Let the number of raspberry bushes in one garden = x
And the number of raspberry bushes in second garden = y
Garden one has 5 times as many raspberry bushes as second garden,
So the equation will be,
x = 5y -------(1)
If 22 bushes were transplanted from garden one to the second, number of bushes in both the garden becomes same,
Therefore, (x - 22) = (y + 22)
x - y = 22 + 22
x - y = 44 ------(2)
Substitute the value of x from equation (1) to equation (2)
5y - y = 44
4y = 44
y = 11
Substitute the value of 'y' in equation (1),
x = 5(11)
x = 55
Therefore, Number of bushes in garden one were 55 and in second garden 11 originally.
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<u>Find Radius:</u>
Radius = Diameter ÷ 2
Radius = 4 ÷ 2 = <u>2 in</u>
.
<u>Find Surface Area:</u>
Surface Area = 2πr² + 2πrh
Surface Area = 2π(2)² + 2π(2)(3)
Surface Area = 8π + 12π
Surface Area = 2<u>0π in²</u>
.
Answer: 20π in²
.
Answer:
Step-by-step explanation:
c.d. d=11-18=4-11=-7
y=18+(n-1)(-7)
or y=18-7n+7
or y=25-7n
n=1,2,3, ...
Answer:
8) (a) total throws = 200
(b)
Experimental probability of throwing an even number (2,4 or 6) = 9/20
Experimental probability of throwing a prime number (2, 3, or 5) = 23/40
Step-by-step explanation:
The question is not well expressed. It should have read the <em>experimental</em> probability is 0.225, because then we can related x to the number of throws. The theoretical probably (again assuming a fair die, not mentioned in the question) is 1/6.
If the experimental probablility is 0.225, then
x / (25+30+x+28+40+32) = 0.225
or
x-0.225x = 0.225(25+30+28+40+32)
0.775x = 0.225(155)
x = 45
(a) Total number of throws = 155+45 = 200
(b)
Experimental probability of throwing an even number (2,4 or 6)
= (30+28+32)/200 = 90/200 = 9/20 (= 45%)
Experimental probability of throwing a prime number (2, 3, or 5)
= (30+45+40)/200
= 115/200
= 23/40
(= 57.5%)
Answer:
9.42 cubic inches
Step-by-step explanation:
We are to find the volume of the cone.
The formula is given as:
1/3πr²h
From the question,
Diameter = 3 inches, Radius = D/2 = 3/2
r = 1.5 inches
h = 4 inches
Hence, the volume of the cone =
1/3 × π × 1.5² × 4
= 9.42 cubic inches
Therefore, the filter can hold 9.42 cubic inches of liquid without overflowing.
