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earnstyle [38]
2 years ago
14

Convert 5.4•10^3 to standard form

Mathematics
1 answer:
hjlf2 years ago
5 0

Answer:

5400

Step-by-step explanation:

The ten is being raised to the 3rd power so you move the decimal 3 times.

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How many three-digit even numbers can be formed using digits from the set {1, 2, 3, 4, 5, 6, 7}
castortr0y [4]
The unit numbers can be 2,4,6.Suppose 2 as the unit number we can form 20 numbers the same way for 4 and 6 so totally 60 numbers.
6 0
2 years ago
According to his car odometer,Alfonso drove a total of 10 kilometers from his house to the movie theater and back again approxim
Brums [2.3K]

Answer:

It is approximately 3.10 miles from Alfonso's house to the movie theater

Step-by-step explanation:

First, identify what you know:

1) Alfonso drove from his house TO the movie theater, and then BACK to his house.

2) Alfonso's odometer shows him that the ENTIRE trip was 10 miles.

If the entire trip was 10 miles, and he only went to the move theater, AND took the same route back, then we simply divide 10 in half! This will give us the distance from his house the the movie theater, or rather, half of the trip, in kilometers.

10/2 = 5 kilometers.

Now, what is 5 kilometers in miles? Well, one mile is equal to 1.60934 kilometers, and one kilometer is equal to 0.621371 miles. So, simply multiply!

5 km  x 0.621371 miles = 3.106855

The approximate distance from his house to the theater is 3.10 miles!

Hope this helps! :)

3 0
3 years ago
12 1/2 as a fraction in the simplest form
djverab [1.8K]
12 = 12/1 = 24/2, 24/2+1/2=25/2 so 25/2, and it's in simplest form because 2 can't go into 25
8 0
3 years ago
In the library on a university campus, there is a sign in the elevator that indicates a limit of 16 persons. Furthermore, there
Vedmedyk [2.9K]

Answer:

a) 151lb.

b) 6.25 lb

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}}.

In this problem, we have that:

\mu = 151, \sigma = 25, n = 16

So

a) The expected value of the sample mean of the weights is 151 lb.

(b) What is the standard deviation of the sampling distribution of the sample mean weight?

This is s = \frac{\sigma}{\sqrt{n}} = \frac{25}{\sqrt{16}} = 6.25

8 0
3 years ago
WE NEED ANSWERS RIGHT NOW
Kisachek [45]

2

it like trying to imagine re-folding back a cardboard box

6 0
2 years ago
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