Step-by-step explanation:Let number of 49 cents =A
Let number of 33 cents=B
<u>A+B=45 equation 1</u>
1 $=100 cents
49 cents =.49 dollar
33 cents=.33 dollar
so .<u>49A+.33B=17.89 equation 2</u>
Add equation 1 and equation 2
since these are simultaneous equations
.49A+.33B=17.89
A+ B=45
so divide equation 2 by 100 we get
49A+33B=1789
A+B =45 multiply equation 1 by 33 on both the sides we get
<u>33A+33B=1485 equation 3</u>
so subtract equation 3 from equation 2 we get
49A+33B=1789 equation 2
<u>33A+33B=1485 equation 3</u>
<u>16A+0 =304</u>
16A=304
A=304/16
A=19
putting nalue of A in equation 1 we get
A+B=45
19+B=45
B=45-19
B=26
so number of 49 cent stamps are 19
and number of 33 cent stamps are 26
Answer:
C, 6
Step-by-step explanation:
31-13 is 18, 18/3 is 6.
Mid term :
Q1 = (88 + 85)/2 = 86.5
Q2 = (92 + 95)/2 = 93.5
Q3 = 100
IQR = Q3 - Q1 = 100 - 86.5 = 13.5
final exams :
Q1 = (65 + 78)/2 = 71.5
Q2 = (88 + 82)/2 = 85
Q3 = (95 + 93)/2 = 94
IQR = Q3 - Q1 = 94 - 71.5 = 22.5
so the final exams has the largest IQR
Answer: 254.714285714 or rounded to 255
Step-by-step explanation:
Have a good day:)
It's [0, 68]. To find the product of these, perform the following: [(3*4)+(-1*2)+(-2*5)]=0 and [(5*4)+(4*2)+(8*5)]=68
