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puteri [66]
3 years ago
12

James is selecting a marble. James chooses a marble at random and then replaces it. He then selects a second marble at random. W

hat is the probability that James

Mathematics
1 answer:
tresset_1 [31]3 years ago
7 0

Answer:

The probability of selecting a solid black marbles both times;

P = 9/100

Attached is the completed question;

Step-by-step explanation:

Number of solid black marbles = 3

Total number of marbles = 10

The probability of selecting a solid black marble;

P1 = 3/10

With the assumption that the marbles are replaced before next selection.

The probability of selecting a solid black marbles both times;

P = P1 × P1 = 3/10 × 3/10 = 9/100

P = 9/100

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Naddika [18.5K]
One that could be close is Letter B?
6 0
2 years ago
<img src="https://tex.z-dn.net/?f=%5Csf%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%20%5Ccfrac%7B%5Csqrt%7Bx-1%7D-2x%20%7D%7Bx-7%7D" id=
BARSIC [14]
<h3>Answer:  -2</h3>

======================================================

Work Shown:

\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{x-1}-2x }{ x-7 }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \frac{1}{x}\left(\sqrt{x-1}-2x\right) }{ \frac{1}{x}\left(x-7\right) }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \frac{1}{x}*\sqrt{x-1}-\frac{1}{x}*2x }{ \frac{1}{x}*x-\frac{1}{x}*7 }\\\\\\

\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x^2}}*\sqrt{x-1}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x^2}*(x-1)}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x}-\frac{1}{x^2}}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \frac{ \sqrt{0-0}-2 }{ 1-0 }\\\\\\\displaystyle L = \frac{-2}{1}\\\\\\\displaystyle L = -2\\\\\\

-------------------

Explanation:

In the second step, I multiplied top and bottom by 1/x. This divides every term by x. Doing this leaves us with various inner fractions that have the variable in the denominator. Those inner fractions approach 0 as x approaches infinity.

I'm using the rule that

\displaystyle \lim_{x\to\infty} \frac{1}{x^k} = 0\\\\\\

where k is some positive real number constant.

Using that rule will simplify the expression greatly to leave us with -2/1 or simply -2 as the answer.

In a sense, the leading terms of the numerator and denominator are -2x and x respectively. They are the largest terms for each, so to speak. As x gets larger, the influence that -2x and x have will greatly diminish the influence of the other terms.

This effectively means,

\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{x-1}-2x }{ x-7 } = \lim_{x\to\infty} \frac{ -2x }{ x} = -2\\\\\\

I recommend making a table of values to see what's going on. Or you can graph the given function to see that it slowly approaches y = -2. Keep in mind that it won't actually reach y = -2 itself.

5 0
2 years ago
Use the information provided to write the equation of the circle Center: (10,-9) Point on Circle: (12,-13)
Zielflug [23.3K]

Answer:

(x - 10)² + (y + 9)² = 20

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r is the radius

The radius is the distance from the centre to a point on the circle

To calculate r use the distance formula

r = √(x₂ - x₁ )² + (y₂ - y₁ )²

with (x₁, y₁ ) = (10, - 9) and (x₂, y₂ ) = (12, - 13)

r = \sqrt{(12-10)^2+(-13+9)^2}

  = \sqrt{4+16} = \sqrt{20}, hence

(x - 10)² + (y - (- 9))² = (\sqrt{20})², that is

(x - 10)² + (y + 9)² = 20

7 0
3 years ago
I need the answer to this.​
Kamila [148]

Answer:

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4 0
2 years ago
∠E and ∠F are vertical angles with m∠E=9x+12 and m∠F=3x+24. What is the value of x?
statuscvo [17]

Answer:

  • x = 2

Step-by-step explanation:

<em>Vertical angles are equal as per definition.</em>

  • ∠E and ∠F are vertical angles
  • m∠E = 9x+12 and m∠F = 3x+24
  • x = ?

<u>Solution</u>

  • 9x+12 = 3x+24
  • 9x - 3x = 24 - 12
  • 6x = 12
  • x = 2

4 0
3 years ago
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