Answer:
<B = 47°
<C = 28°
b = AC = 28.0
Step-by-step explanation:
Given:
∆ABC
AB = c = 18
BC = a = 37
<A = 105°
Required:
Length of AC = b
measure of angle B and angle C
SOLUTION:
==>Use the sine rule, sin A/a = sinC/c to find the angle of C:
SinA = sin(105) = 0.9659
a = 37
sinC = ?
c = 18
0.9659/37 = sinC/18
Cross multiply
0.9659*18 = 37*sinC
17.3862 = 37*sinC
Divide both sides by 37
17.3862/37 = sinC
0.4699 = sinC
sinC = 0.4699
C = Sin-¹(0.4699)
C = 28.0° (nearest tenth)
==>Find angle B using sum of angles in a triangle:
Angle B = 180 - (105+28)
Angle B = 180 - 133
Angle B = 47°
==>Find length of b using sine rule, b/sinB = c/sinC:
SinC = sin(28) = 0.4695
SinB = sin(47) = 0.7314
c = 18
b = ?
b/0.7314 = 18/0.4695
Cross multiply
b*0.4695 = 18*0.7314
b*0.4695 = 13.1652
Divide both sides by 0.4695
b = 13.1652/0.4695
b = 28.0 (nearest tenth)
In a 2-way frequency table, you take the row entry and divide it by its row total to get a decimal.
For your example:
It starts out like this:
sunny snowy
Male: 17 11
Female: 15 16
Add the rows up.
sunny snowy total
Male: 17 11 28
Female: 15 16 31
Then divide the number the by the total:
sunny snowy total
Male: 17/28 11/28 28
Female: 15/31 16/31 31
Yours says to round to the 1000th place, so it ends up like:
sunny snowy
Male: .607 .393
Female: .484 .515
Hope that helps!!
Answer:
0.7026
Step-by-step explanation:
Let X denote the lifetime of light bulb. Given
where the mean is
.
Recall that,


3.5x-10>-3 add 10 to both sides
3.5x>7 divide both sides by 3.5
x>2
... now for the other inequality:
8x-9<39 add 9 to both sides
8x<48 divide both sides by 3
x<6
So we have x>2 and x<6, so the compound inequality is:
2<x<6 and this means that the solution set is:
x=(2, 6)
Step-by-step explanation:
let necklace : x
let earrings : y
x+y=144 -1st eqn
6x+5y=825 -2nd eqn
y=144-X -3rd eqn
substitute eqn into 2nd eqn
6x+5(144-x)= 825
6x +720-5X =825 (expanded)
X = 825-720
X =105
substitute "X=105" into 1st eqn
(105)+y=144
y= 144-105
y=39
1 necklace = 105$
1 pair of earrings = 39$
(to check back: substitute the value into the original eqn)