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3 years ago

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Mr.Lee wants to survey 20 of the 210 students in seventh grade about the hours they slerp each night. He finds 2 of the 20 stude

nts surveyed sleep for 9 hours each night. About how many of the 210 seventh graders sleeo for 9 hours each night?

1 answer:

allsm [11]3 years ago

6 0

21 students sleep for 9 hours each night

<em><u>Solution:</u></em>

Given that, Mr.Lee wants to survey 20 of the 210 students in seventh grade about the hours they sleep each night

He finds 2 of the 20 students surveyed sleep for 9 hours each night

Therefore, 2 out of 20 students sleep for 9 hours each night

Find out the percentage of 2 out of 20

$\rightarrow \frac{2}{20} \times 100 = 10 \%$

Thus 10 % of 210 students sleep for 9 hours each night

Find out 10 % of 210

$\rightarrow 10\% \text{ of } 210\\\\\rightarrow 10 \% \times 210\\\\\rightarrow \frac{10}{100} \times 210\\\\\rightarrow 21$

Thus 21 students sleep for 9 hours each night

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Answer:

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calculate the areas of 2 rectangles

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one traingle with lenghts 5 and 7

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A news item is spreading by word of mouth through a population of size 12,000 people. After t days, the number of people (in tho

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Answer:

a) 2.303

b) $f'(t)=\frac{432e^{-0.48t}}{(1+75e^{-0.48t})^{2}}$

c) 1.3609

e) $\frac{dy}{dx}=0.04y(12-y)$

f) 4.5 thousand people(smaller)

7.5 thousand people (larger)

e) t=7.93 days (smaller)

t=10.06 days (larger)

g) 1.44

Step-by-step explanation:

The given function is $y=f(t)=\frac{12}{1+75e^{-0.48t}}$

To find how many thousand people have heard the news after 6 days, we substitute to get:

$f(6)=\frac{12}{1+75e^{-0.48*6}}=2.303$ thousands.

b) We rewrite to get: $f(t)=12(1+75e^{-0.48t})^{-1}$

We differentiate using the chain rule to obtain:

$f'(t)=-12(1+75e^{-0.48t})^{-2}\cdot 75e^{-0.48t}\cdot -0.48$

This simplifies to:

$f'(t)=432e^{-0.48t}(1+75e^{-0.48t})^{-2}$

We rewrite as positive index to get:

$f'(t)=\frac{432e^{-0.48t}}{(1+75e^{-0.48t})^{2}}$

c) To find the rate at which the news is spreading after 8 days, we substitute t=8 into $f'(t)=\frac{432e^{-0.48t}}{(1+75e^{-0.48t})^{2}}$ to get:

$f'(8)=\frac{432e^{-0.48\cdot 8}}{(1+75e^{-0.48\cdot 8})^{2}}=1.3609$

d) The solution to the logistic differential equation:

$\frac{dy}{dt}=ky(M-y)$

is

$y=\frac{M}{1+be^{-Mkt}}$

By comparison, M=12, and $Mk=0.48$

$12k=0.48\\k=0.04$

The differential equation is:

$\frac{dy}{dx}=0.04y(12-y)$

e) To how many people have heard the news when its rate of spread is 1.35 thousand per day, we equate the differential equation to 1.35 and so for t first.

$\frac{432e^{-0.48t}}{(1+75e^{-0.48t})^{2}}=1.35$

This gives us: t=10.06 ot t=7.93

We substitute the times into the function to get:

$f(7.93)=\frac{12}{1+75e^{-0.48\cdot7.93}}$ =4.5 thousand: smaller value

$f(10.06)=\frac{12}{1+75e^{-0.48\cdot10.06}}=7.5$ thousand: Larger value

f) The two times that the news is spreading at rate 1.35 thousand people per day are:

$t=7.93$ days: Smaller value

$t=10.06$ : Larger value

g) To find the fastest rate at which the news spread we take the second derivative and equate it to zero.

$f''(t)=0$

This corresponds to where the horizontal line is tangent to

$f'(t)=\frac{432e^{-0.48t}}{(1+75e^{-0.48t})^{2}}$

From the graph the point of tangency is:

(8.99,1.44)

Therefore the fastest rate at which the news spread is 1.44

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