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3 years ago

CHứng minh rằng trong hệ g - phân với 2

Mathematics

1 answer:

ruslelena [56]3 years ago

3 0

Jejejwjehejjejekekekeke

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What is 1+1. 100 points to whoever gets it first. Answer fast!

Anni [7]

Answer:

da answer iz two cause wen u add one plus one u get answer two

Step-by-step explanation:

let mi hope dat it useful

4 0

3 years ago

Read 2 more answers

Assume a jar has five red marbles and three black marbles. Draw out two marbles with and without replacement. Find the requested

Doss [256]

Answer:

For probabilities with replacement

$P(2\ Red) = \frac{25}{64}$

$P(2\ Black) = \frac{9}{64}$

$P(1\ Red\ and\ 1\ Black) = \frac{15}{32}$

$P(1st\ Red\ and\ 2nd\ Black) = \frac{15}{64}$

For probabilities without replacement

$P(2\ Red) = \frac{5}{14}$

$P(2\ Black) = \frac{3}{28}$

$P(1\ Red\ and\ 1\ Black) = \frac{15}{28}$

$P(1st\ Red\ and\ 2nd\ Black) = \frac{15}{56}$

Step-by-step explanation:

Given

$Marbles = 8$

$Red = 5$

$Black = 3$

For probabilities with replacement

(a) P(2 Red)

This is calculated as:

$P(2\ Red) = P(Red)\ and\ P(Red)$

$P(2\ Red) = P(Red)\ *\ P(Red)$

So, we have:

$P(2\ Red) = \frac{n(Red)}{Total} \ *\ \frac{n(Red)}{Total}\\$

$P(2\ Red) = \frac{5}{8} * \frac{5}{8}$

$P(2\ Red) = \frac{25}{64}$

(b) P(2 Black)

This is calculated as:

$P(2\ Black) = P(Black)\ and\ P(Black)$

$P(2\ Black) = P(Black)\ *\ P(Black)$

So, we have:

$P(2\ Black) = \frac{n(Black)}{Total}\ *\ \frac{n(Black)}{Total}$

$P(2\ Black) = \frac{3}{8}\ *\ \frac{3}{8}$

$P(2\ Black) = \frac{9}{64}$

(c) P(1 Red and 1 Black)

This is calculated as:

$P(1\ Red\ and\ 1\ Black) = [P(Red)\ and\ P(Black)]\ or\ [P(Black)\ and\ P(Red)]$

$P(1\ Red\ and\ 1\ Black) = [P(Red)\ *\ P(Black)]\ +\ [P(Black)\ *\ P(Red)]$

$P(1\ Red\ and\ 1\ Black) = 2[P(Red)\ *\ P(Black)]$

So, we have:

$P(1\ Red\ and\ 1\ Black) = 2*[\frac{5}{8} *\frac{3}{8}]$

$P(1\ Red\ and\ 1\ Black) = 2*\frac{15}{64}$

$P(1\ Red\ and\ 1\ Black) = \frac{15}{32}$

(d) P(1st Red and 2nd Black)

This is calculated as:

$P(1st\ Red\ and\ 2nd\ Black) = [P(Red)\ and\ P(Black)]$

$P(1st\ Red\ and\ 2nd\ Black) = P(Red)\ *\ P(Black)$

$P(1st\ Red\ and\ 2nd\ Black) = \frac{n(Red)}{Total} *\ \frac{n(Black)}{Total}$

So, we have:

$P(1st\ Red\ and\ 2nd\ Black) = \frac{5}{8} *\frac{3}{8}$

$P(1st\ Red\ and\ 2nd\ Black) = \frac{15}{64}$

For probabilities without replacement

(a) P(2 Red)

This is calculated as:

$P(2\ Red) = P(Red)\ and\ P(Red)$

$P(2\ Red) = P(Red)\ *\ P(Red)$

So, we have:

$P(2\ Red) = \frac{n(Red)}{Total} \ *\ \frac{n(Red)-1}{Total-1}$

We subtracted 1 because the number of red balls (and the total) decreased by 1 after the first red ball is picked.

$P(2\ Red) = \frac{5}{8} * \frac{4}{7}$

$P(2\ Red) = \frac{5}{2} * \frac{1}{7}$

$P(2\ Red) = \frac{5}{14}$

(b) P(2 Black)

This is calculated as:

$P(2\ Black) = P(Black)\ and\ P(Black)$

$P(2\ Black) = P(Black)\ *\ P(Black)$

So, we have:

$P(2\ Black) = \frac{n(Black)}{Total}\ *\ \frac{n(Black)-1}{Total-1}$

We subtracted 1 because the number of black balls (and the total) decreased by 1 after the first black ball is picked.

$P(2\ Black) = \frac{3}{8}\ *\ \frac{2}{7}$

$P(2\ Black) = \frac{3}{4}\ *\ \frac{1}{7}$

$P(2\ Black) = \frac{3}{28}$

(c) P(1 Red and 1 Black)

This is calculated as:

$P(1\ Red\ and\ 1\ Black) = [P(Red)\ and\ P(Black)]\ or\ [P(Black)\ and\ P(Red)]$

$P(1\ Red\ and\ 1\ Black) = [P(Red)\ *\ P(Black)]\ +\ [P(Black)\ *\ P(Red)]$

$P(1\ Red\ and\ 1\ Black) = [\frac{n(Red)}{Total}\ *\ \frac{n(Black)}{Total-1}]\ +\ [\frac{n(Black)}{Total}\ *\ \frac{n(Red)}{Total-1}]$

So, we have:

$P(1\ Red\ and\ 1\ Black) = [\frac{5}{8} *\frac{3}{7}] + [\frac{3}{8} *\frac{5}{7}]$

$P(1\ Red\ and\ 1\ Black) = [\frac{15}{56} ] + [\frac{15}{56}]$

$P(1\ Red\ and\ 1\ Black) = \frac{30}{56}$

$P(1\ Red\ and\ 1\ Black) = \frac{15}{28}$

(d) P(1st Red and 2nd Black)

This is calculated as:

$P(1st\ Red\ and\ 2nd\ Black) = [P(Red)\ and\ P(Black)]$

$P(1st\ Red\ and\ 2nd\ Black) = P(Red)\ *\ P(Black)$

$P(1st\ Red\ and\ 2nd\ Black) = \frac{n(Red)}{Total} *\ \frac{n(Black)}{Total-1}$

So, we have:

$P(1st\ Red\ and\ 2nd\ Black) = \frac{5}{8} *\frac{3}{7}$

$P(1st\ Red\ and\ 2nd\ Black) = \frac{15}{56}$

7 0

3 years ago

After reaching the top of a cliff a rock climber descended the rock face using a 65 ft rope the distance to the base of cliff wa

suter [353]

Answer:

The required number of ropes is 9.

Step-by-step explanation:

The length of rope is 65 ft.

The distance to the base of cliff was 585 ft.

Let the number of ropes or same lengths did it take climber to reach descent be x.

$65\times x=585$

Divide both sides of 65.

$x=\frac{585}{65}$

$x=9$

Therefore the required number of ropes is 9.

4 0

3 years ago

Can someone please help me with this math question? I am desperate!!

stiks02 [169]

Hello from MrBillDoesmath!

The questions are a bit unclear but here's my best shot

Answer:

a. "c" = 0

As x=0 is a root of f(x) = ax^2 + bx + c (I think this is the equation you had in mind. Please correct me if I'm wrong)

a(0)^2 + b(0) + c = f(0) = 0.

As any number times 0 is 0 this is equivalent to

0 + 0 + c = 0. So c = 0!

b. From part a (above) f(x) = ax^2 + bx. Suppose x is an extremely large number (positive or negative). If "a" is positive then f(x) is a large positive number so f(x) is large and looks like the letter "U". But if "a" is negative and x large (positive or negative), then f(x) is a large negative number, meaning the function looks like an upside-down "u". IN short, f(x) is a parabola that opens upward if a > 0 and opens downward if a < 0.

Given that f(x) = ax^2 + bx = x(ax+b), f(x) = 0 when x = 0 or (ax + b) = 0. The latter happens when ax = -b or x = - (b/a)

c. ax^2 + bx = 0

Ragards, Mr B.

8 0

3 years ago

Write the equation for a parabola that has x-intercepts (−4.5, 0) and (−2.8, 0) and y-intercept (0, 37.8).

BARSIC [14]

Answer:

$y = 3(x + 4.5)(x + 2.8)$ .

Step-by-step explanation:

Start with the two $x$ -intercepts. The two zeros of the quadratic equation for this parabola are:

$x_1 = -4.5$ , and
$x_2 = 2.8$ .

(These are the $x$ -coordinates of the two $x$ -intercepts.)

By the factor theorem, $x = k$ (where $k$ is a real number) is a zero of a polynomial if and only if $(x - k)$ is a factor of that polynomial.

A quadratic equation is also a polynomial. In this case, the two zeros would correspond to the two factors

$(x - (-4.5)) = (x + 4.5)$ .
$(x - (-2.8)) = (x + 2.8)$

A parabola could only have up to two factors. As a result, the power of these two factor should both be one. Hence, the equation for the parabola would be in the form

$y = a \, (x + 4.5)(x + 2.8)$ ,

where $a$ is the leading coefficient that still needs to be found. Calculate the value of $a$ using the $y$ -intercept of this parabola. (Any other point on this parabola that is not one of the two $x$ -intercepts would work.)