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aliya0001 [1]
3 years ago
8

PLZZZ ANSWER!! Answers for points will be reported.

Mathematics
1 answer:
Lisa [10]3 years ago
7 0

Answer:

  • 272.6

Step-by-step explanation:

This is an isosceles triangle with base of 21 and sides of 28.

<u>Area formula</u>

  • A = 1/2bh

<u>Find height:</u>

  • h = \sqrt{28^2 - (21/2)^2}  = \sqrt{673.75} = 25.96

<u>Area:</u>

  • A = 1/2(25.96)(21) = 272.6 (rounded)
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Answer:

2. Find the five boundary values

Step-by-step explanation:

When drawing a dot, the steps are;

i) The data set (the list of numbers to be presented in the dot plot) are arranged in increasing order from left to right, with thee lowest value on the left and the highest value located at the right of the list

ii) The five boundary values are listed as follows;

The Minimum and Maximum Values

1) The minimum value is the lowest number in the list, which is the first number from the left

2) The maximum value is the final number on the right of the list

3) The Median, Q₂

The median number is the number which divides the data arranged in increasing number, into two equal parts

The First Quartile and Third Quartile

The numbers are divided from the median, such that the Lower half is to the left of the Median, and the Upper half is to the right of median

4) The First Quartile, Q₁ = The median of the Lower half

5) The Third Quartile, Q₂ = The median of the Upper half

iii) The box plot is then drawn with the five boundary values found above

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Write 2% as a reduced fraction
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2% as a reduced fraction is 2/100 = 1/50
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When Colton commutes to work, the amount of time it takes him to arrive is normally distributed with a mean of 41 minutes and a
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Answer:

(34, 48)

Step-by-step explanation:

According to the Empirical Rule, 95% of normally distributed data lie within two standard deviations of the mean.  That, in turn, means 95% of the data in this problem lie within 2(3.5 min), or 7 min, of the mean:

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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
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