Answer:
$50
Step-by-step explanation:
5= 26+4=30
6=30+4=34
7=34+4=38
8=38+4=42
9=42+4=46
10=46+4=50
Answer:
a) C(x) = 15000/x + 6x +80
b) Domain of C(x) { R x>0 }
Step-by-step explanation:
We have:
Enclosed area = 1500 ft² = x*y from which y = 1500 / x (a) where x is perpendicular to the river
Cost = cost of sides of fenced area perpendicular to the river + cost of side parallel to river + cost of 4 post then
Cost = 10*y + 2*3*x + 4*20 and accoding to (a) y = 1500/x
Then
C(x) = 10* ( 1500/x ) + 6*x + 80
C(x) = 15000/x + 6x +80
Domain of C(x) { R x>0 }
From the box plot, it can be seen that for grade 7 students,
The least value is 72 and the highest value is 91. The lower and the upper quartiles are 78 and 88 respectively while the median is 84.
Thus, interquatile range of <span>the resting pulse rate of grade 7 students is upper quatile - lower quartle = 88 - 78 = 10
</span>Similarly, from the box plot, it can be seen that for grade 8 students,
The
least value is 76 and the highest value is 97. The lower and the upper
quartiles are 85 and 94 respectively while the median is 89.
Thus, interquatile range of the resting pulse rate of grade 8 students is upper quatile - lower quartle = 94 - 85 = 9
The difference of the medians <span>of the resting pulse rate of grade 7 students and grade 8 students is 89 - 84 = 5
Therefore, t</span><span>he difference of the medians is about half of the interquartile range of either data set.</span>
Answer:
x=9 Hope this helps :)
Step-by-step explanation:
Step 1: Simplify both sides of the equation.
Step 2: Subtract 3x from both sides.
Step 3: Subtract 6 from both sides.
Step 4: Multiply both sides by 3/(-8).
And you get x=9
