The solution is A) (11, 1).
As given,
Loan amount is = $45000
Rate of interest = 8.5%
So, Tony's mortgage will attract an interest of:
= $3825 (this is yearly)
And for 1st month it will be =
= $318.75
As given, the first month's payment is $390.60 and this covers the interest Additional amount ($390.60 - $318.75 = $71.85) is a payment against the principle.
Hence, the new principle after the 1st month is $71.85 less than $45000
= 45000-71.85 = $44928.15
Hence, the last option $44928.15 is the correct answer.
<span>gross =768
fed = 68
ss = 6.2/100 times 765
med = 1.45/100 times 765
state = 22/100 times 68
gross - ( sum of all deductions) = net pay
</span><span>Subtract 6.2% of $765 = 0.062 * $765 = 48.67
</span><span>Subtract 1.45% of $765 = 0.0145 * $765 = 11,0925
</span><span>Subtract 22% of $68 = 0.222 * $65 = 14.43
</span>Social security tax = 765 * 6.2% = 47.43
Medicare tax = 765 * 1.45% = 11.09
State tax (22% of federal tax) = 68 * 22% = 14.96
765 - 68 - 47.43 - 11.09 - 14.96 = 623.52
net pay is 623.52
Answer:
The proportion of the chocolate bars produced that have mass within one standard deviation of the mean is 0.6826 = 68.26%.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
A consumer organization finds that the mass of the chocolate bar is normally distributed with a mean of 200.4 g and a standard deviation of 0.05 g.
This means that 
a) Find the proportion of the chocolate bars produced that have mass within one standard deviation of the mean.
pvalue of Z when X = 200.4 + 0.05 = 200.9 subtracted by the pvalue of Z when X = 200.4 - 0.05 = 199.9.
X = 200.9



has a pvalue of 0.8413
X = 199.9



has a pvalue of 0.1587
0.8413 - 0.1587 = 0.6826
The proportion of the chocolate bars produced that have mass within one standard deviation of the mean is 0.6826 = 68.26%.
The answer is 162. Hope this helps