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3 years ago

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A soccer is having a car wash. The team spent $55 on supplies. They earned $275, including tips. The team profit is the amount t

he team made after paying for supplies.

1 answer:

Vikki [24]3 years ago

3 0

The profit is:

275-55= 220$

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Which best represents the center of the data set below ? <br> mean median range mode

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Answer:{}

B. Median

Their is no real explanation.

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In a state lottery, 28 balls are numbered 1-28. Eight of them are randomly selected without replacement. To win, you only need t

podryga [215]

Answer:

2/7

Step-by-step explanation:

Each time you play, you get 1 number. Since 8/28 numbers win, you have a 8/28 chance to win if you buy one ticket. 8/28 = 2/7

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2 years ago

What is the value of y in the equation 2(3y + 7 + 5) = 196 − 16? (1 point) Group of answer choices 13 14 26 28

Tema [17]

Answer:

Option c) is correct.

The value of y in the given equation is 26 ie., y=26

Step-by-step explanation:

Given equation is

$2(3y+7+5)=196-16$

Now to the find the value of y in the above equation.

To find y simplify the above equation

$2(3y+7+5)=196-16$

$2(3y+12)=180$

Now apply distributive property

$6y+24=180$

$6y=180-24$

$=\frac{156}{6}$

Therefore y=26

Option c) is correct.

The value of y in the given equation is 26 ie., y=26

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3 years ago

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A rectangular barn measuring 14 feet b 10 feet is located in one corder of a fenced area. Robert ties his cow to the corner of t

Dennis_Churaev [7]

For this problem you would be solving for the area of 3/4 of a circle with the radius being 24ft. So area of a circle is a= PI R squared so plug in 3.14(24)squared order of operations says to square first so 24 squared is 576. Then PI or 3.14 times 576 equals 1808.64
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3 years ago

The population of mosquitoes in a certain area increases at a rate proportional to the current pop-ulation, and in the absence o

ollegr [7]

Answer:

The population of mosquitoes in the area at any time <em>t</em> is:

$P(t)=201977.31-1977.31\times 2^{t}$

Step-by-step explanation:

The rate of growth of mosquitoes can be expressed as:

$\frac{dP}{dt}=kP$

$\frac{dP}{P}=k\ dt$

Integrate the above expression as follows:

$\int {\frac{dP}{P}} \, =\int {k\ dt} \, \\\ln|P|=kt+c\\e^{\ln|P|}=e^{kt+c}\\P=Ce^{kt}$

$\Rightarrow P=P_{0}e^{kt}$

It is provided that the population doubles every day.

Compute the value of <em>k</em> as follows:

$2=1\times e^{k\times1}\\2=e^{k}\\k=\ln (2)$

It is also provided that every day 20,000 mosquitoes are eaten.

The rate of growth per week can be expressed as:

$\frac{dP}{dt}=\ln(2)P-14000\\\frac{dP}{dt}-\ln(2)P=14000$

The integrating factor for this is:

$e^{\int {\ln(2)dP}}=e^{\ln(2)\int {dt}}=e^{\ln(2)t}$

Then,

$P(t)\ e^{-\ln(2)t}=\int {e^{-\ln(2)t}}-14000\, dt\\=-14000\int {e^{-\ln(2)t}}\, dt\\=-14000\times \frac{e^{-\ln(2)t}}{-\ln(2)}+C\\P(t)=(e^{-\ln(2)t})\times [-14000\times \frac{e^{-\ln(2)t}}{-\ln(2)}+C]\\=\frac{14000}{\ln(2)}+Ce^{-\ln(2)t}$

The initial population is 200,000.

Compute the value of <em>C</em> as follows:

$P(t)=\frac{140000}{\ln(2)}+Ce^{-\ln(2)t}\\200000=\frac{14000}{\ln(2)}+Ce^{-\ln(2)(0)}\\C=200000-\frac{140000}{\ln(2)}\\C=-1977.31$

Now substitute <em>C</em> in P (t),

$P(t)=\frac{140000}{\ln(2)}+Ce^{\ln(2)t}\\P(t)=201977.31-1977.31\times 2^{t}$

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2 years ago