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Aloiza [94]
2 years ago
4

The following data pairs represent the average temperature x (in degrees Fahrenheit) and electricity costs y (in dollars) for di

fferent homes: (65, 147), (59, 141), (71, 176), (78, 189), (82, 183), (85, 211), (88, 231), (91, 227), (84, 198), (79, 188), (63, 152), (59, 126) Which of the following values is most likely the closest to the correlation coefficient of the data?
Mathematics
2 answers:
Mariana [72]2 years ago
6 0

Answer:

E R=1

Step-by-step explanation:

KengaRu [80]2 years ago
3 0

Answer:

0.97

Step-by-step explanation:

To find the correlation coefficient, we must plot all of the data values in lists in the graphing calculator.  The x-values go into the first list and the y-values go into the second list.

From here, we bring up the scatter plot.  We then calculate the linear regression; when we do this, we see that r = 0.97.

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An executive for a large national restaurant chain with multiple locations in each of 513 counties wants to personally sample th
ohaa [14]

Answer:

D. Obtain a cluster sample by randomly selecting 20 counties and visiting every restaurant within those counties.

Step-by-step explanation:

Since this executive wants to minimize cost and expenses he should use a cluster sample of 20 counties and using the selected counties as a representative of the whole. Cluster sampling is a good and useful way of doing market research. It is more economical than other sampling methods and as the executive cannot get information on the whole population, he would be able to get information from the clusters.

4 0
2 years ago
A 400 gallon tank initially contains 100 gal of brine containing 50 pounds of salt. Brine containing 1 pound of salt per gallon
posledela

Answer:

The amount of salt in the tank when it is full of brine is 393.75 pounds.

Step-by-step explanation:

This is a mixing problem. In these problems we will start with a substance that is dissolved in a liquid. Liquid will be entering and leaving a holding tank. The liquid entering the tank may or may not contain more of the substance dissolved in it. Liquid leaving the tank will of course contain the substance dissolved in it. If Q(t) gives the amount of the substance dissolved in the liquid in the tank at any time t we want to develop a differential equation that, when solved, will give us an expression for Q(t).

The main equation that we’ll be using to model this situation is:

Rate of change of <em>Q(t)</em> = Rate at which <em>Q(t)</em> enters the tank – Rate at which <em>Q(t)</em> exits the tank

where,

Rate at which Q(t) enters the tank = (flow rate of liquid entering) x

(concentration of substance in liquid entering)

Rate at which Q(t) exits the tank = (flow rate of liquid exiting) x

(concentration of substance in liquid exiting)

Let y<em>(t)</em> be the amount of salt (in pounds) in the tank at time <em>t</em> (in seconds). Then we can represent the situation with the below picture.

Then the differential equation we’re after is

\frac{dy}{dt} = (Rate \:in)- (Rate \:out)\\\\\frac{dy}{dt} = 5 \:\frac{gal}{s} \cdot 1 \:\frac{pound}{gal}-3 \:\frac{gal}{s}\cdot \frac{y(t)}{V(t)}  \:\frac{pound}{gal}\\\\\frac{dy}{dt} =5\:\frac{pound}{s}-3 \frac{y(t)}{V(t)}  \:\frac{pound}{s}

V(t) is the volume of brine in the tank at time <em>t. </em>To find it we know that at time 0 there were 100 gallons, 5 gallons are added and 3 are drained, and the net increase is 2 gallons per second. So,

V(t)=100 + 2t

We can then write the initial value problem:

\frac{dy}{dt} =5-\frac{3y}{100+2t} , \quad y(0)=50

We have a linear differential equation. A first-order linear differential equation is one that can be put into the form

\frac{dy}{dx}+P(x)y =Q(x)

where <em>P</em> and <em>Q</em> are continuous functions on a given interval.

In our case, we have that

\frac{dy}{dt}+\frac{3y}{100+2t} =5 , \quad y(0)=50

The solution process for a first order linear differential equation is as follows.

Step 1: Find the integrating factor, \mu \left( x \right), using \mu \left( x \right) = \,{{\bf{e}}^{\int{{P\left( x \right)\,dx}}}

\mu \left( t \right) = \,{{e}}^{\int{{\frac{3}{100+2t}\,dt}}}\\\int \frac{3}{100+2t}dt=\frac{3}{2}\ln \left|100+2t\right|\\\\\mu \left( t \right) =e^{\frac{3}{2}\ln \left|100+2t\right|}\\\\\mu \left( t \right) =(100+2t)^{\frac{3}{2}

Step 2: Multiply everything in the differential equation by \mu \left( x \right) and verify that the left side becomes the product rule \left( {\mu \left( t \right)y\left( t \right)} \right)' and write it as such.

\frac{dy}{dt}\cdot \left(100+2t\right)^{\frac{3}{2}}+\frac{3y}{100+2t}\cdot \left(100+2t\right)^{\frac{3}{2}}=5 \left(100+2t\right)^{\frac{3}{2}}\\\\\frac{dy}{dt}\cdot \left(100+2t\right)^{\frac{3}{2}}+3y\cdot \left(100+2t\right)^{\frac{1}{2}}=5 \left(100+2t\right)^{\frac{3}{2}}\\\\\frac{dy}{dt}(y \left(100+2t\right)^{\frac{3}{2}})=5\left(100+2t\right)^{\frac{3}{2}}

Step 3: Integrate both sides.

\int \frac{dy}{dt}(y \left(100+2t\right)^{\frac{3}{2}})dt=\int 5\left(100+2t\right)^{\frac{3}{2}}dt\\\\y \left(100+2t\right)^{\frac{3}{2}}=(100+2t)^{\frac{5}{2} }+ C

Step 4: Find the value of the constant and solve for the solution y(t).

50 \left(100+2(0)\right)^{\frac{3}{2}}=(100+2(0))^{\frac{5}{2} }+ C\\\\100000+C=50000\\\\C=-50000

y \left(100+2t\right)^{\frac{3}{2}}=(100+2t)^{\frac{5}{2} }-50000\\\\y(t)=100+2t-\frac{50000}{\left(100+2t\right)^{\frac{3}{2}}}

Now, the tank is full of brine when:

V(t) = 400\\100+2t=400\\t=150

The amount of salt in the tank when it is full of brine is

y(150)=100+2(150)-\frac{50000}{\left(100+2(150)\right)^{\frac{3}{2}}}\\\\y(150)=393.75

6 0
2 years ago
When the chef turned the new freezer on, the temperature on the internal thermometer was 2 c .
Elan Coil [88]

-13 c I think.  It is this way because if it went down 15, then you subtract 2-15.  2-2 = 0, and then 0-13= -13 c.  Hope this helps

8 0
2 years ago
To find the distance between (-2,-3) and (4,1), Peggy's teacher drew a diagram and marked blue lines to represent the legs of a
zhuklara [117]
Thtas what you must mean. 
The distance beteen th epoints will be sqrt(4^2 + 6^2) = sqrt 52  = 2 sqrt13
3 0
2 years ago
Read 2 more answers
How many groups of 1/2 days are in 1 week
SSSSS [86.1K]

Answer:14

Step-by-step explanation:

7 0
2 years ago
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