Answer:
guy7p9;y
Step-by-step explanation:
Given : The addition 21.4 + 15 + 17.17 + 4.003 by the rules for significant figures.
Solution: According to given numbers, let us count number of significant figures in each of the given numbers.
21.4 has three significant figures.
15 has two significant figures.
17.17 has four significant figures.
4.003 has four significant figures.
We have to round the final answer to the nearest tens place because 15 has only accurate to the ones place.
We can see that other numbers are accurate to the tenths, hundredths and thousandths places, respectively.
Let us add them as usually first,
21.4 + 15 + 17.17 + 4.003.
It gives 57.573.
Now, we need to round it to the ones place.
We get 58 as the final answer.
37000 (total price of the car) - 17000 (the target value) = 20,000 (the difference)
20,000 (difference) / 2000 (how much it depreciates per year) = 10 (the amount of years it will take)
Verify:
17000 + (2000 * 10) = 37000? Yes.
Answer:
2.063X10^-4
0.999
0.833
3.26X10^-3
9.85x10^-3
Step-by-step explanation:
B="At least one component needs service during the warranty period"
C="All three components need service during the warranty period"
D="Only the receiver needs service during the warranty period"
E="Exactly one of the three components needs service during the warranty period"
P(A1)=0.0595
P(A2)=0.0598
P(A3)=0.058
a) P(A1∩A2∩A3)=P(A1)P(A2)P(A3)=0.0595*0.0598*0.058=2.063X10^-4
b) P(B)=1-P(A1∩A2∩A3)=1-2.063X10^-4=0.999
c) P(C)=P(A1'∩A2'∩A3')=P(A1')P(A2')P(A3')=0.9405*0.9402*0.942=0.833
d) P(D)=P(A1'∩A2∩A3)=P(A1')P(A2)P(A3)=0.9405*0.0598*0.058=3.26X10^-3
e) P(E)=P(A1'∩A2∩A3)+P(A1∩A2'∩A3)+P(A1∩A2∩A3')=3.26X10^-3 + 0.0595*0.9402*0.058+0.0595*0.0598*0.942=9.85x10^-3
There is 100 cm in a meter. 536•100=536,000 cm
