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3 years ago

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The scores for a particular examination are normally distributed with a mean of 68.5% and a standard deviation of 8.2%. What is

the probability that a student who wrote the examination had a mark between 80% and 100%? Give your answer to the nearest hundredth.

1 answer:

stepladder [879]3 years ago

6 0

Answer:

$P(80/100$

Step-by-step explanation:

We are given that

Mean, $\mu=68.5$ %=68.5/100

Standard deviation, $\sigma=8.2$ %=8.2/100

We have to find the probability that a student who wrote the examination had a mark between 80% and 100%.

$P(80/100$

$P(80/100$

We know that

$P(a$

Using the formula

$P(80/100$

$P(80/100$

$P(80/100$

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Use pemdas (18 - 14) x (4 x 2) write 3 steps

Deffense [45]

Answer:

32

Step-by-step explanation:

Parentheses:

(18 - 14) = 4

(4 x 2) = 8

4 x 8

Multiplication:

4 x 8 = 32

<em>good luck hope this helps :)</em>

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3 years ago

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Solve the following system of equations. Write each of your answers as a fraction reduced to lowest terms. In other words, write

morpeh [17]

Answer:

x = 57/28

y = -95/84

z = 97/168

Step-by-step explanation:

Use the application in the next link: https://www.zweigmedia.com/RealWorld/tutorialsf1/scriptpivotold.html

Start with the expanded array:

$\left[\begin{array}{cccc}1&5&8&1\\3&2&2&5\\-2&-7&2&5\\\end{array}\right]$

then using the tool provided, make row operations until you find the solution:

r2 = r2-3r1

$\left[\begin{array}{cccc}1&5&8&1\\0&-13&-22&2\\-2&-7&2&5\\\end{array}\right]$

r3 = r3+2r1

$\left[\begin{array}{cccc}1&5&8&1\\0&-13&-22&2\\0&3&18&7\\\end{array}\right]$

r2 = r2*(-1/13)

$\left[\begin{array}{cccc}1&5&8&1\\0&1&22/13&-2/13\\0&3&18&7\\\end{array}\right]$

r1 = r1- r2*5

$\left[\begin{array}{cccc}1&0&-6/13&23/13\\0&1&22/13&-2/13\\0&3&18&7\\\end{array}\right]$

r3 = r3+ r2*-3

$\left[\begin{array}{cccc}1&0&-6/13&23/13\\0&1&22/13&-2/13\\0&0&168/13&97/13\\\end{array}\right]$

r3 = r3*13/168

$\left[\begin{array}{cccc}1&0&-6/13&23/13\\0&1&22/13&-2/13\\0&0&1&97/168\\\end{array}\right]$

r2 = r2- r3*22/13

$\left[\begin{array}{cccc}1&0&-6/13&23/13\\0&1&0&-95/84\\0&0&1&97/168\\\end{array}\right]$

r2 = r2+ r3*6/13

$\left[\begin{array}{cccc}1&0&0&57/28\\0&1&0&-95/84\\0&0&1&97/168\\\end{array}\right]$

Here you have a reduced array an therefore the answers to each variable are on each row:

$\left[\begin{array}{c}x\\y\\z\end{array}\right]$

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3 years ago

Find the sum of<br>−3x2−x−10 and 10x^2+x-10

Andru [333]

<em>Look at the attached picture⤴</em>

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3 years ago

Fill in the Blanks:

Damm [24]

Answer:

Unlike fractions: Fractions with different denominators are called, unlike fractions.

In case of two equivalent fractions, the product of the numerator of one fraction and denominator of the second fraction is equal to the product of the denominator of the first fraction and numerator of the second fraction.

A fraction is said to be in form if its numerator and denominator are 5. To add or subtract like fractions, we add or subtract the keeping the same.

Step-by-step explanation:

4 POINT I DONT NO

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2 years ago

Consider the differential equation x2y′′ − 9xy′ + 24y = 0; x4, x6, (0, [infinity]). Verify that the given functions form a funda

pantera1 [17]

Answer:

The functions satisfy the differential equation and linearly independent since W(x)≠0

Therefore the general solution is

$y= c_1x^4+c_2x^6$

Step-by-step explanation:

Given equation is

$x^2y'' - 9xy+24y=0$

This Euler Cauchy type differential equation.

So, we can let

$y=x^m$

Differentiate with respect to x

$y'= mx^{m-1}$

Again differentiate with respect to x

$y''= m(m-1)x^{m-2}$

Putting the value of y, y' and y'' in the differential equation

$x^2m(m-1) x^{m-2} - 9 x m x^{m-1}+24x^m=0$

$\Rightarrow m(m-1)x^m-9mx^m+24x^m=0$

$\Rightarrow m^2-m-9m+24=0$

⇒m²-10m +24=0

⇒m²-6m -4m+24=0

⇒m(m-6)-4(m-6)=0

⇒(m-6)(m-4)=0

⇒m = 6,4

Therefore the auxiliary equation has two distinct and unequal root.

The general solution of this equation is

$y_1(x)=x^4$

and

$y_2(x)=x^6$

First we compute the Wronskian

$W(x)= \left|\begin{array}{cc}y_1(x)&y_2(x)\\y'_1(x)&y'_2(x)\end{array}\right|$

$= \left|\begin{array}{cc}x^4&x^6\\4x^3&6x^5\end{array}\right|$

=x⁴×6x⁵- x⁶×4x³

=6x⁹-4x⁹

=2x⁹

≠0

The functions satisfy the differential equation and linearly independent since W(x)≠0

Therefore the general solution is

$y= c_1x^4+c_2x^6$

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3 years ago