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forsale [732]
2 years ago
12

Guys I need help in this ​

Mathematics
1 answer:
tigry1 [53]2 years ago
8 0

Answer:

For Table 1, the answer is Not Proportional (B).

For Table 2, the answer is proportional, with y is 3 times x.

Step-by-step explanation:

In table 1, for the first value of x, y is equal to 32, and 32/8 is 4. Remember that, because in order for x and y to be proportional, y/x always has to equal the same number. Let's try the next value. 50/10 = 5, which is obviously not 4, so in table 1, x and y are not proportional. (we didn't even have to try the last one, yay!!!)

In table 2, for the first values 21 and 7, 21/7 is 3. 30/10, the second values, also equals to 3, and so does the last, 39/13 = 3. Therefore, in table 2, x and y are proportional, and y is 3 times x. (7 * 3 = 21, 10 * 3 = 30, and 13 * 3 = 39.)

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Aleksandr-060686 [28]

Answer:

Collin: about $401 thousand

Cameron: about $689 thousand

Step-by-step explanation:

A situation in which doubling time is constant is a situation that can be modeled by an exponential function. Here, you're given an exponential function, though you're not told what the variables mean. That function is ...

P(t)=P_0(2^{t/d})

In this context, P0 is the initial salary, t is years, and d is the doubling time in years. The function gives P(t), the salary after t years. In this problem, the value of t we're concerned with is the difference between age 22 and age 65, that is, 43 years.

In Collin's case, we have ...

P0 = 55,000, t = 43, d = 15

so his salary at retirement is ...

P(43) = $55,000(2^(43/15)) ≈ $401,157.89

In Cameron's case, we have ...

P0 = 35,000, t = 43, d = 10

so his salary at retirement is ...

P(43) = $35,000(2^(43/10)) ≈ $689,440.87

___

Sometimes we like to see these equations in a form with "e" as the base of the exponential. That form is ...

P(t)=P_{0}e^{kt}

If we compare this equation to the one above, we find the growth factors to be ...

2^(t/d) = e^(kt)

Factoring out the exponent of t, we find ...

(2^(1/d))^t = (e^k)^t

That is, ...

2^(1/d) = e^k . . . . . match the bases of the exponential terms

(1/d)ln(2) = k . . . . . take the natural log of both sides

So, in Collin's case, the equation for his salary growth is

k = ln(2)/15 ≈ 0.046210

P(t) = 55,000e^(0.046210t)

and in Cameron's case, ...

k = ln(2)/10 ≈ 0.069315

P(t) = 35,000e^(0.069315t)

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