Answer:
5.2
Step-by-step explanation:
distance(Mercury, Sun) = 5.8 x 10^7
distance(Venus, Sun) = 1.1 x 10^8
distance(Venus, Mars) = 1.1 x 10^8 - 5.8 x 10^7
The subtraction is a little easier if you borrow a power of 10 from the 10^8, multiplying 1.1 by 10.
simply subtract 5.8 from 11.0.
This gives us a result of 5.2 x 10^7.
The 3 is .3, or 3 tenths, because the 3 is in the tenths place
The area of the rectangular garden given the length and width is 16 1/2 square feet
<h3> Area of the garden</h3><h3 />
- Length = 5 1/2 feet
- Width = 3 feet
Area of a rectangular garden = length × width
= 5 1/2 feet × 3 feet
= 11/2 × 3
= 33/2
= 16 1/2 square feet
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Answer:
> a<-rnorm(20,50,6)
> a
[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905
[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501
Then we can find the mean and the standard deviation with the following formulas:
> mean(a)
[1] 50.72451
> sqrt(var(a))
[1] 7.470221
Step-by-step explanation:
For this case first we need to create the sample of size 20 for the following distribution:
And we can use the following code: rnorm(20,50,6) and we got this output:
> a<-rnorm(20,50,6)
> a
[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905
[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501
Then we can find the mean and the standard deviation with the following formulas:
> mean(a)
[1] 50.72451
> sqrt(var(a))
[1] 7.470221