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Paladinen [302]
3 years ago
8

Urgently plsss A randomly generated password contains four characters. Each of the four characters is either a lowercase letter

or a digit from 0–9. Each character in the password cannot be used more than once.
What is the approximate probability that exactly one of the four characters will be a number?

1%
11%
28%
44%
Mathematics
1 answer:
Flauer [41]3 years ago
3 0

Answer:

Step-by-step explanation:

Sample space is 36C4

Now, we want to know all of the combinations that have 1 digit in it.

So, we can have one here:

1XXX

X1XX

XX1X

XXX1

But we have 10 different digits to choose from. So, we need to introduce the combination term, nCr, where n is a list of all digits and r is how many we want.

Since we only want one, we will need 10C1 for the number of digits. But we need to choose three lowercases, so it becomes 10C1 × 26C3

Since it's a probability question, we need to divide that by our sample space, 36C4, and our percentage becomes 44%

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horrorfan [7]
B. I did this test too.
7 0
2 years ago
A student used a compass and a straightedge to construct <img src="https://tex.z-dn.net/?f=%5Coverline%7BCE%7D" id="TexFormula1"
Luba_88 [7]

Answer:

2) ∠ACE ≅ ∠BCE

Step-by-step explanation:

Considering the complete question attached in Figure below, geometry shows that ∠ACB is bisected into two angles ∠ACE and ∠BCE and we know that bisection means to divide an angle into two equal angles. According to this the resultant angles ∠ACE and ∠BCE must be equal and congruent.

7 0
3 years ago
One of our brainliest, Konrad509, made this:
Reil [10]

\dfrac{B_x \sqrt{74_x}}{1D_x}+J_x51_x=4G3_x

A=10, B=11, C=12, etc.

\dfrac{11\cdot x^0\cdot \sqrt{7\cdot x^1+4\cdot x^0}}{1\cdot x^1+13\cdot x^0}+19\cdot x^0\cdot (5\cdot x^1+1\cdot x^0)=4\cdot x^2+16\cdot x^1+3\cdot x^0\\\\\dfrac{11\sqrt{7x+4}}{x+13}+19(5x+1)=4x^2+16x+3\\\\\dfrac{11\sqrt{7x+4}}{x+13}+95x+19=4x^2+16x+3\\\\11\sqrt{7x+4}+95x(x+13)+19(x+13)=(4x^2+16x+3)(x+13)\\\\11\sqrt{7x+4}+95x^2+1235x+19x+247=4x^3+52x^2+16x^2+208x+3x+39\\\\11\sqrt{7x+4}=4x^3-27x^2-1043x-208\\\\121(7x+4)=(4x^3-27x^2-1043x-208)^2

121(7x+4)=(4x^3-27x^2-1043x-208)^2\\\\847x+484=16 x^6 - 216 x^5 - 7615 x^4 + 54658 x^3 + 1099081 x^2 + 433888 x + 43264\\\\16 x^6 - 216 x^5 - 7615 x^4 + 54658 x^3 + 1099081 x^2 + 433041 x +42780=0

Now, the "only" thing that remains to do is solving the above equation.

While making this problem I only made sure it has a solution. I didn't try to solve it myself and I didn't know it will end up with such "convoluted" polynomial. Sorry to everyone who tried to solve it... m(_ _)m

I think the best way to approach it is using the rational root theorem since we know that x\in\mathbb{N}. Moreover we can deduce that x\geq19 since there is J and J=19.

After you succesfully solve it, you should get the answer x=20.

7 0
3 years ago
Which choice is the approximate measure of ∠E?<br><br> 64.80<br> 61.93<br> 28.07<br> 25.20
Advocard [28]
The answer is 28.07. I just took this test
4 0
3 years ago
Maddy bought 2 ⅔ pounds of grapes and 1 ⅕ pounds of apples. If she bought bananas that weighed 1 ¼ times as much as the grapes,
spin [16.1K]

Answer:

3 whole number, 1/3

Step-by-step explanation:

To solve this, Multiply the value of the banana to the weight of the Grapes.

Please find the attached file for a comprehensive Solution.

Maths is fun !

4 0
3 years ago
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