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4 years ago

6

the formula below changes degrees Fahrenhiet to degrees Celsius 5/9(F-32). what is the temperature in degrees Celsius, for -4 Fa

hrenheit?

1 answer:

lesya [120]4 years ago

5 0

Substitute -4 for F and do the arithmetic.
.. C = (5/9)*(-4-32) = (5/9)(-36) = -20

-4 °F corresponds to -20 °C.

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How do i do 15, 16???

a_sh-v [17]

It's very long so I'm not giving answers but here's how i would do it:

use Pythagoras $a^2 +b^2 = c^2$ where c is the longest side
use sine rule $\frac{a}{sin(A)} = \frac{b}{sin(B)} = \frac{c}{sin(C)}$ where lower cases are the sides, and capitals are their opposite angles.

e.g for 15:

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2 years ago

I WILL GIVE BRAINLIEST TO WHOEVER ANSWERS BEST AND POINTS!!

pogonyaev

Answer:

ok, Im no exactly sure this is right becasue fractions aren't suppused to ever be a thing but here we go

Step-by-step explanation:

you want to get it to a same denominator.... so

2/3 times 5=10/15

3/5 times 5= 9/15

its asking about ATUL so, 10/15 divide 2=

ANSWER=5/15

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6 0

3 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

P and Q are points on the line y= 2 -4x complete the coordinates of P and Q, P(0, ) Q( ,0)

docker41 [41]

(d): y=2-4x
P(0;y) belongs to (d)
=> y=2-4*0
= 2
So P(0;2)
Q(x;0) belongs to (d)
=> 0=2-4*x
<=> x=1/2
So Q(1/2;0)

4 0

4 years ago

What is the slope passing through the points (6,5) and (5,3)

Rus_ich [418]

Answer:

$m=2$

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Algebra I</u>

Slope Formula: $m=\frac{y_2-y_1}{x_2-x_1}$

Step-by-step explanation:

<u>Step 1: Define</u>

Point (6, 5)

Point (5, 3)

<u>Step 2: Find slope </u><em><u>m</u></em>

Substitute [SF]: $m=\frac{3-5}{5-6}$
Subtract: $m=\frac{-2}{-1}$
Divide: $m=2$

5 0

3 years ago