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SVEN [57.7K]
3 years ago
10

A phone company took a survey of its customers and found that 103 of 150 people said that they were pleased with their service t

he phone company has 72,000 customers what is w good predictions of the number of customers that are satisfied
Enter only the number of customers
Mathematics
2 answers:
Anarel [89]3 years ago
6 0
103 : 150
x : 72000

150x = 72000 \times 103 \\ 150x = 7416000 \\ x = 49440

Answer : Number of customers that are satisfied = 49,440

Hope this helps. - M
tino4ka555 [31]3 years ago
5 0
103 ÷ 150 = 0.68666666666

<span>0.68666666666 </span>× 72,000 = 49,440

Answer = 49,440

Hope this helped☺☺
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A device produces random 64-bit integers at a rate of one billion per second. After how many years of running is it unavoidable
nikitadnepr [17]

Answer:

3171 × 10^(44) years

Step-by-step explanation:

For each bit, since we are looking how many years of running it is unavoidable that the device produces an output for the second time, the possible integers are from 0 to 9. This is 10 possible integers for each bit.

Thus, total number of possible 64 bit integers = 10^(64) integers

Now, we are told that the device produces random integers at a rate of one billion per second (10^(9) billion per second)

Let's calculate how many it can produce in a year.

1 year = 365 × 24 × 60 × 60 seconds = 31,536,000 seconds

Thus, per year it will produce;

(10^(9) billion per second) × 31,536,000 seconds = 3.1536 × 10^(16)

Thus;

Number of years of running is it unavoidable that the device produces an output for the second time is;

(10^(64))/(3.1536 × 10^(16)) = 3171 × 10^(44) years

6 0
3 years ago
Maria studied the traffic trends in India. She found that the number of cars on the road increases by 10% each year. If there we
lord [1]

Answer:

b. 8,800,000.

Step-by-step explanation:

In maria study the number of cars on the road increases by 10% each year.

Total number of cars in the first year =80 million cars.

We will add 10% of 80 million to 80 milion to enable us get the number of cars for the second year.

For year 2 we have;

10% of 80,000,000+80,000,000

\frac{10}{100}*80,000,000+80,000,000

8,000,000 + 80,000,000

88,000,000

number of cars in year 2 = 88 million

We will add 10% of 88 million to 88 milion to enable us get the number of cars for the third year.

\frac{10}{100}*88,000,000+88,000,000

8,800,000+88,000,000

96,800,000

the number of cars on the road in year 3 compared to year 2 will just be the increase which is in BOLD. That is 8,800,000

<em>or simply subtract the number of cars in year 2 from that in year 3;</em>

96,800,000-88,000,000 = 8,800,000

8 0
4 years ago
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8 is the greatest component.
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Solve the equation for the unknown variable.<br><br> z³ = 10,648
Svetllana [295]
<span>z³ = 10,648
</span>z³ = 22³
z = 22
7 0
3 years ago
The number of charter schools is 4 less than twice the number of alternative schools and there are 48 charter schools how many a
jarptica [38.1K]
Let x = the number of alternative schools.

Therefore, 2x - 4 = 48.
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Therefore, there are 26 alternative schools(You could check this answer by substituting x=26 into the equation above.).
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3 years ago
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