ax+r=7
Move ax to the other side. +positive ax changes to -negative ax
ax-ax+r=7-ax
r=7-ax
Answer: r=7-ax or r=-ax+7 ( same thing) but rearranged
Answer:
Diameter = 19.33
Step-by-step explanation:
Imaging a radius line from O to the endpoints of the 7. call this line R.
Label the part of the vertical line from O to the 90 degree intersection y.
Now you have a right triangle.
Using the Pythagorean theorem:
R² = 7² + y²
also
y = R - 3
substitute for y:
R² = 49 + (R-3)²
R² = 49 + R² - 6R + 9
simplify:
0 = 58 - 6R
6R = 58
R = 9.6667
Diameter = 2(9.6667) = 19.33

a. "Chocolate" and "Adults" (whatever those mean) will be independent as long as

"Chocolate" has the marginal distribution given by the second column, with a total probability of

. Similarly, "Adults" has the marginal distribution described by the third row, so that

. Then

Meanwhile, the joint probability of "Chocolate" and "Adults" is given by the cell in the corresponding row/column, with

.
The probabilities match, so these events are indeed independent.
Parts (b) and (c) are checked similarly.
b. Yes;


c. Yes;

Answer:
1. S.A. = 4350 cm²
2. S.A. = 864 cm²
3. S.A. = 240 cm²
4. S.A. = 224 m²
5. S.A. = 301.6 in.²
6. S.A. = 6,082.1 cm²
7. S.A. = 923.6 in.²
Step-by-step explanation:
1. Surface area of the rectangular prism = 2(LW + LH + WH)
L = 45 cm
W = 25 cm
H = 15 cm
S.A. = 2(45*25 + 45*15 + 25*15)
S.A. = 4350 m²
2. Surface area of the cube = 6a²
a = 12 cm
S.A. = 6(12²)
S.A. = 864 cm²
3. Surface area of triangular prism = bh + (s1 + s2 + s3)*H
b = 4 cm
h = 6 cm
s1 = 4 cm
s2 = 7 cm
s3 = 7 cm
H = 12 cm
Plug in the values
S.A. = 4*6 + (4 + 7 + 7)*12
S.A. = 24 + (18)*12
S.A. = 24 + 216
S.A. = 240 cm²
4. Surface area of the square based pyramid = area of the square base + 4(area of 1 triangular face)
S.A. = (8*8) + 4[(8*10)/2]
S.A. = 64 + 4(40)
S.A. = 64 + 160
S.A. = 224 cm²
5. Surface area of the cone = πrl + πr²
r = 6 in.
l = 10 in.
S.A. = π*6*10 + π*r²
S.A. = 60π + 36π
S.A. = 301.592895
S.A. = 301.6 in.² (nearest tenth)
6. Surface area of the sphere = 4πr²
r = 22 cm
S.A. = 4*π*22²
S.A. = 1,936π
S.A. = 6,082.1 cm² (nearest tenth)
7. Surface area of the cylinder = 2πrh + 2πr²
r = 7 in.
h = 14 in.
S.A. = 2*π*7*14 + 2*π*7²
S.A. = 923.6 in.² (nearest tenth)
Answer:
we are given
part-A:
Since, x is number of tablets
C(x) is cost of producing x tablets
so, vertical box is C(x)
so, we write in vertical box is "cost of producing x tablets"
Horizontal box is x
so, we write in horizontal box is "number of tablets"
part-B:
we have to find average on [a,b]
we can use formula
we are given point as
a: (15 , 395)
a=15 and C(15)=395
b: (20, 480)
b=20 , C(20)=480
now, we can plug values
...........Answer
part-c:
we have to find average on [b,c]
we can use formula
we are given point as
b: (20, 480)
b=20 , C(20)=480
c:(25,575)
c=25 , C(c)=575
now, we can plug values