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3 years ago

What is p-7/12>3/10 please help

1 answer:

5 0

We have such inequality
$p- \frac{7}{12}> \frac{3}{10}$ /+ $\frac{7}{12}$
$p> \frac{3}{10}+ \frac{7}{12} \\ p> \frac{3*6}{10*6}+ \frac{7*5}{12*5} \\ p> \frac{18}{60}+ \frac{35}{60} \\ p> \frac{18+35}{60} \\ p> \frac{53}{60}$ - its the answer

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Please help with Part A and B. Thank you.

Jet001 [13]

Hello!

Part A:
The equation for a circle is (x - a)^2 + (y - b)^2 = r^2

We can say that the center of the earth is point (0, 0)

This gives us the equation

x^2 + y^2 = 3960^2

Square the number

x^2 + y^2 = 15681600

The answer is C
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Part B

The answer is C) 69 miles

Hope this helps!

6 0

3 years ago

Linear Functions look like what?

algol [13]

Answer: C

Step-by-step explanation:

Its either C or A I would go with C though

6 0

3 years ago

Y-2x^2=-3 graph of the function?

blsea [12.9K]

Answer:

We cant make graphs on here but if you put the quadiraletal to the right and drag the left numbers toward the front.

4 0

3 years ago

Draw the unit circle with center (0,0)and plot the point P=(6,2). Observe there are TWO lines tangent to the circle passing thro

scZoUnD [109]

Answer:

Step-by-step explanation:

Given that a tangent is drawn from P (6,2) to unit circle with center at the origin.

The tangent passing through (6,2) would have equaiton of the form

$y-6 = m(x-2)\\y =mx-2m+6$ for suitable m.

Because this line is tangent, distance of centre of circle namely origin is the radius 1.

i.e. $|\frac{-2m+6}{\sqrt{1+m^2} } |=1\\(-2m+6 )^2 = 1+m^2\\3m^2-24m+35=0\\\\m=1.918, 6.082$

the coordinates are the intersection of the tangent with the circle

They are (-0.162, 0.987) and (0.462,-0.887)

8 0

4 years ago

The n candidates for a job have been ranked 1, 2, 3,..., n. Let X 5 the rank of a randomly selected candidate, so that X has pmf

jeka57 [31]

Question:

The n candidates for a job have been ranked 1, 2, 3,..., n. Let x = rank of a randomly selected candidate, so that x has pmf:

$p(x) = \left \{ {{\frac{1}{n}\ \ x=1,2,3...,n} \atop {0\ \ \ Otherwise}} \right.$

(this is called the discrete uniform distribution).

Compute E(X) and V(X) using the shortcut formula.

[Hint: The sum of the first n positive integers is $\frac{n(n +1)}{2}$ , whereas the sum of their squares is $\frac{n(n +1)(2n+1)}{6}$

Answer:

$E(x) = \frac{n+1}{2}$

$Var(x) = \frac{n^2 -1}{12}$ or $Var(x) = \frac{(n+1)(n-1)}{12}$

Step-by-step explanation:

Given

PMF

$p(x) = \left \{ {{\frac{1}{n}\ \ x=1,2,3...,n} \atop {0\ \ \ Otherwise}} \right.$

Required

Determine the E(x) and Var(x)

E(x) is calculated as:

$E(x) = \sum \limits^{n}_{i} \ x * p(x)$

This gives:

$E(x) = \sum \limits^{n}_{x=1} \ x * \frac{1}{n}$

$E(x) = \sum \limits^{n}_{x=1} \frac{x}{n}$

$E(x) = \frac{1}{n}\sum \limits^{n}_{x=1} x$

From the hint given:

$\sum \limits^{n}_{x=1} x =\frac{n(n +1)}{2}$

So:

$E(x) = \frac{1}{n} * \frac{n(n+1)}{2}$

$E(x) = \frac{n+1}{2}$

Var(x) is calculated as:

$Var(x) = E(x^2) - (E(x))^2$

Calculating: $E(x^2)$

$E(x^2) = \sum \limits^{n}_{x=1} \ x^2 * \frac{1}{n}$

$E(x^2) = \frac{1}{n}\sum \limits^{n}_{x=1} \ x^2$

Using the hint given:

$\sum \limits^{n}_{x=1} \ x^2 = \frac{n(n +1)(2n+1)}{6}$

So:

$E(x^2) = \frac{1}{n} * \frac{n(n +1)(2n+1)}{6}$

$E(x^2) = \frac{(n +1)(2n+1)}{6}$

So:

$Var(x) = E(x^2) - (E(x))^2$

$Var(x) = \frac{(n+1)(2n+1)}{6} - (\frac{n+1}{2})^2$

$Var(x) = \frac{(n+1)(2n+1)}{6} - \frac{n^2+2n+1}{4}$

$Var(x) = \frac{2n^2 +n+2n+1}{6} - \frac{n^2+2n+1}{4}$

$Var(x) = \frac{2n^2 +3n+1}{6} - \frac{n^2+2n+1}{4}$

Take LCM

$Var(x) = \frac{4n^2 +6n+2 - 3n^2 - 6n - 3}{12}$

$Var(x) = \frac{4n^2 - 3n^2+6n- 6n +2 - 3}{12}$

$Var(x) = \frac{n^2 -1}{12}$

Apply difference of two squares

$Var(x) = \frac{(n+1)(n-1)}{12}$

3 0

3 years ago