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Mathematics

1 answer:

SOVA2 [1]2 years ago

4 0

Exact answer as a fraction = 1/3

Approximate answer in decimal form = 0.333

Approximate answer as a percentage = 33.3%

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Explanation:

Check out the attached image below. It shows the various dice sums that add to either 4, 5 or 6. We have 12 occurrences total out of 6*6 = 36 overall giving 12/36 = 1/3 as the probability. This is in fraction form. As a decimal, 1/3 = 0.333 approximately which covers over to 33.3%

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The expression 5n+2 can be used to find the total cost in dollars of bowling where n us is the number of games bowled and 2 repr

sukhopar [10]

Plug in the known variables.

5n + 2 n = number of games bowled

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2 years ago

A rumor spreads through a small town. Let y(t) be the fraction of the population that has heard the rumor at time t and assume t

sladkih [1.3K]

Answer:

The answer is shown below

Step-by-step explanation:

Let y(t) be the fraction of the population that has heard the rumor at time t and assume that the rate at which the rumor spreads is proportional to the product of the fraction y of the population that has heard the rumor and the fraction 1−y that has not yet heard the rumor.

$\frac{dy}{dt}\ \alpha\ y(1-y)$

$\frac{dy}{dt}=ky(1-y)$

where k is the constant of proportionality, dy/dt = rate at which the rumor spreads

$\frac{dy}{dt}=ky(1-y)\\\frac{dy}{y(1-y)}=kdt\\\int\limits {\frac{dy}{y(1-y)}} \, =\int\limit {kdt}\\\int\limits {\frac{dy}{y}} +\int\limits {\frac{dy}{1-y}} =\int\limit {kdt}\\\\ln(y)-ln(1-y)=kt+c\\ln(\frac{y}{1-y}) =kt+c\\taking \ exponential \ of\ both \ sides\\\frac{y}{1-y} =e^{kt+c}\\\frac{y}{1-y} =e^{kt}e^c\\let\ A=e^c\\\frac{y}{1-y} =Ae^{kt}\\y=(1-y)Ae^{kt}\\y=\frac{Ae^{kt}}{1+Ae^{kt}} \\at \ t=0,y=10\%\\0.1=\frac{Ae^{k*0}}{1+Ae^{k*0}} \\0.1=\frac{A}{1+A} \\A=\frac{1}{9} \\$