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3 years ago

5

The average score of 100 students taking a statistics final was 70 with a standard deviation of 7. Assuming a normal distributio

n, what test score separates the top 25% of the students from the lower 75% of students?

1 answer:

san4es73 [151]3 years ago

8 0

Answer:

74.72

Step-by-step explanation:

Given

mean $\left ( \mu \right )=70$

Standard deviation $\left ( \sigma \right )=7$

For Top 25%

$P\left ( Z$

and Z value for 75 % is 0.6744

Thus score which separates the Top 25% of students is

$=\mu +\sigma \cdot Z$

$=70+7\cdot 0.6744=74.7208$

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Please, I need help in this ??

nignag [31]

Answer:

$\int\frac{x^{4}}{x^{4} -1}dx = x + \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1)-\frac{1}{2} arctanx + c$

Step-by-step explanation:

$\int\frac{x^{4}}{x^{4} -1}dx$

Adding and Subtracting 1 to the Numerator

$\int\frac{x^{4} - 1 + 1}{x^{4} -1}dx$

Dividing Numerator seperately by $x^{4} - 1$

$\int 1 + \frac{1}{x^{4}-1 }\, dx$

Here integral of 1 is x +c1 (where c1 is constant of integration

$x + c1 + \int\frac{1}{(x-1)(x+1)(x^{2}+1)}\, dx$ ----------------------------------(1)

We apply method of partial fractions to perform the integral

$\frac{1}{(x-1)(x+1)(x^{2}+1)}$ = $\frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{x^{2} + 1}$ ------------------------------------------(2)

$\frac{1}{(x-1)(x+1)(x^{2}+1)} = \frac{A(x+1)(x^{2} +1) + B(x-1)(x^{2} +1) + C(x-1)(x+1)}{(x-1)(x+1)(x^{2} +1)}$

1 = $A(x+1)(x^{2} +1) + B(x-1)(x^{2} +1) + C(x-1)(x+1)$ -------------------------(3)

Substitute x= 1 , -1 , i in equation (3)

1 = A(1+1)(1+1)

A = $\frac{1}{4}$

1 = B(-1-1)(1+1)

B = $-\frac{1}{4}$

1 = C(i-1)(i+1)

C = $-\frac{1}{2}$

Substituting A, B, C in equation (2)

$\int\frac{x^{4}}{x^{4} -1}dx$ = $\int\frac{1}{4(x-1)} - \frac{1}{4(x+1)} -\frac{1}{2(x^{2}+1) }$

On integration

Here $\int \frac{1}{x}dx = lnx and \int\frac{1}{x^{2}+1 } dx = arctanx$

$\int\frac{x^{4}}{x^{4} -1}dx$ = $\frac{1}{4} ln(x-1)$ - $\frac{1}{4} ln(x+1)$ - $\frac{1}{2} arctanx$ + c2---------------------------------------(4)

Substitute equation (4) back in equation (1) we get

$x + c1 + \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1) - \frac{1}{2} arctanx + c2$

Here c1 + c2 can be added to another and written as c

Therefore,

$\int\frac{x^{4}}{x^{4} -1}dx = x + \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1)-\frac{1}{2} arctanx + c$

4 0

4 years ago

Write the numbers in the correct order from least to greatest -4, 2 1/5, -2 3/4, 1/10

Mandarinka [93]

1: -4
2: -2 3/4
3: 1/10
4: 2 1/5

8 0

4 years ago

What is the perimeter of a Polygon with vertices at (3,2),(3,9),(7,12),(11,9), and (11,2)?

jeka57 [31]

Answer:

The perimeter is 32.

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Express the sum of the polymonial 3x^2+15x-56 and the square of the binomial (x-8) as a polynomial in standard form.

tester [92]

Given:

Polynomial is $3x^2+15x-56$ .

To find:

The sum of given polynomial and the square of the binomial (x-8) as a polynomial in standard form.

Solution:

The sum of given polynomial and the square of the binomial (x-8) is

$3x^2+15x-56+(x-8)^2$

$=3x^2+15x-56+x^2-2(x)(8)+8^2$ $[\because (a-b)^2=a^2-2ab+b^2]$

$=3x^2+15x-56+x^2-16x+64$

On combining like terms, we get

$=(3x^2+x^2)+(15x-16x)+(-56+64)$

$=4x^2-x+8$

Therefore, the sum of given polynomial and the square of the binomial (x-8) as a polynomial in standard form is $4x^2-x+8$ .

7 0

3 years ago

Pam and her sister have a pattern of saving money. Pam saved 40 dollars and her brother saved 100 dollars.

Keith_Richards [23]

Answer:

true

Step-by-step explanation:

Pam saved 40

her bro saves 100

let us take this as ration 40 is to 100 which can also be written in fractional for like 40/100 now we simplify it how by dividing both by 20 so (40/20)/(100/20) which will give 2/5 which can be written as 2 is to 5 so this is true

8 0

3 years ago

Read 2 more answers