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11111nata11111 [884]

2 years ago

12

The numbers on a certain combination lock start with 0 and go through 10. If a combination consists of 3 numbers, how many diffe

rent lock combinations are possible?

2 answers:

NikAS [45]2 years ago

5 0

There would be 165 combinations

Ganezh [65]2 years ago

5 0

the answer isnt that^ its not an option.

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What is the solution to the equation 4x-6= 10x-3?

Lelechka [254]

Answer:

To solve this equation we need to get x by itself.

4x-6=10x-3, (the first step would be adding 6 to both sides.)

4x=10x+3 (the next step is the subtract ten from both sides.)

-6x=3 (Last step is the divide both sides by -6.)

x=-1/2

Hope this helps ;)

6 0

3 years ago

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Multiply then write the product in the simplest form <br> 3/1

Ganezh [65]

Answer:

9

Step-by-step explanation:

3/1*3/1=

3*3= 9

Answer: 9

9 is the most simplified version.

(I think this is what it means)

6 0

3 years ago

Orlando says that he can find the length of the diagonal, c, of the rectangular prism by

Zanzabum

Answer:

Step-by-step explanation:

6 0

2 years ago

Fifty-three percent of employees make judgements about their co-workers based on the cleanliness of their desk. You randomly sel

GarryVolchara [31]

Answer:

The unusual $X$ values for this model are: $X = 0, 1, 2, 7, 8$

Step-by-step explanation:

A binomial random variable $X$ represents the number of successes obtained in a repetition of $n$ Bernoulli-type trials with probability of success $p$ . In this particular case, $n = 8$ , and $p = 0.53$ , therefore, the model is ${8 \choose x} (0.53) ^ {x} (0.47)^{(8-x)}$ . So, you have:

$P (X = 0) = {8 \choose 0} (0.53) ^ {0} (0.47) ^ {8} = 0.0024$

$P (X = 1) = {8 \choose 1} (0.53) ^ {1} (0.47) ^ {7} = 0.0215$

$P (X = 2) = {8 \choose 2} (0.53)^2 (0.47)^6 = 0.0848$

$P (X = 3) = {8 \choose 3} (0.53) ^ {3} (0.47)^5 = 0.1912$

$P (X = 4) = {8 \choose 4} (0.53) ^ {4} (0.47)^4} = 0.2695$

$P (X = 5) = {8 \choose 5} (0.53) ^ {5} (0.47)^3 = 0.2431$

$P (X = 6) = {8 \choose 6} (0.53) ^ {6} (0.47)^2 = 0.1371$

$P (X = 7) = {8 \choose 7} (0.53) ^ {7} (0.47)^ {1} = 0.0442$

$P (X = 8) = {8 \choose 8} (0.53)^{8} (0.47)^{0} = 0.0062$

The unusual $X$ values for this model are: $X = 0, 1, 7, 8$

6 0

3 years ago

Read 2 more answers

Brainliest, but explain your answer for once? Thanks in advance, appreciate ya!

GenaCL600 [577]

Answer:

c. 1/6

Step-by-step explanation:

I hope it helps you...

4 0

2 years ago

Read 2 more answers