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3 years ago

11

A combination lock has 25 numbers on it. How many different 3-digit lock combinations are possible if no digit can be repeated?

1 answer:

IgorC [24]3 years ago

7 0

Answer:

The number of ways is 13,800 ways

Step-by-step explanation:

In this question, we are tasked with calculating the number of ways to form 3-digits codes from 25 different numbers given that no repetition is allowed.

Now, for the first digit, we have all the 25 numbers completely wanting to take a spot. Now, in how many ways can we choose a single out of 25, that would be 25 ways

Now, we have 24 numbers left, and we are trying to pick one, the number of ways this can be done is 24 ways also

for the last digit, we have 23 numbers and we are selecting just one, the number of ways this can be done is 23 ways too

Thus, the cumulative number of ways would be 25 * 24 * 23 = 13,800 ways

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Answer:

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3 years ago

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victus00 [196]

Answer:

$P(A\ or\ B)=\frac{7}{9}$

Step-by-step explanation:

We need to use the formula to calculate the probability of (A or B) where

A=Probability a student likes pepperoni

B=Probability a student likes olive

A and B =Probability a student likes both toppings in a pizza

A or B =Probability a student likes pepperoni or olive (and maybe both), a non-exclusive or

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$P(A\ or\ B)=P(A)+P(B)-P(A\ and\ B)$

Since 6 students like pepperoni out of 9:

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Since 4 students like olive out of 9:

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Then we have

$P(A\ or\ B)=\frac{6}{9}+\frac{4}{9}-\frac{3}{9}$

$P(A\ or\ B)=\frac{7}{9}$

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3 years ago

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Lilit [14]

Answer:

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3 years ago

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$\begin{gathered} (y-y_0)=m(x-x_0) \\ y-1=-9(x-1) \\ y-1=-9x+9 \\ y=-9x+9+1 \\ y=-9x+10 \end{gathered}$

2) Note that we had to add 1 to both sides, to keep the "y "on the left.

3) Therefore, y=-9x+10 is the answer.

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1 year ago