Answer:
?????
Step-by-step explanation:
Answer:
36°
Step-by-step explanation:
Like your other question, the angles of the triangle must add up to 180. The tangent line is perpendicular to the center, so the angle must be 90°.
90° + 54° + 36° = 180°
Find the median of each set:-
Median is middle number of a data set. If a data set has an odd number of numbers then the median is the middle number when ordered form least to greatest but if its an even number you have to find the mean for the middle 2 numbers when ordered for least to greatest.
A.
1.2, 2.4, 3.2, 3.2, 3.6, 4.0, 4.1, 4.7
Even numbers = 8
3.2 + 3.6 = 6.8
6.8 ÷ 2 =
Median = 3.4
So this shows that A isn't the answer because the median of A is 3.4, not 3.2.
B.
1.6, 2.8, 2.9, 3.1, 3.3, 3.6, 4.2, 4.5
Even numbers = 8
3.1 + 3.3 = 6.4
6.4 ÷ 2 = 3.2
Median = 3.2
<span>So this shows that B is the answer because the median of B is 3.2.
C.
1.8, 2.0, 2.0, 2.2, 3.2, 4.7, 4.8, 4.9
</span>
Even numbers = 8
2.2 + 3.2 = 5.4
5.4 ÷ 2 = 2.7
Median = 2.7
<span>So this shows that C isn't the answer because the median of C is 2.7, not 3.2.
</span>
D.
1.4, 1.7, 2.9, 3.0, 3.1, 3.2, 3.2, 3.2, 4
Odd numbers = 9
Median = 3.1
<span>So this shows that D isn't the answer because the median of D is 3.1, not 3.2.
</span>
The stem and leaf plot which median is 3.2 is B.
A = P(1 + rt)
Where:
<span>A = Total Accrued Amount (principal + interest)
P = Principal AmountI = Interest Amount
r = Rate of Interest per year in decimal;
r = R/100
R = Rate of Interest per year as a percent;
R = r * 100
<span>t = Time Period involved in months or years
</span></span>Calculation:
First, converting R percent to r a decimal
r = R/100 = 6.5%/100 = 0.065 per year,
putting time into years for simplicity,
30 months ÷ 12 months/year = 2.5 years,
then, solving our equation
<span>A = 1800(1 + (0.065 × 2.5)) = 2092.5 </span>
A = $ 2,092.50
The total amount accrued, principal plus interest,
from simple interest on a principal of $ 1,800.00
at a rate of 6.5% per year
<span>for 2.5 years (30 months) is $ 2,092.50.</span>
