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grandymaker [24]

2 years ago

7

A brochure from the department of public safety in a northern state recommends that motorists should carry 12 items (flashlights

, blankets, and so forth) in their vehicles for emergency use while driving in winter. The following data give the number of items out of these 12 that were carried in their vehicles by 15 randomly selected motorists.
5.3 78 0 105 12 10 7 6 7 119
Find the mean, median, and mode for these data. Round your answers to two decimal places, where appropriate.
Mean items
Median items
Mode - items

1 answer:

scoray [572]2 years ago

6 0

Answer:

$Mean = 6.07$

$Median = 7$

$Mode = 7$

Step-by-step explanation:

Given

$Data: 5\ 3\ 7\ 8\ 0\ 1\ 0\ 5\ 12\ 10\ 7\ 6\ 7\ 11\ 9$

$n = 15$

Solving (a): The mean

Mean is calculated as:

$Mean = \frac{\sum x}{n}$

This gives:

$Mean = \frac{5+ 3+ 7+ 8+ 0+ 1+ 0+ 5+ 12+ 10+ 7+ 6+ 7+ 11+ 9}{15}$

$Mean = \frac{91}{15}$

$Mean = 6.07$

Solving (b): The median

Sort the data in ascending order:

$Data: 5\ 3\ 7\ 8\ 0\ 1\ 0\ 5\ 12\ 10\ 7\ 6\ 7\ 11\ 9$

$Sorted: 0\ 0\ 1\ 3\ 5\ 5\ 6\ 7\ 7\ 7\ 8\ 9\ 10\ 11\ 12$

The median is:

$Median = \frac{n + 1}{2}th$

$Median = \frac{15 + 1}{2}th$

$Median = \frac{16}{2}th$

$Median = 8th$

The 8th item on the sorted dataset is 7; So:

$Median = 7$

Solving (c): The mode

$Mode = 7$

Because it has a frequency of 3 (more than any other element of the dataset).

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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

If ABCD is a square and AB = 10, what is the measure of CE to the nearest hundredth?

andrey2020 [161]

Answer:

$AC=14.14\ units$

Step-by-step explanation:

<em><u>The correct question is </u></em>

If ABCD is a square and AB = 10, what is the measure of AC? (rounded to the nearest hundredth)

see the attached figure to better understand the problem

we know that

All four sides of the square are congruent and all four interior angles are equal to 90°.

so

AB=BC=CD=AD

In the right triangle ACD

Applying the Pythagorean Theorem

$AC^2=CD^2+AD^2$

we have that

$CD=AD=AB=10\ units$

substitute the given value

$AC^2=10^2+10^2$

$AC^2=200$

$AC=\sqrt{200}\ units$

$AC=14.14\ units$

6 0

2 years ago

3 values that would make this inequality true. 28+ x > 42

Mumz [18]

24, 36, 42 these aare answers that you can add to 28 that will give you something bigger then 42

4 0

2 years ago

Carl has new bottle of laundry detergent he uses 0.4 cups of laundry detergent while washing one load of laundry and 0.7 coupon

oee [108]

Answer:

-1.1 cups

Step-by-step explanation:

To find out how much was spent on the main bottle of detergent, we must add the expenses that were incurred in the wash.

The statement tells us that first there is an expense of 0.4 cups of laundry detergent while washing one load of laundry and 0.7 coupon another load, therefore this means that the change in the bottle would be:

(-0.4) + (-0.7) = -0.4 - 0.7 = -1.1 cups.

In other words, the main bottle would have 1.1 cups less, that is, -1.1 cups.

4 0

2 years ago

Read 2 more answers

Fruit juice comes in two types of containers. The first is a rectangular prism that is 9 inches tall and has a rectangular base

My name is Ann [436]

The volume of a rectangular prism:

$V_p=Bh\\\\B=wl\\\\w=2\ in;\ l=3\ in;\ h=9\ in$

substitute

$V_p=2\cdot3\cdot9=54\ in^3$

The volume of a cylinder:

$V_c=\pi r^2h\\\\r=1.5\ in;\ h=8\ in$

substitute:

$V_c=\pi\cdot(1.5)^2\cdot8=\pi\cdot2.25\cdot8=18\pi\ in^3\approx18\cdot3.14=56.52\ in^3$

Answer: C. The cylinder has the greater volume.

3 0

2 years ago