T)  =  P (Q and T ) / P(T)  =   .12/ .40  = 0.3

But

P( T l Q )  = P(Q and T) / P(Q)  = .12 / .30  = 0.4

First one :

a_n  = 1500 (1.15)^n-1

Second one....note the terms

7 (-4)^n-1

7, -28,  112 , -448, etc...

an   =  -4 a_n-1  , a₁  = 7