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valkas [14]
3 years ago
5

A gardener found that he was able to plant 1/4 of a packet of flower seeds in 1/5 of a garden.At this rate how much of the garde

n would he cover with the entire packet of seeds
Mathematics
1 answer:
telo118 [61]3 years ago
3 0

Answer: \frac{4}{5} of garden.


Step-by-step explanation:

Given: The area covered by \frac{1}{4} of a packet of flower seeds=\frac{1}{5} of a garden.

The area covered in 1 packet=4\times\frac{1}{5}

⇒The area covered in 1 packet=\frac{4}{5} of garden.

Thus, the entire packet of seeds would cover \frac{4}{5} of garden.

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A box contains 17 balls numbered 1 through 17. Two balls are drawn in succession without replacement. If the second ball has the
Brilliant_brown [7]

Answer:

1 ; 7 /17

Step-by-step explanation:

17 balls numbered 1 through 17

Picking without replacement :

If the second ball picked = 4

P(first ball has a smaller number)

Numbers less Than = (3, 2, 1)

P(number less than second ball Given 4 is drawn for second ball) :

= (3/17 * 1/16) ÷ (3/17 * 1/16) +) 14/17 * 0)

= (3 / 272) / (3 /272) * 0

= 3 / 272 * 272 / 3

= 1

2.)

(8/17 * 7/8) ÷ (8/17 * 7/8) + (9/17 * 1/17)

7/17 ÷ (7 /17) + 10/17

7 /17 ÷ 17/17

7/17 ÷ 1

7 /17

7 0
3 years ago
Work out x most dedicated not rushed gets brainiest
Softa [21]

Answer:54

Step-by-step explanation:

Pentagon has 5 sides

n=5

Sum of interior angles=180(n-2)

Sum of interior angles=180(5-2)

Sum of interior angles=180x3

Sum of interior angles=540

Size of each interior angle =540/n

Size of each interior angle =540/5

Size of each interior angle=108

x=108/2

x=54

6 0
3 years ago
In this figure, ∠a and ∠b are
ELEN [110]

angles a and b are supplementary because they form a straight line

5 0
3 years ago
Read 2 more answers
I need help with this
slega [8]
Y is greater than or equal to 5. It says no less than, making the equal to possible.
6 0
3 years ago
Please help me :)))))
svetoff [14.1K]
Part a)

Answer: 5*sqrt(2pi)/pi

-----------------------

Work Shown:

r = sqrt(A/pi)
r = sqrt(50/pi)
r = sqrt(50)/sqrt(pi)
r = (sqrt(50)*sqrt(pi))/(sqrt(pi)*sqrt(pi))
r = sqrt(50pi)/pi
r = sqrt(25*2pi)/pi
r = sqrt(25)*sqrt(2pi)/pi
r = 5*sqrt(2pi)/pi

Note: the denominator is technically not able to be rationalized because of the pi there. There is no value we can multiply pi by so that we end up with a rational value. We could try 1/pi, but that will eventually lead back to having pi in the denominator. I think your teacher may have made a typo when s/he wrote "rationalize all denominators"

============================================================

Part b)

Answer: 3*sqrt(3pi)/pi

-----------------------

Work Shown:

r = sqrt(A/pi)
r = sqrt(27/pi)
r = sqrt(27)/sqrt(pi)
r = (sqrt(27)*sqrt(pi))/(sqrt(pi)*sqrt(pi))
r = sqrt(27pi)/pi
r = sqrt(9*3pi)/pi
r = sqrt(9)*sqrt(3pi)/pi
r = 3*sqrt(3pi)/pi

Note: the same issue comes up as before in part a)

============================================================

Part c)

Answer: sqrt(19pi)/pi

-----------------------

Work Shown:

r = sqrt(A/pi)
r = sqrt(19/pi)
r = sqrt(19)/sqrt(pi)
r = (sqrt(19)*sqrt(pi))/(sqrt(pi)*sqrt(pi))
r = sqrt(19pi)/pi
8 0
3 years ago
Read 2 more answers
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