Answer:
i think a
Step-by-step explanation:
Step-by-step explanation:
1st year interest calculated = 3.25/100 × $4500/1 = $146.25
2nd year principal = 1st year principal + 1st year interest = $4500 + $146.25 = $4646.25. 2nd year interest calculated = 3.25/100 ×$4646.25/1 =$151
3rd year principal = 2nd principal + 2nd year interest = $4646.25 + $151 = $4797.25. 3rd year interest calculated = 3.25/100 × $4797.25/1 = $155.91
4th year principal = 3rd year principal + 3rd year interest =$4797.25 + $155.91 =$4953.16. 4th year principal calculated = 3.25/100 × 4953.16/1 =$160.98
5th year principal = 4th year principal + 4 year interest = &4953.16 +$160.98 = $5114.14. 5th year interest calculated = 3.25/100 × 5114.14/1 =$166.21
6th year principal =5th year principal + 5th year interest = $5114.14 + $166.21 =$5280.35. 6th year interest calculated = 3.25/100 × 5280.35/1 =$171.61
Amount = main principal + compound interest =$4500 +&146.25 +$151+$155.91+$160.98+$166.21+$171.61 =$5451.96
Step-by-step explanation:
a)2×30=60J
b) 1.25×0.28 = 0.35J
Answer:
Step-by-step explanation:
Corresponding gasoline consumption when radial tires is used and gasoline consumption when regular belted tires is used form matched pairs.
The data for the test are the differences between the gasoline consumption when radial tires is used and gasoline consumption when regular belted tires is used.
μd = the gasoline consumption when radial tires is used minus the gasoline consumption when regular belted tires is used.
For the null hypothesis
H0: μd ≥ 0
For the alternative hypothesis
H1: μd < 0
The resulting p-value was .0152.
Since alpha, 0.05 > than the p value, 0.0152, then we would reject the null hypothesis. Therefore, at 5% significance level, we can conclude that the gasoline consumption when regular belted tires is used is higher than the gasoline consumption when radial tires is used.
Answer:
6
Step-by-step explanation:
n(AUB) = n(A) + n(B) - n(AnB)
n(AUB) can have the minimum number of elements if n(AnB) has the maximum number of elements.
n(AnB) maximum = 3
so n(AUB) = 3+6-3 = 6
