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2 years ago

11. The sales tax rate is 6.9%

Mathematics

1 answer:

Norma-Jean [14]2 years ago

7 0

Answer:

40.1%

Step-by-step explanation:

I am assuming that 192 is in 100%.

100% = 192

I then represent the value that we are looking for with $x$ .

x% = 77

By dividing both equations (100% = 192 and x% = 77) and remembering that both left hand sides of BOTH equations have the percentage unit (%).

$\frac{100}{x} = \frac{77}{192}$

Now, of course, we take the reciprocal, or inverse, of both sides:

$\frac{x}{100} = \frac{7}{192}$

x = 40.1%

Thus making the answer: 40.1% of 192 is 77.

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the required condition for using an anova procedure on data from several populations for mean comparison is that the

Anuta_ua [19.1K]

Answer:

The sampled populations have equal variances

Step-by-step explanation:

ANOVA which is fully known as Analysis of variances can be defined as the collection of statistical models as well as their associated estimation procedures which enables easily and effectively analyzis of the differences among various group means in a sample reason been that ANOVA is a total variance in which the observed variance in a specific variable is been separated into components which are attributable to various sources of variation which is why ANOVA help to provides a statistical test to check whether two or more population means are equal.

Therefore the required condition for using an ANOVA procedure on data from several populations for mean comparison is that THE SAMPLED POPULATION HAVE EQUAL VARIANCE.

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2 years ago

A recent poll of 124 randomly selected residents of a town with a population of 310 showed that 93 of them are opposed to a new

tester [92]

The sample size is 124.
93 of them are opposed to new shopping center.

So,
n = 124
p = $\frac{93}{124}=0.75$

The point estimate of the population proportion = p = 0.75
q = 1 - p = 0.25

Margin of error (E) can be calculated by:

$E= Z_{c} \sqrt{ \frac{pq}{n} }$

Using the values, we get:

$E=1.645 \sqrt{ \frac{0.75*0.25}{124} }=0.06$

Therefore, the margin of error is approximately 0.06 or 6%.

6 0

2 years ago

Read 2 more answers

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 Q(t) = Rate at which Q(t) enters the tank – Rate at which Q(t) 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(t) be the amount of salt (in pounds) in the tank at time t (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 t. 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 P and Q 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

The population p of a small community on the outskirts of a city grows rapidly over a 20-year period: t05101520p1002004509502000

myrzilka [38]

Answer:

The population of the small community, 5 years into the future, after the initial 20-year period = 4268.

Step-by-step explanation:

t | 0 | 5 | 10 | 15 | 20

p | 100 | 200 | 450 | 950 | 2000

The exponential function will look like

p = aeᵏᵗ

where a and k are constants.

Take the natural logarithms of both sides

In p = In aeᵏᵗ

In p = In a + In eᵏᵗ

In p = In a + kt

In p = kt + In a.

We then use linear regression to fit the data of In p against t to obtain k and In a.

t | 0 | 5 | 10 | 15 | 20

p | 100 | 200 | 450 | 950 | 2000

In p | 4.605 | 5.298 | 6.109 | 6.856 | 7.601

In p = kt + In a.

y = mx + b

m = k and b = In a

Performing a linear regression analysis on the now-linear relationship between In p and t and also plotting a graph of the variables.

The regression equation obtained is

y = 0.151x + 4.584

The first attached image shows the equations necessary for the estimation of the linear regression parameters.

The second attached image shows the use of regression calculator and the plot of the function In p versus t.

Comparing

y = 0.151x + 4.584

With

In p = kt + In a.

y = In p

k = 0.151

x = t

In a = 4.584

a = 97.905

The exponential function relating p and t,

p = aeᵏᵗ now becomes

p = 97.905 e⁰•¹⁵¹ᵗ

So, to predict the population 5 years into the future, that is 5 years after the 20 year period.

we need p at t=25 years.

0.151 × 25 = 3.775

p(t=25) = 97.905 e³•⁷⁷⁵ = 4268.41 = 4268.

Hope this Helps!!!

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2 years ago

Write 11,760,825 in word form and expanded form

tankabanditka [31]

11,000,000 + 700,000 + 60,000 + 800 + 20 + 5 (expanded form)

eleven million seven hundred sixty thousand eight hundred twenty-five (word form)

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2 years ago