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tamaranim1 [39]
4 years ago
7

solve the question and show your work so julie checked the prices of her favorite CDs at Music Madness. What is the median for $

14,$12,$15$12,$17
Mathematics
2 answers:
Pavlova-9 [17]4 years ago
6 0
<span>To find the Median, place the numbers you are given in order and find the middle number. 12,12,14,15,17 its 14</span>
Phoenix [80]4 years ago
4 0
Original set = {14, 12, 15, 12, 17}

Sort the set from smallest to largest and you get {12, 12, 14, 15, 17}

The middle most value is in slot three, which is the value 14.

The median price is 14 dollars
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What's the difference when five squared is subtracted from four cubed?
elena55 [62]
4^3-5^2= 4\cdot 4 \cdot 4-5\cdot 5 = 64-25 = 39


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4 years ago
On the way to school the temperature was 20 degrees it dropped 31 degrees by the end of the day what was the temperature at yhe
serious [3.7K]
It started at 20.....then dropped 31 degrees

20 - 31 = - 11 <== its 11 below zero
5 0
3 years ago
The population of mosquitoes in a certain area increases at a rate proportional to the current population, and in the absence of
Alexandra [31]

Answer:

Population of mosquitoes in the area at any time t is:

P(t) =504,943.26  -104,943.26e^{0.693t}

Step-by-step explanation:

assume population at any time t = P(t)

population increases at a rate proportional to the current population:

⇒dP/dt ∝ P

 \implies \frac{dP}{dt} =kP----(1)

where k is constant rate at which population is doubled

solving (1)

ln|P(t)|=kt +C\\P(t)= e^{kt+C}\\P(t)=Ce^{kt}

t=0\\P(0) = P_{o}\\\implies C= P_{o}\\P(t) =P_{o}e^{kt}\\ ---- (2)

initial population = 400,000

population is doubled every week

                                                 ⇒P(1)=2P(0)

Using (2)

                                 P_{o}e^{k(1)} = 2P_{o} e^{k(0)}\\

                                            e^{k} =2\\k=ln|2|\\

In presence of predators amount is decreased by 50,000 per day

Then amount decreased per week = 350,000

In this case (1) becomes

\frac{dP}{dt}=kP-350,000\\\frac{dP}{dt} - kP=-350,000\\ ---(3)

solving (3) by calculating integrating factor

                                          I.F=e^{\int-k dt}

Multiplying I.F with all terms of (3)

e^{-kt}\frac{dP}{dt} - ke^{-kt}P =-350,000 e^{-kt}\\\frac{d}{dt}(e^{-kt}P) =  -350,000 e^{-kt}

Integrating w.r.to t

                         e^{-kt}P(t)= \frac{350,000e^{-kt}}{k} +C

                         P(t) =\frac{350,000}{k} +Ce^{kt}\\

                                          k=ln|2| =0.693

                          P(t) =504,943.26 + Ce^{0.693t}\\

at t=0

                        P(0) =504,943.26 + Ce^{0.693(0)}

                        400,000 =504,943.26 + C

                           C = -104,943.26

So, population of mosquitoes in the area at any time t is

                  P(t) =504,943.26  -104,943.26e^{0.693t}

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Answer:

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Step-by-step explanation:

4 0
3 years ago
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The population mean annual salary for environmental compliance specialists is about ​$63 comma 500. A random sample of 31 specia
Nat2105 [25]

Answer:

0.0035289

Step-by-step explanation:

From the question;

mean annual salary = $63,500

n = sample size = 31

Standard deviation = $6,200

Firstly, we calculate the z-score of $60,500

Mathematically;

z-score = x-mean/SD/√n = (60500-63500)/6200/√(31) = -2.6941

So we want to find the probability that P(z < -2.6941)

We can get this from the standard normal table

P( z < -2.6941) = 0.0035289

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