Answer:
The number of times the variability in the heights of the sixth graders is the variability in the heights of the seventh graders is approximately 1.4
Step-by-step explanation:
From the question, the mean absolute deviation (MAD) of the sixth graders = 1.2 and that of the seventh graders = 1.7
The variability in the heights of the sixth graders = 1.2
The variability in the heights of the seventh graders = 1.7
To calculate how many times the variability in the heights of the sixth graders is the variability in the heights of the seventh graders, we will divide the variability of the seventh graders by the variability of the sixth graders
That is, 1.7/ 1.2 = 1.4167 ≅ 1.4
Hence, the number of times the variability in the heights of the sixth graders is the variability in the heights of the seventh graders is approximately 1.4
There are bunch of factors of both. Here are the GCF and LCM.
Greatest Common Factor (GCF) = 12
Least Common Factor (LCM) = 120
Answer:
the 7th
0.007805 or 59 / 756
Step-by-step explanation:
To determine the Friday that he needed more pounds, determine the number of pounds he needed per dog for each Friday.
to do this, divide the total pounds by the number of dogs
on the 14th
3/8 ÷ 35 =
3/8 x 1/35 = 3/280
on the 7th
1/2 ÷ 27 =
1/2 x 1/27 = 1/54
to determine which fraction is larger multiply each by 100 to determine its percent
3/280 x 100 = 1.071
1/54 x 100 = 1.852
1/54 is larger. Thus, more pounds was needed on Friday the 7th
the difference between these fractions should be determined
- 
= 
= 59 / 756
Answer:
Step-by-step explanation:
7 friends plus himself = 8 persons
Round trip flight cost = &389x8= & 3112
Hotel bill = &249 x 4 rooms 3 nights = &
& 2988
Cost at Observatory desk = & 18.5 x 8= & 148
Cost at base ball viewing venue = 38 x 8 = & 148
Total budget apart from feeding should be = & 6552
