Answer:
2.828 i think
Step-by-step explanation:
Answer:6908 (miles) ^2
Step-by-step explanation:
diameter=2200 miles
π=3.14
Circumference=π x diameter
Circumference=3.14 x 2200
Circumference=6908 (miles)^2
6. One variable only so pretty straightforward.
length-x+4
width-x
x+x+4=80
2x=76
x=38
x+4=42
answer: length 42cm and width 38cm
7. Another money problem!
n-# of nickels
q-# of quarters
n=3+q
0.05n+0.25q=2.85
Substitution works like a charm!
0.05(3+q)+0.25q=2.85
0.15+0.05q+0.25q=2.85
0.3q=2.7
q=9
n=3+q
n=3+9
n=12
answer: 9 nickels and 12 quarters
8. One variable situation again.
Ann's money-2b+9
Betty's money-b
b+2b+9=60
3b=51
b=17
2b+9=43
answer: Ann has $43 and Betty has $17.
9. # of red m&m's-x+1
# of blue m&m's-x
x+1+x=13
x=6
x+1=7
answer: 6 blue and 7 red m&m's
10. a-number of adult tickets
s-number of student tickets
a+s=785 ----> a=785-s
5a+2s=3280
5(785-s)+2s=3280
-3s=-645
s=215
a+s=785
a+215=785
a=570
answer: 215 children tickets and 570 adult tickets
Answer:
(8, 0 )
Step-by-step explanation:
under a reflection in the line y = - x
a point (x, y ) → (- y, - x ) , then
(0, - 8 ) → (8, 0 )
Answer:
Step-by-step explanation:
Suppose we think of an alphabet X to be the Event of the evidence.
Also, if Y be the Event of cheating; &
Y' be the Event of not involved in cheating
From the given information:
Thus, 
P(Y') = 1 - 0.01
P(Y') = 0.99
The probability of cheating & the evidence is present is = P(YX)
The probabilities of not involved in cheating & the evidence are present is:
(b)
The required probability that the evidence is present is:
P(YX or Y'X) = 0.006 + 0.000099
P(YX or Y'X) = 0.006099
(c)
The required probability that (S) cheat provided the evidence being present is:
Using Bayes Theorem
