in a grocery store steak cost $3.85 per pound if you buy a three pound steak and pay for it with a $20 bill how much change will you get

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

Answer:

Step-by-step explanation:

A man divided $9,000 among his wife, son and daughter.

The wife received twice as much as the daughter.

Let the daughter received d amount.

Then the wife received = 2d

and son received $1,000 more than the daughter.

The son received the amount = 1000+d

So the expression will be = d + 2d +(1000+d) = 9,000

3d + (1000+d) = 9000

4d = 9000 - 1000

4d = 8000

d =  [tex]\frac{8000}{4}[/tex]

d = 2000

Daughter received $2,000

Wife received 2d = 2 × 2000 = $4,000

Son received 1000 + d = 1000 + 2000 = $3,000

If x is the amount the wife received, then the expression represents the amount received by the son :

S = 1000 + (x/2)

The probability that the archer hits the target on exactly three out of five shots is 0.3087, or 30.87%, calculated by using the binomial probability formula.

The probability that an archer hits a target on a given shot is 0.7 and the goal is to calculate the probability that the archer hits the target on exactly three out of five shots. This is a binomial probability problem, as each shot can end in either a success (hitting the target) with a probability of 0.7, or a failure (missing the target) with a probability of 0.3.

To calculate the probability of exactly three successes (hits) out of five, we use the binomial probability formula:

P(X=k) = (n choose k) * (p)^k * (1-p)^(n-k)

Where:

n = total number of trials (5 shots)

k = number of successes (3 hits)

p = probability of success on a single trial (0.7)

Applying the formula, we get:

P(3 hits out of 5) = (5 choose 3) * (0.7)^3 * (0.3)^2

= 10 * (0.343) * (0.09)

= 10 * 0.03087

= 0.3087

Therefore, the probability that the archer hits the target on exactly three out of five shots is 0.3087, or 30.87%.

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.

12/1 / 4/5=

12/1 * 5/4 =

60/4 = 15

 they can make 15 bowls

9/12= 0.75

8.00 * 0.75 = 6.00

 the 9" costs $6.00

Answer:

It's (7,5)

Step-by-step explanation:

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].

For more details, refer to the llink:

https://brainly.com/question/15739221

y = k/x

7=k/-2

k = 7/-2 = -3.5

y =-3.5/7 =-0.5

y=-0.5

Answer:

X = In4-1    C on edge, just took the test

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 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.

Learn more about Joint probability distribution here:https://brainly.com/question/17279418

#SPJ6

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]