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natta225 [31]
3 years ago
13

The number of unique visitors to the college website can be approximated by the formula N(t)=410(1.32)t where t represents the n

umber of years after 1997 when the website was created. Approximate to the nearest integer the number of unique visitors to the college website in the year 2020.
Mathematics
1 answer:
Reika [66]3 years ago
3 0

Answer:

243212

Step-by-step explanation:

Substitute the given value of t into the given formula. To find t, subtract 1997 from 2020.

2020−1997=23

Now substitute 23 into the equation for t and calculate.

N(t)N(23)==≈410(1.32)t410(1.32)23243,212

The number of unique visitors to the college website in the year 2020 was approximately 243,212.

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The equation 3/x=5/7 is equivalent to the equation
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Answer:

Step-by-step explanation:

3/x=5/7 is the same as A 21=5x Both equal to 21/5 or 4 1/5

7 0
3 years ago
The work of a student to solve the equation 2(5y − 2) = 12 + 6y is shown below:
jolli1 [7]
We know that Step 1 is correct, because it is just a restatement of the equation. Therefore, we can eliminate Step 1:

2(5y – 2) = 12 + 6y

In Step 2, the student tried using the Distributive Property. The Distributive Property can be written as one of the two following formulas:

a(b + c) = ab + ac
a(b – c) = ab – ac

In this case, we'll use the second formula. Substitute any known values into the equation above and simplify:

2(5y – 2) = 2(5y) – 2(2)
2(5y – 2) = 10y – 4

In Step 2, the student calculated 2(5y – 2) to equal 7y – 4. However, we have just proven that 2(5y – 2) is equal to 10y – 4.

The student first made an error in Step 2, and the correct step is:

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I hope this helps!
8 0
3 years ago
One urn contains one blue ball (labeled b1) and three red balls (labeled r1, r2, and r3). a second urn contains two red balls (r
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Answer:

12 possibilities

Step-by-step explanation:

In the first urn, we have 4 balls, and all of them are different, as they have different labels, so the group of two red balls r1 and r2 is different from the group of red balls r2 and r3.

The same thing occurs in the second urn, as all balls have different labels.

The problem is a combination problem (the group r1 and r2 is the same group r2 and r1).

For the first urn, we have a combination of 4 choose 2:

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For the second urn, we also have a combination of 4 choose 2, so 6 possibilities.

In total we have 6 + 6 = 12 possibilities.

3 0
3 years ago
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