7 x 106 = 742 3.5 x 10 = 35 742-35=701.it is 701 time larger
70. Bobbie wants to make 2.00 dollars using dollars with the value of half dollars and quarters. Now, how many way he can do it?
=> half dollars = 0.5
=> quarter dollars = 0.25
Possible ways:
=> 0.5 + 0.5 + 0.5 + 0.5 = 2 dollars (4 half dollars)
=> .25 + .25 + .25 +. 25 + .25 + .25 + .25 + .25 = 2 dollars (8 quarter dollars)
=> 0.5 + 0.5 + 0.25 + 0.25 + 0.25 + 0.25 = 2 dollars
=> 0.5 + 0.5 + 0.5 + 0.25 + 0.25 = 2 dollars
=> 0.5 + 0.25 + .25 + .25 + .25 + .25 + .25 = 2 dollars
Answer:
Mean = 90, Median = 93, Mode = 90, Range = 6
Step-by-step explanation:
Mean:
96 + 90 + 94 + 93 + 90 = 463
463 ÷ 5 = 92.5
92.5 to nearest tenth = 90
Mean = 90
Median:
<em>90, 90</em>, <u>93</u>, <em>94 ,96</em>
Median = 93
Mode
<em>96, </em><u>90</u><em>, 94, 93, </em><u>90</u><em> </em>
Mode = 90
Range:
96 - 90 = 6
Range = 6
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
Answer:
A) +5000
Step-by-step explanation:
I'm not sure of my answer but
hope that help