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Alisiya [41]
3 years ago
14

What is 3x5x6x19x500

Mathematics
2 answers:
alisha [4.7K]3 years ago
8 0
3x5x6x19x500 is 855,000
kakasveta [241]3 years ago
7 0
The answer is 855,000. Hope this helps :)
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Last Saturday, the MoviePlex Iheater sold 1,456 tickets. The theater sold 6 times as many regular price tickets as it sold disco
blagie [28]
Well, one divides by 7, actually, because the total number of tickets is 6+1 times the # of discount tickets. That is, 1,456/7. You don't need to actually perform this division.
6 0
3 years ago
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Write two sets of 5 different numbers that have the same mean but different standard deviations
Zina [86]

Answer:

The sets could be 5, 5, 5, 5, 5 and 0, 10, 20, -10, 5

Step-by-step explanation:

Standard deviations are a measure of how far the average point is away from the mean. In the first example, each term is exactly the mean. In the second set, some are quite a distance from the mean and therefore, it would have a bigger standard deviation.

6 0
3 years ago
A person walks 264 feet per minute. Which of the following is the correct conversion of this quantity into miles per hour? (Note
saul85 [17]
The answer is <span>c. 3
proof 

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5 0
3 years ago
Rodrigo made 6/8 of a pound of trail mix. If he puts 1/4 of a pound into each bag, how many bags can Rodrigo fill?
ValentinkaMS [17]

Answer:

He can fill 3 bags

Step-by-step explanation:

You would first make them have a common denominator so you would have 2/8 instead of 1/4 then you would do 6/8 divided by 2/8 to get 3

1/4 = 2/8

6/8 divided by 2/8 = 3

7 0
3 years ago
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The probabilities that a batch of 4 computers will contain 0, 1, 2, 3, and 4 defective computers are 0.4096, 0.4096, 0.1536, 0.0
Zinaida [17]

Answer:

E(X) = \sum_{i=1}^n X_i P(X_i)

And if we replace we got:

E(X) = 0*0.4096 +1*0.4096+ 2*0.1536+ 3*0.0256 +4*0.0016 = 0.8

So we expect about 0.8 defective computes in a batch of 4 selected.

Step-by-step explanation:

Previous concepts

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

Solution to the problem

For this case we have the following distribution given:

X           0             1               2               3             4

P(X)  0.4096    0.4096    0.1536    0.0256    0.0016

And we satisfy that P(X_i) \geq 0 and \sum P(X_i) =1 so we have a probability distribution. And we can find the expected value with the following formula:

E(X) = \sum_{i=1}^n X_i P(X_i)

And if we replace we got:

E(X) = 0*0.4096 +1*0.4096+ 2*0.1536+ 3*0.0256 +4*0.0016 = 0.8

So we expect about 0.8 defective computes in a batch of 4 selected.

5 0
4 years ago
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