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3 years ago

6.73 - 2x5 to the 0 power divided by 2

Mathematics

1 answer:

djyliett [7]3 years ago

8 0

If you mean $\frac{6.73 - 2}{2}\$ , then the answer is going to be

6.73 - 2 = 4.73

4.73/2 = 2.365 (though it may be 2.37 because you have to round it)

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More 5th grade work c’mon

fgiga [73]

Answer: honestly for the first one, i think the answer would be 2.

they both have 5 at the end so you know its divisible :)

Step-by-step explanation:

6 0

3 years ago

It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute

NemiM [27]

Answer:

a) 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.

b) 99.22% probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

Mildly obese

Normally distributed with mean 373 minutes and standard deviation 67 minutes. So $\mu = 373, \sigma = 67$

A) What is the probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes?

So $n = 5, s = \frac{67}{\sqrt{5}} = 29.96$

This probability is 1 subtracted by the pvalue of Z when X = 410.

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

$Z = \frac{410 - 373}{29.96}$

$Z = 1.23$

$Z = 1.23$ has a pvalue of 0.8907.

So there is a 1-0.8907 = 0.1093 = 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.

Lean

Normally distributed with mean 526 minutes and standard deviation 107 minutes. So $\mu = 526, \sigma = 107$

B) What is the probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes?

So $n = 5, s = \frac{107}{\sqrt{5}} = 47.86$

This probability is 1 subtracted by the pvalue of Z when X = 410.

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

$Z = \frac{410 - 526}{47.86}$

$Z = -2.42$

$Z = -2.42$ has a pvalue of 0.0078.

So there is a 1-0.0078 = 0.9922 = 99.22% probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes.

7 0

3 years ago

What is 0.49,0.5 and 1/5 from least to greatest?

brilliants [131]

1/5 (0.2)
0.49
0.5

Hope this helped!! xx

7 0

3 years ago

Which postulate or theorem could be used to prove triangleLMQ is congruent to triangleLNP

Sliva [168]

The answer to this query is AA similarity postulate. 

Because the triangles given are only similar in angle but dissimilar in sides which makes it incongruent with respect to the sides, AA similarity postulate is the exact answer.

SAS ASA are not possible answers.

6 0

3 years ago

Read 2 more answers

How do you solve 1-5+1×(4×4-31)×8

Alja [10]

Hey there!

In order to solve this, remember PEMDAS. First, start with solving whatever is in parentheses. Then, move on to exponents. Then, complete the multiplication and division. Lastly, add or subtract anything remaining. Omit any steps that aren't present.

Parentheses:

1 – 5 + 1 × (4 × 4 – 31) × 8
1 – 5 + 1 × (–15) × 8

Multiplication/Division (Left to Right):

1 – 5 + (–15) × 8
1 – 5 + (–120)
1 – 5 – 120

Addition/Subtraction (Left to Right):

–4 – 120
–124

Your answer is –124.

Hope this helped you out! :-)

5 0

3 years ago