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3 years ago

There are 9 pies cut into 1/8 pieces, how many pieces would you get ?

Mathematics

1 answer:

Alborosie3 years ago

6 0

Answer:

72 pieces

Step-by-step explanation:

since 1/8 pieces would mean 8 per pie then

8*9 = 72

So 72 pieces

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What word best fits this definition

postnew [5]

Answer:

EXTRAPOLATION.

Step-by-step explanation:

We have been given a crossword puzzle in the figure

And we need to fill the missing word in this which has to be a mathematics word because we need to follow the pattern

And all words are mathematical

Hence, The word that can be filled there is EXTRAPOLATION.

Extrapolation means in mathematics is finding value of the variable on basis of the relationship this variable have with other variable.

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2 years ago

Can someone help with geometry study

Studentka2010 [4]

This question is incomplete.

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3 years ago

Read 2 more answers

I need help with this question ASAP

IRINA_888 [86]

Answer:

you would have to add 3 to 1 cuz there is a one with the x and since 3 is negative that how you would get y=-3x+1

Step-by-step explanation:

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3 years ago

Given P(A)=0.5P(A)=0.5, P(B)=0.71P(B)=0.71 and P(A\cap B)=0.415P(A∩B)=0.415, find the value of P(A|B)P(A∣B), rounding to the nea

Marysya12 [62]

Answer:

0.585

Step-by-step explanation:

P(A|B) = P(A∩B) / P(B)

P(A|B) = 0.415 / 0.71

P(A|B) = 0.585

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3 years ago

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