In a recent local taste contest testing Coke against Pepsi it was found that 3/5 of all people surveyed preferred to taste the Coke. if 7,500 people were in the survey how many chose Pepsi ?

Answer with Step-by-step explanation:

we are given a equation:

e=mc

We have to find which equation is not equivalent to the above formula.

Dividing both sides by c,we get

m=e/c

i.e. m equals e over c

Dividing both sides by m,we get

c=e/m

i.e. c equals e over m

Hence, The equation which is not equivalent to e=mc is:

m equals c over e

9/12= 0.75

8.00 * 0.75 = 6.00

 the 9" costs $6.00

An equation relating W to T is,

W = 700 + 35T

And, Total amount of water after 19 minutes is, 1365 liters

We have to given that,

Owners of a recreation area are filling a small pond with water. They are adding water at a rate of 35 liters per minute.

There are 700 liters in the pond to start.

Now, Let W represent the total amount of water in the pond (in liters), and let T represent the total number of minutes that water has been added.

Hence, an equation relating W to T is,

W = 700 + 35T

So, For T = 19 minutes,

W = 700 + 35  x 19

W = 700 + 665

W = 1365

Therefore, Total amount of water after 19 minutes is, 1365 liters

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Final answer:

To find f/g, divide each term in f(x) by g(x). Resulting in f(x)/g(x) = -4x - 2 with the domain being all real numbers except x = 6/5. Hence, the correct answer is c. -4x - 2; all real numbers except x = 6/5.

Explanation:

To find the function f/g, we divide the function f(x) by g(x). Given f(x) = -20x2 + 14x + 12 and g(x) = 5x - 6, we divide these to get:

f(x)/g(x) = (-20x2 + 14x + 12) / (5x - 6)

Dividing each term in f(x) by g(x):

f(x)/g(x) = -4x - 2

The domain of this function would be all real numbers except where g(x) = 0, since we cannot divide by zero. g(x) = 0 when x = 6/5. Thus, the domain is all real numbers except x = 6/5.

The correct answer to the student's question is therefore c. -4x - 2; all real numbers except x = 6/5.

Answer:

3x+27+3.5x=397.5

6.5x=370.5

x=57 mph for jeep and travels 199.5 miles in 3.5 hours

x+9=66 mph for B and travels 198 miles in 3 hours

Step-by-step explanation:

speed of jeep=x, time is     >>>>>    t+0.5 in hours or 3.5 hours here

speed of B=x+9, time is  >>>>>>>>>>>> 3 hours

distance is >>>>>>>>>    speed time

Hope it Helps.

Answer on: june 11, 2021

The student's question involves integrating a function in a region bounded by two spheres in the first octant, implying the use of spherical coordinates and integration over a sphere with a constant radius.

The question pertains to evaluating an integral within the region bounded by two spheres in the first octant. When dealing with spheres and integrals, the use of spherical coordinates is often beneficial. The question suggests using spheres with a constant radius and spherical coordinates (r, θ, φ), where a typical point in space is represented as (r sin(θ) cos(φ), r sin(θ) sin(φ), r cos(θ)). To integrate over the sphere, we consider the bounds given by the radii of the inner and outer spheres, (r = 3 and r = 5, respectively, since the square roots of 9 and 25 are 3 and 5), and the fact that it is within the first octant which further restricts the limits of θ and φ. The rest of the provided excerpts seem to be unrelated specifically to this problem but are examples of standard integrals and applications of integration in physics and potential theory.

The final answer after evaluating the integral is: [tex]\[\frac{49\pi}{3}\][/tex]. This is the value of the integral over the region between the spheres [tex]\( x^2 + y^2 + z^2 = 9 \) and \( x^2 + y^2 + z^2 = 25 \)[/tex] in the first octant.

To evaluate the given integral over the region between the spheres [tex]\( x^2 + y^2 + z^2 = 9 \)[/tex]and [tex]\( x^2 + y^2 + z^2 = 25 \)[/tex]  in the first octant, we can use spherical coordinates. In spherical coordinates, the volume element is given by [tex]\( r^2 \sin(\phi) \, dr \, d\theta \, d\phi \),[/tex] where r is the radial distance, [tex]\( \theta \)[/tex] is the azimuthal angle, and [tex]\( \phi \)[/tex] is the polar angle.

The limits for the integral are as follows:

[tex]- \( 3 \leq r \leq 5 \) (limits of the radii for the spheres)\\- \( 0 \leq \theta \leq \frac{\pi}{2} \) (first octant)\\- \( 0 \leq \phi \leq \frac{\pi}{2} \) (first octant)[/tex]

The integral to evaluate is not specified, so let's assume it's a simple function like \( f(x, y, z) = 1 \) for the sake of demonstration. The integral would then be:

[tex]\[\iiint_E 1 \, dV = \int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} \int_{3}^{5} r^2 \sin(\phi) \, dr \, d\theta \, d\phi\][/tex]

Now, let's evaluate this integral step by step:

[tex]\[\int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} \int_{3}^{5} r^2 \sin(\phi) \, dr \, d\theta \, d\phi\][/tex]

[tex]\[= \int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} \left[ \frac{1}{3} r^3 \sin(\phi) \right]_{3}^{5} \, d\theta \, d\phi\][/tex]

[tex]\[= \int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} \left( \frac{125}{3} - \frac{27}{3} \right) \sin(\phi) \, d\theta \, d\phi\][/tex]

[tex]\[= \int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} \frac{98}{3} \sin(\phi) \, d\theta \, d\phi\][/tex]

[tex]\[= \int_{0}^{\frac{\pi}{2}} \left[ \frac{98}{3} \theta \right]_{0}^{\frac{\pi}{2}} \, d\phi\][/tex]

[tex]\[= \int_{0}^{\frac{\pi}{2}} \frac{98}{3} \cdot \frac{\pi}{2} \, d\phi\][/tex]

[tex]\[= \frac{98\pi}{6}\][/tex]

[tex]\[= \frac{49\pi}{3}\][/tex]

So, the value of the integral over the specified region is[tex]\( \frac{49\pi}{3} \).[/tex]

Answer:

It's (7,5)

Step-by-step explanation:

The number of  individuals you would expect to handle the message is 2.5.

Let Z represent the number of individuals that handle the message

Table for the possible joint value of X and Y

Z                       Y

                         1          2          3            4         5  

X         1             -          2           3            3         2

          2           2         -           2            3         3

          3           3         2           -             2         3

           4           3         3           2            -          2

           5            2         3           3           2          -

Each cell contain=1/4×1/5=1/20

Hence:

Number of individual=10×2×1/20+10×3×1/20

Number of individual=20×0.05+30×0.05

Number of individual=2.5

Therefore the number of  individuals you would expect to handle the message is 2.5.

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dived 13 by 17

13/17 = 0.7647

  = 76.47%

 if you need to round the number it would be 76%, which is a 76 grade

Answer:

The correct answer is “the point lies directly on the regression line”

Step-by-step explanation:

When you do a regression analysis, then you get a line of regression that best fits it. The data points usually tend to fall in the regression line, but they do not precisely fall there but around it. A residual is the vertical distance between a data point and the regression line. Every single one of the data points had one residual. If one of this residual is equal to zero, then it means that the regression line truly passes through the point.  

The Taylor polynomial [tex]T_3(x)[/tex] will be written as [tex]2x-8x^2+\dfrac{44x^3}{3}+......[/tex].

Given:

The given function is [tex]f(x) = e^{-4x}sin(2x)[/tex].

It is required to find the Tylor polynomial [tex]t_3(x)[/tex] centered at a=0.

Now, the expansion of the function [tex]e^{-4x}[/tex] can be written as,

[tex]e^{-4x}=\sum\dfrac{(-4x)^n}{n!}\\e^{-4x}=1+(-4x)^1+\dfrac{(-4x)^2}{2!}+\dfrac{(-4x)^3}{3!}+.....\\e^{-4x}=1-4x+\dfrac{16x^2}{2}-\dfrac{64x^3}{6}+.....\\e^{-4x}=1-4x+8x^2-\dfrac{32x^3}{3}+.....[/tex]

Similarly, the expansion of the function [tex]sin(2x)[/tex] will be,

[tex]sin(2x)=\sum\dfrac{(-1)^n(2x)^{2n+1}}{(2n+1)!}\\=\dfrac{2x}{1!}+\dfrac{-(2x)^3}{3!}+.....\\=2x-\dfrac{4x^3}{3}+......[/tex]

So, the function [tex]f(x) = e^{-4x}sin(2x)[/tex] will be written as,

[tex]f(x) = e^{-4x}sin(2x)\\f(x)=(1-4x+8x^2-\dfrac{32x^3}{3}+.....)(2x-\dfrac{4x^3}{3}+......)\\f(x)=2x-8x^2+16x^3-\dfrac{4x^3}{3}+.......\\f(x)=2x-8x^2+\dfrac{(48-4)x^3}{3}+......\\f(x)=2x-8x^2+\dfrac{44x^3}{3}+......[/tex]

Therefore, the Taylor polynomial [tex]T_3(x)[/tex] will be written as [tex]2x-8x^2+\dfrac{44x^3}{3}+......[/tex].

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12/1 / 4/5=

12/1 * 5/4 =

60/4 = 15

 they can make 15 bowls