Answer:
a = $31.23 per month
b = $20.83 per month
c = 249.98 = $249.98 interest charges
= $624-249.98 = 374.02 profit part decrease 8% inc.
d = 35.35%
e = Answer is in the name; basic payment is a contract which means whilst account remains open charges are requested without fail. Should balance be less or on zero, charges are still applied.
Step-by-step explanation:
2500 x 1.2499 = interest only for start month 13 =3124.75
3124.75/ 100 x 1.08 = 8% of this = 249.98 each year.
We only have to divide each by 12 to work out monthly individual charges and subtract to find out payments.
3124.75 - 249.98 = 2874.77 = Total after charges each year.
249.98/12 = 20.83 = monthly charges.
3124.75- 2500 = 624.75 payments each year
624.75/12 = 52.06 month 1 payment before charge
52.06-20.83 =31.23 total minimum payment
2500 + 249.98
Percentage = 200:600 = 1/3 33% + (comparing to ratio 10:25 closer to 40%)
We find ratio 200:600= 33.33 + 49.995/24.75 = 2.02
33+2.02 = 35.35%
Answer:
I think the table is linear. The total amount after one year is 248 ants.
Step-by-step explanation:
Answer:
z= 0.278
Step-by-step explanation:
Given data
n1= 60 ; n2 = 100
mean 1= x1`= 10.4; mean 2= x2`= 9.7
standard deviation 1= s1= 2.7 pounds ; standard deviation 2= s2 = 1.9 lb
We formulate our null and alternate hypothesis as
H0 = x`1- x`2 = 0 and H1 = x`1- x`2 ≠ 0 ( two sided)
We set level of significance α= 0.05
the test statistic to be used under H0 is
z = x1`- x2`/ √ s₁²/n₁ + s₂²/n₂
the critical region is z > ± 1.96
Computations
z= 10.4- 9.7/ √(2.7)²/60+( 1.9)²/ 100
z= 10.4- 9.7/ √ 7.29/60 + 3.61/100
z= 0.7/√ 0.1215+ 0.0361
z=0.7 /√0.1576
z= 0.7 (0.396988)
z= 0.2778= 0.278
Since the calculated value of z does not fall in the critical region so we accept the null hypothesis H0 = x`1- x`2 = 0 at 5 % significance level. In other words we conclude that the difference between mean scores is insignificant or merely due to chance.
For this case we have the following polynomial:
The first thing to do is to place the variables on the same side of the equation.
We have then:
We complete the square by adding the term (b / 2) ^ 2 on both sides of the equation.
We have then:
Rewriting we have:
Therefore, the solutions are:
Answer:
the solution set of the equation is:
