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3 years ago

WILL MARK BRAINLEIST PLEASE HELP HURRY

Mathematics

1 answer:

Zina [86]3 years ago

7 0

Answer:

Hello, Your answer will be D.

Step-by-step explanation:

The data is symmetric for Shop A but not for Shop B ( note the values 10 and 11 for Shop B which are a lot lower than the other values).

Mean for Shop A and Median for Shop B. Have a Great Day!

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3+ (-5) explain please i need it asap so... hurry! pls

Anastasy [175]

Answer:

-2

Step-by-step explanation:

3 + -5

Adding -5 is like subtracting 5

3 -5

-2

6 0

3 years ago

Read 2 more answers

Please help and thank you

Ksivusya [100]

Answers: Angle A is 36 degrees, angle C is 54 degrees

=================================================================

Work Shown:

The angles A and C are complementary, meaning they add to 90 degrees.

(Angle A) + (Angle C) = 90

(2x-4) + (2x+14) = 90

(2x+2x)+(-4+14) = 90

4x+10 = 90

4x+10-10 = 90-10 .... subtract 10 from both sides

4x = 80

4x/4 = 80/4 ..... divide both sides by 4

x = 20

If x = 20, then,

angle A = 2x-4 = 2*20-4 = 40-4 = 36 degrees

angle C = 2x+14 = 2*20+14 = 40+14 = 54 degrees

Note how A+C = 36+54 = 90 which helps confirm our answers.

8 0

3 years ago

3y''-6y'+6y=e*x sexcx

Simora [160]

From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

$3r^2-6r+6=0\iff r^2-2r+2=0$

which has roots at $r=1\pm i$ . This admits the two fundamental solutions

$y_1=e^x\cos x$
$y_2=e^x\sin x$

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

$y_p=u_1y_1+u_2y_2$

where

$u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$

and $W(y_1,y_2)$ is the Wronskian of the fundamental solutions. We have

$W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}$

and so

$u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx$
$u_1=\dfrac13\ln|\cos x|$

$u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx$
$u_2=\dfrac13x$

Therefore the particular solution is

$y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

so that the general solution to the ODE is

$y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

7 0

3 years ago

There are 7 acts in a talent show.

lara31 [8.8K]

Answer:

Event A: 1/35

Event B: 1/840

Explanation:

Event A

For the event A, the order of the first 4 acts does not matter.

The number of different four acts taken from a set of seven acts, when the order does not matter, is calculated using the concept of combinations.

Thus, the number of ways that the first four acts can be scheduled is:

$C(m,n)=\dfrac{m!}{n!(m-n)!}$