The numbers of blocked intrusion attempts on each day during the first two weeks of

2 answers:

4 0

(a) Construct a 95% confidence interval for the difference between the average number of intrusion attempts per day before and after the change of firewall setting (assume equal variance)

rodikova [14]2 years ago

4 0

here is your answer hope it's helpful for you thankyou

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janelle wishes to finance a car for $33,000. the bank's annual interest rate is 3.5%, and she can choose between durations of fi

Harman [31]

Answer:

For a duration of 5 years, Monthly Payment =$600.42

For a duration of 6 years, Monthly Payment =$508.83

Step-by-step explanation:

$P=\dfrac{ar (1+r)^n}{(1+r)^n-1} \\$

where a= Amount to Finance=$33,000

Annual interest rate = 3.5%=0.035

r=Monthly Interest Rate= 0.035 ÷ 12 = $\frac{7}{2400}$

n=number of months to pay

For a duration of 5 years

n=5X12=60 months

$P=\dfrac{ar (1+r)^n}{(1+r)^n-1} \\\\P=\dfrac{33000 X\frac{7}{2400} (1+\frac{7}{2400} )^{60}}{(1+\frac{7}{2400})^{60}-1} \\=\dfrac{96.25 (1.1909)}{1.1909-1}\\=\dfrac{96.25 (1.1909)}{0.1909}\\=\dfrac{114.62}{0.1909}=\$600.42$

For a duration of 6 years

n=6X12=72 months

$P=\dfrac{ar (1+r)^n}{(1+r)^n-1} \\\\P=\dfrac{33000 X\frac{7}{2400} (1+\frac{7}{2400} )^{72}}{(1+\frac{7}{2400})^{72}-1} \\=\dfrac{96.25 (1.2333)}{1.2333-1}\\=\dfrac{96.25 (1.2333)}{0.2333}\\=\dfrac{118.71}{0.2333}=\$508.83$