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2 years ago

Least to greatest 63.37,62.95,63.7

Mathematics

1 answer:

Anit [1.1K]2 years ago

3 0

Answer:

1. 62.95

2. 63.37

3. 63.7

Step-by-step explanation:

List the numbers by lining up the decimals and filling in zeros in the empty spaces to make all the numbers have the same number of digits, then compare whole numbers first then the decimals. There are two 63's so then you look at the decimal side one is .37 and one is .70 (add the zero so all numbers have the same number of digits)

63.37

62.95

63.70 (added the zero)

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2 years ago

Read 2 more answers

Question Help Suppose that the lifetimes of light bulbs are approximately normally distributed, with a mean of 5656 hours and a

koban [17]

Answer:

a)3.438% of the light bulbs will last more than 6262 hours.

b)11.31% of the light bulbs will last 5252 hours or less.

c) 23.655% of the light bulbs are going to last between 5858 and 6262 hours.

d) 0.12% of the light bulbs will last 4646 hours or less.

Step-by-step explanation:

Normally distributed problems can be solved by the z-score formula:

On a normaly distributed set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a value X is given by:

$Z = \frac{X - \mu}{\sigma}$

After we find the value of Z, we look into the z-score table and find the equivalent p-value of this score. This is the probability that a score will be LOWER than the value of X.

In this problem, we have that:

The lifetimes of light bulbs are approximately normally distributed, with a mean of 5656 hours and a standard deviation of 333.3 hours.

So $\mu = 5656, \sigma = 333.3$

(a) What proportion of light bulbs will last more than 6262 hours?

The pvalue of the z-score of $X = 6262$ is the proportion of light bulbs that will last less than 6262. Subtracting 100% by this value, we find the proportion of light bulbs that will last more than 6262 hours.