What is 1.65 divided by 100?

Alchen [17]

$\bf \cfrac{x^2-x-6}{4x^3}\cdot \cfrac{2x^2+2x}{x^2+5x+6}\implies \cfrac{(x-3)\underline{(x+2)}}{\underline{2x}(2x^2)}\cdot \cfrac{\underline{2x}(x+1)}{(x+3)\underline{(x+2)}}
\\\\\\
\cfrac{x-3}{2x^2}\cdot \cfrac{x+1}{x+3}\implies \cfrac{\stackrel{FOIL}{x^2-2x-3}}{2x^3+6x^2}$

3 0

3 years ago

The University of Washington claims that it graduates 85% of its basketball players. An NCAA investigation about the graduation

Nonamiya [84]

Probabilities are used to determine the chances of events

The given parameters are:

Sample size: n = 20
Proportion: p = 85%

<h3>(a) What is the probability that 11 out of the 20 would graduate? </h3>

Using the binomial probability formula, we have:

$P(X = x) = ^nC_x p^x(1 - p)^{n -x}$

So, the equation becomes

$P(x = 11) = ^{20}C_{11} \times (85\%)^{11} \times (1 - 85\%)^{20 -11}$

This gives

$P(x = 11) = 167960 \times (0.85)^{11} \times 0.15^{9}$

$P(x = 11) = 0.0011$

Express as percentage

$P(x = 11) = 0.11\%$

Hence, the probability that 11 out of the 20 would graduate is 0.11%

<h3>(b) To what extent do you think the university’s claim is true?</h3>

The probability 0.11% is less than 50%.

Hence, the extent that the university’s claim is true is very low

<h3>(c) What is the probability that all 20 would graduate? </h3>

Using the binomial probability formula, we have:

$P(X = x) = ^nC_x p^x(1 - p)^{n -x}$

So, the equation becomes

$P(x = 20) = ^{20}C_{20} \times (85\%)^{20} \times (1 - 85\%)^{20 -20}$

This gives

$P(x = 20) = 1 \times (0.85)^{20} \times (0.15\%)^0$

$P(x = 20) = 0.0388$

Express as percentage

$P(x = 20) = 3.88\%$

Hence, the probability that all 20 would graduate is 3.88%

<h3>(d) The mean and the standard deviation</h3>

The mean is calculated as:

$\mu = np$

So, we have:

$\mu = 20 \times 85\%$

$\mu = 17$

The standard deviation is calculated as:

$\sigma = np(1 - p)$

So, we have:

$\sigma = 20 \times 85\% \times (1 - 85\%)$

$\sigma = 20 \times 0.85 \times 0.15$

$\sigma = 2.55$

Hence, the mean and the standard deviation are 17 and 2.55, respectively.

Read more about probabilities at:

brainly.com/question/15246027

8 0

2 years ago

Some help me please help me please help me please

kodGreya [7K]

1_ 210+90= 300

2_ it's prime:

[1,19]

good luck

4 0

3 years ago

How do I do 8b(ii) ? Please help me thank you!

Mila [183]

Step One
======
Find the length of FO (see below)

All of the triangles are equilateral triangles. Label the center as O
FO = FE = sqrt(5) + sqrt(2)

Step Two
======
Drop a perpendicular bisector from O to the midpoint of FE. Label the midpoint as J. Find OJ

Sure the Pythagorean Theorem. Remember that OJ is a perpendicular bisector.

FO^2 = FJ^2 + OJ^2
FO = sqrt(5) + sqrt(2)
FJ = 1/2 [(sqrt(5) + sqrt(2)] \
OJ = ??

[Sqrt(5) + sqrt(2)]^2 = [1/2(sqrt(5) + sqrt(2) ] ^2 + OJ^2
5 + 2 + 2*sqrt(10) = [1/4 (5 + 2 + 2*sqrt(10) + OJ^2
7 + 2sqrt(10) = 1/4 (7 + 2sqrt(10)) + OJ^2 Multiply through by 4
28 + 8* sqrt(10) = 7 + 2sqrt(10) + 4 OJ^2 Subtract 7 + 2sqrt From both sides
21 + 6 sqrt(10) = 4OJ^2 Divide both sides by 4
21/4 + 6/4* sqrt(10) = OJ^2
21/4 + 3/2 * sqrt(10) = OJ^2 Take the square root of both sides.
sqrt OJ^2 = sqrt(21/4 + 3/2 sqrt(10) )
OJ = sqrt(21/4 + 3/2 sqrt(10) )

Step three
find h
h = 2 * OJ
h = 2* sqrt(21/4 + 3/2 sqrt(10) ) <<<<<< answer.

7 0

3 years ago

Each letter represents a question answer

evablogger [386]

A. 2x+3x-6=10+4x

5x-6=10+4x

5x=16+4x

X=16

5 0

3 years ago