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Mashcka [7]
2 years ago
10

What is two decimal that are equivalent to 7.06

Mathematics
2 answers:
ycow [4]2 years ago
4 0

Answer:7.060 and 7.0600


Step-by-step explanation:0 doest not mean nothing


Thepotemich [5.8K]2 years ago
4 0

Answer: 7.060 and 7.0600


they are the same number just with an added digit if that is what you meant

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282 Hours after 6 weeks

Step-by-step explanation:

Multiply the 6 weeks and the 48 hour a week adverage and theres your answer!

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What’s the answer to this question?
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Step-by-step explanation:

its x^4

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Use the list below to find the upper quartile. 27, 5, 11, 13, 10, 8, 14, 18, 7
Alinara [238K]

Find the  median first. The middle of all the numbers.

5, 7, 8, 10, 11, 13, 14, 18, 27

11 is the median.

Then find the median of the upper quartile.

The upper quartile consists of numbers...

5, 7, 8, 10

Since it is an even set of numbers add the two in the middle and divided by two. So 7 plus 8=  15/2.

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bogdanovich [222]

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Step-by-step explanation:

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I arrive at a bus stop at a time that is normally distributed with mean 08:00 and SD 2 minutes. My bus arrives at the stop at an
Nimfa-mama [501]

Answer:

0.0485 = 4.85% probability that you miss the bus.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

When two normal distributions are subtracted, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of the variances.

In this question:

We have to find the distribution for the difference in times between when you arrive and when the bus arrives.

You arrive at 8, so we consider the mean 0. The bus arrives at 8:05, 5 minutes later, so we consider mean 5. This means that the mean is:

\mu = 0 - 5 = -5

The standard deviation of your arrival time is of 2 minutes, while for the bus it is 3. So

\sigma = \sqrt{2^2 + 3^2} = \sqrt{13}

The bus remains at the stop for 1 minute and then leaves. What is the chance that I miss the bus?

You will miss the bus if the difference is larger than 1. So this probability is 1 subtracted by the pvalue of Z when X = 1.

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

Z = \frac{1 - (-5)}{\sqrt{13}}

Z = \frac{6}{\sqrt{13}}

Z = 1.66

Z = 1.66 has a pvalue of 0.9515

1 - 0.9515 = 0.0485

0.0485 = 4.85% probability that you miss the bus.

5 0
3 years ago
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