<span><span><span>Year 1 $642.00
Year 2 $686.94
Year 3 $735.03
Year 4 $786.48
Year 5 $841.53
</span></span></span>
Answer:the maximum number of ccf Jake can use if he wants his bill to be no more than $58.00 is $15.3
Step-by-step explanation:
Let x represent the number of ccf of water that Jake uses in a month.
Jake's water bill is $24.40 per month plus $2.20 per ccf (hundred cubic feet) of water. This means that if she uses x ccf of water in a month, her total water bill would be
24.40 + 2.2x
Therefore, the maximum number of ccf Jake can use if he wants his bill to be no more than $58.00 would be
24.40 + 2.2x ≤ 58
2.2x ≤ 58 - 24.4
2.2x ≤ 33.6
x ≤ 33.6/2.2
x ≤ 15.3
Answer:
1. 7
2. 15
3. 181
Step-by-step explanation:
Just make sure that you make your denominator a whole number
1. 9.8/1.4....,you multiply by 10...which only removes the comma but it keeps the structure of the question
Always remember that what you do to the denominator...you also do to the numerator
: 9.8/1.4 × 10/10 = 98/14 = 7
2. 40.5/2.7....,you multiply by 10
: 40.5/2.7 × 10/10 = 405/27 = 15
3. 3.62/0.02...., you multiply by 100
: 3.62/0.02 × 100/100 = 362/2 = 181
Answer:
0.625
Step-by-step explanation:
Given :
Ambell company uses or procures batteries from two manufacturers.
The life of a battery in critical tool is at = 32 hours
Ambell uses 70% of its batteries from manufacturer 1 and out of that 90% batteries lasts for about 40 hours.
Similarly, 75% of the batteries procured from the manufacturer 2 lasts for about 40 hours.
Therefore, the probability that a battery is form manufacturer 1 is :
= 0.625
Answer:
b. x>1
Step-by-step explanation:
2/3+x/3>1
Multiply the inequality by 3 to get rid of the fractions
3(2/3+x/3)>1*3
2 +x >3
Subtract 2 from each side
2+x-2 >3-2
x>1
