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yKpoI14uk [10]
3 years ago
10

What is 16×500×60=? i need help

Mathematics
1 answer:
pychu [463]3 years ago
6 0

Answer:

480000

Step-by-step explanation:

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45+ 46+47+ 48+ ... 113 + 114Gauss approach
Doss [256]

Answer:

5565

Step-by-step explanation:

1+2+3+4...+113+114=\frac{114*115}{2} =6555

1+2+3+4 ... +43+44=\frac{44*45}{2} =990

Thus, 45+46+47+48...+113+114= 6555-990=5565

3 0
3 years ago
ALGEBRA 2- If f(x)= √x-3 , which inequality can be used to find the domain of f(x)?
Fed [463]
B, because you can't get the square root of a negative number
8 0
3 years ago
Read 2 more answers
Annual starting salaries in a certain region of the U. S. for college graduates with an engineering major are normally distribut
defon

Answer:

0.8665 = 86.65% probability that the sample mean would be at least $39000

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean $39725 and standard deviation $7320.

This means that \mu = 39725, \sigma = 7320

Sample of 125:

This means that n = 125, s = \frac{7320}{\sqrt{125}} = 654.72

The probability that the sample mean would be at least $39000 is about?

This is 1 subtracted by the pvalue of Z when X = 39000. So

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

By the Central Limit Theorem

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

Z = \frac{39000 - 39725}{654.72}

Z = -1.11

Z = -1.11 has a pvalue of 0.1335

1 - 0.1335 = 0.8665

0.8665 = 86.65% probability that the sample mean would be at least $39000

4 0
3 years ago
What is the common ratio in this geometric series?
Lerok [7]
For a geometric sequence or geometric series, the common ratio is the ratio of a term to the previous term. This ratio is usually indicated by the variable r. Example: The geometric series 3, 6, 12, 24, 48, . . . has common ratio r = 2.
4 0
3 years ago
Ynnette can wash 959595 cars in 555 days.<br> How many cars can Lynnette wash in 111111 days?
gavmur [86]

Answer:

192,110,919

Step-by-step explanation:

You would first need to divide 959595 by 555 which would equal 1729. Then you would need to multiply 1729*111111 which would equal 192,110,919.

Hope this helps and pls do mark me brainliest if you can:)

4 0
3 years ago
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