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3 years ago

14

hey thereee! please help me with this, or tell me what topic this is under because I haven't learn this but i would like to try

it out..

1 answer:

Jlenok [28]3 years ago

6 0

Answer:

B.)70

Step-by-step explanation:

I'm honestly guessing here but it seems like your best chance

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Find the proportion of test worth 50 points; test score of 40%

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I THINK it is 20% i used a calculator to help me find m answer

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3 years ago

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I just need to find this data set out, please anyone?

rewona [7]

Answer:

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Mean: 68.76

Median: 72

Range: 85

Step-by-step explanation:

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3 years ago

Which sentence can represent the inequality 2.4 (6.2 minus x) greater-than negative 4.5?

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Answer: Choice B

Two and four tenths multiplied by the difference of six and two tenths and a number is more than negative four and five tenths.

======================================================

Explanation:

2.4 = 2 + 0.4

2.4 = 2 and 4/10

2.4 = 2 and 4 tenths

2.4 = two and four tenths

-------------------

Through similar reasoning,

6.2 = six and two tenths

And also,

-4.5 = negative four and five tenths

---------------------

Notice how 6.2 - x translates into "difference of six and two tenths and a number"

We then multiply that by 2.4, aka two and four tenths.

So that's how we get the phrasing "Two and four tenths multiplied by the difference of six and two tenths and a number"

All of this is greater than -4.5 aka negative four and five tenths.

This points us to Choice B as the final answer.

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A given field mouse population satisfies the differential equation dp dt = 0.5p − 410 where p is the number of mice and t is the

ohaa [14]

Answer:

a) $t = 2 *ln(\frac{82}{5}) =5.595$

b) $t = 2 *ln(-\frac{820}{p_0 -820})$

c) $p_0 = 820-\frac{820}{e^6}$

Step-by-step explanation:

For this case we have the following differential equation:

$\frac{dp}{dt}=\frac{1}{2} (p-820)$

And if we rewrite the expression we got:

$\frac{dp}{p-820}= \frac{1}{2} dt$

If we integrate both sides we have:

$ln|P-820|= \frac{1}{2}t +c$

Using exponential on both sides we got:

$P= 820 + P_o e^{1/2t}$

Part a

For this case we know that p(0) = 770 so we have this:

$770 = 820 + P_o e^0$

$P_o = -50$

So then our model would be given by:

$P(t) = -50e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=-50 e^{1/2 t} +820$

$\frac{820}{50} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(\frac{82}{5}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(\frac{82}{5}) =5.595$

Part b

For this case we know that p(0) = p0 so we have this:

$p_0 = 820 + P_o e^0$

$P_o = p_0 -820$

So then our model would be given by:

$P(t) = (p_o -820)e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=(p_o -820)e^{1/2 t} +820$

$-\frac{820}{p_0 -820} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(-\frac{820}{p_0 -820}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(-\frac{820}{p_0 -820})$

Part c

For this case we want to find the initial population if we know that the population become extinct in 1 year = 12 months. Using the equation founded on part b we got:

$12 = 2 *ln(\frac{820}{820-p_0})$

$6 = ln (\frac{820}{820-p_0})$

Using exponentials we got:

$e^6 = \frac{820}{820-p_0}$

$(820-p_0) e^6 = 820$

$820-p_0 = \frac{820}{e^6}$

$p_0 = 820-\frac{820}{e^6}$

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3 years ago

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Answer:

The slope of the function is 4040 which represents the monthly fee to remain a member.

Step-by-step explanation:

i promise its right

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