40s+2bs+200=number of books needed
if we want 12 books and 8 stalls
b=12
s=8
solve
40(8)+2(12)(8)+200=
320+192+200=
712
needs 712 books
Answer:
a. 0.48
b. 0.14
c. 0.47
Step-by-step explanation:
Data provided in the question
0 - 2 0.48
3 - 5 0.26
6 - 8 0.12
9- 11 0.09
12- 14 0.05
Based on the above information
a. The probability for fewer than three phobias is
= P( x < 3)
= 0.48
b. The probability for more than eight phobias is
= P( x >8)
= 0.09 + 0.05
= 0.14
c. Probability between 3 and 11 phobias is
= P(3 < x < 11)
= 0.26 + 0.12 + 0.09
= 0.47
Step-by-step explanation:
Sn= a+[n-1]d
S2= a+[2-1]d
8.5 = a+d _______Equation 1
S5 = a+[5-1]d
13=a+4d ________Equation 2
Subtract Equation 1 from Equation 2
13=a+4d
-
8.5= a+d
________
4.5= 3d
d=1.5
Substituting d=1.5 in equation 1
8.5=a+1.5
a=8.5-1.5
a=7
Sum of terms of an A.P =
Sn= n/2[2a+(n-1)d]
292= n/2[2×7+(n-1)d]
292=n/2[14+(n-1)1.5
292×2=n[14+(n-1)1.5]
584=n[14+1.5n-1.5]
584= 14n+1.5n²-1.5n
584= 1.5n²+12.5n
1.5n²+12.5n-584
1.5n²-24n+36.5n-584
1.5n(n-16)+36.5(n-16)
(1.5n+36.5)(n-16)
**n-16=0
n=16
Answer:
<h3>
3n + 3</h3>
Step-by-step explanation:
The consecutive integers are increasing by 1
n ← the smallest integer
n+1 ← the middle integer
n+1+1 = n+2 ← the largest integer
The sum:
S = n + n+1 + n+2
S = 3n + 3
