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3 years ago

8

Use a calculator to find the standard deviation of this data set: 8, 14, 13, 10, 17. Round your answer to the nearest tenth.

2 answers:

quester [9]3 years ago

8 0

Answer:

3.5 Apex Trust

Step-by-step explanation:

Buddy kdjejhehejehheheheieieiwiww

AlladinOne [14]3 years ago

5 0

Answer:

3.5

Step-by-step explanation:

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A camera manufacturer spends $2250 each day for overhead expenses plus $6 per camera for labor and materials. The cameras sell f

mart [117]

16x = 2250 + 6x
x = 225
So he must sell 225 cameras

16(275) = 4400
6(275) = 1650

4400 - (1650 + 2250) = 500
$500 profit

6 0

4 years ago

Read 2 more answers

What is the distance between (-5,5) and (2,6) rounding to the nearest hundred

ExtremeBDS [4]

$\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ -5}}\quad ,&{{ 5}})\quad 
% (c,d)
&({{ 2}}\quad ,&{{ 6}})
\end{array}\qquad 
% distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
d=\sqrt{[2-(-5)]^2+[6-5]^2}\implies d=\sqrt{(2+5)^2+(6-5)^2}
\\\\\\
d=\sqrt{7^2+1^2}\implies d=\sqrt{50}$

5 0

4 years ago

HURRY PLEASE HELP

Mrac [35]

Answer:

The answer is D

Step-by-step explanation:

I tested this in apex learning and got 100 so its true.

8 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

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

6 0

3 years ago

12. The width of a rectangle is 16.55 inches. The length of the rectangle is half its width.

zepelin [54]

The answer is D

16.55 divide by 2= 8.275
8.275+8.275+16.55+16.55=49.65

4 0

4 years ago