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LiRa [457]
2 years ago
13

Diseases tend to spread according to the exponential growth model. In the early days of AIDS, the growth factor (i.e. common rat

io; growth multiplier) was around 2.0. In 1983, about 1800 people in the U.S. died of AIDS. If the trend had continued unchecked, how many people would have died from AIDS in 2003?
Mathematics
1 answer:
weqwewe [10]2 years ago
5 0

Answer:

1,887,436,800

Step-by-step explanation:

f(t) = a·b^t

f (t) = number of cases at year t

a = starting value = 1800 in 1983

b = growth factor = 2

t = years since 1983 = 2003 - 1983 = 20

f(t) = 1800·(2)^20 =

1,887,436,800

wyzant

philip p

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Scott buys candy that costs $7 per kilogram. He will spend at least $35 on candy. What are the possible numbers of kilograms he
slega [8]

Answer:

<em>p ≥ 5</em>

<em>Scott will buy at least 5 kilograms of candy.</em>

Step-by-step explanation:

<u>Inequalities</u>

The candy Scott buys cost $7 per kilogram.

Let's set p=number of kilograms of candy Scott will buy.

The money spent to buy p kilograms of candy is 7p dollars.

The condition states he will spend at least $35 on candies, thus the following inequality is formed:

7p ≥ 35

Dividing by 7:

p ≥ 35/7

Operating:

p ≥ 5

Scott will buy at least 5 kilograms of candy.

3 0
3 years ago
5.1(3 + 2.2x) &gt; –14.25 – 6(1.7x + 4)?
MaRussiya [10]

Answer:

x > -2.5

Step-by-step explanation:

1) 5.1 (3 + 2.2x) > -14.25 - 6 (1.7x + 4)

2) 15.3 + 11.22x > -14.25 - 10.2x - 24

3) 15.3 + 11.22x > -38.25 - 10.2x

4) 11.22x + 10.2x > -38.25 - 15.3

5) 21.42x > -53.55

6) x > -2.5

Alternate forms

x > -5/2x or x > -2 1/2

5 0
3 years ago
Evaluate the function f(x) = 6x - 5 at f(1)
Likurg_2 [28]
Remove the parentheses around the expression 1
5 0
3 years ago
Can somebody prove this mathmatical induction?
Flauer [41]

Answer:

See explanation

Step-by-step explanation:

1 step:

n=1, then

\sum \limits_{j=1}^1 2^j=2^1=2\\ \\2(2^1-1)=2(2-1)=2\cdot 1=2

So, for j=1 this statement is true

2 step:

Assume that for n=k the following statement is true

\sum \limits_{j=1}^k2^j=2(2^k-1)

3 step:

Check for n=k+1 whether the statement

\sum \limits_{j=1}^{k+1}2^j=2(2^{k+1}-1)

is true.

Start with the left side:

\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}\ \ (\ast)

According to the 2nd step,

\sum \limits_{j=1}^k2^j=2(2^k-1)

Substitute it into the \ast

\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}=2(2^k-1)+2^{k+1}=2^{k+1}-2+2^{k+1}=2\cdot 2^{k+1}-2=2^{k+2}-2=2(2^{k+1}-1)

So, you have proved the initial statement

4 0
3 years ago
A rectangle is 36 inches x 48 inches.<br><br> what is the perimeter.?<br><br> what is its area.?
olga nikolaevna [1]
For area, just multiply the length x width
So therefore, multiply 36 inches x 48 inches and there you have it.

Now for the perimeter, it is a little different
You'd have to add the sides up together to get your answer for the perimeter.
So, 36 in + 36 in + 48 in + 48 in = answer
3 0
3 years ago
Read 2 more answers
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