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4 years ago

How many 1/4 teaspoon doeses are in 7/8

Mathematics

2 answers:

Elden [556K]4 years ago

8 0

There are 3 1/2 doses

sveticcg [70]4 years ago

3 0

namely, how many times does 1/4 go into 7/8?

$\bf \cfrac{7}{8}\div \cfrac{1}{4}\implies \cfrac{7}{8}\cdot \cfrac{4}{1}\implies \cfrac{7}{1}\cdot \cfrac{4}{8}\implies \cfrac{7}{1}\cdot \cfrac{1}{2}\implies \cfrac{7}{2}\implies 3\frac{1}{2}$

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Answer:

30 cm

Step-by-step explanation:

add all sides and use 7 for the bottom line because 5+2=7 and use 8 for the side because 5+3=8.

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3 years ago

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sergejj [24]

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3 years ago

Assume that there is a 4% rate of disk drive failure in a year. a. If all your computer data is stored on a hard disk drive wit

kap26 [50]

Answer:

a) 99.84% probability that during a year, you can avoid catastrophe with at least one working drive

b) 99.999744% probability that during a year, you can avoid catastrophe with at least one working drive

Step-by-step explanation:

For each disk drive, there are only two possible outcomes. Either it works, or it does not. The disks are independent. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

4% rate of disk drive failure in a year.

This means that 96% work correctly, $p = 0.96$

a. If all your computer data is stored on a hard disk drive with a copy stored on a second hard disk drive, what is the probability that during a year, you can avoid catastrophe with at least one working drive?

This is $P(X \ geq 1)$ when $n = 2$

We know that either none of the disks work, or at least one does. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

We want $P(X \geq 1)$ . So

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{2,0}.(0.96)^{0}.(0.04)^{2} = 0.0016$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0016 = 0.9984$

99.84% probability that during a year, you can avoid catastrophe with at least one working drive

b. If copies of all your computer data are stored on four independent hard disk drives, what is the probability that during a year, you can avoid catastrophe with at least one working drive?

This is $P(X \ geq 1)$ when $n = 4$