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MrMuchimi
3 years ago
12

Observe randomly the weight of 100 chickens at farm A. The sample mean is 3.4 kg and sample standard deviation is 0.5. Observe r

andomly the weight of 200 chickens at farm B. The sample mean is 4.1 kg and sample standard deviation is 0.9. With the significance level of 2%, can we conclude that the true average weight of farm A is something other than farm B ?
Mathematics
1 answer:
attashe74 [19]3 years ago
4 0

Answer:

40 percent goes to farmers so hipe it

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437.445 yards

Step-by-step explanation:

Convert: its 437.445 yards.

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2 years ago
Find angle PRS in the image above
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142°

Step-by-step explanation:

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3 years ago
What is( -10a^5-2a^5+14a^5)(12a^5-6a-10a^3)<br><br> using distributive and the area model.
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(-10a^5-2a^5+14a^5)(12a^5-6a-10a^3) \\ = 2a^5 (12a^5-6a-10a^3) \\ = 24a^{10} - 12a^6 - 20a^8
8 0
3 years ago
45 points! Please help!
Tju [1.3M]

Answer:

Sequence One

Step-by-step explanation:

S = a₁ / (1 − r) is the infinite sum of a geometric series where |r| < 1.

Sequence One: r = 1/5

Sequence Two: r = 2

Sequence Three: not geometric

Sequence Four: r = -3/2

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5 0
3 years ago
A round wading pool has a diameter of 115cm deep. A hose is filling the pool at a rate of 34000cm^3, cubed per minute. How long
inn [45]

Answer:

It will take 6.118 minutes to fill up the pool to a depth of 20 cm

Step-by-step explanation:

The first step is to calculate the volume of the wading pool.

we will assume it is a cylinder, hence the volume will be = \pi r^{2}h

<em>Where r= radius of the pool = 115/2 = 57.5cm</em>

<em>h = depth of the pool =20 cm</em>

The volume of the pool will be \pi \times57.5^{2} \times 20 =2.08 \times 10^{5} cm ^{3}

We are filling a pool of 208,000cm ^{3} at the rate of 34000cm^3, cubed per minute.

To get the time it will take to fill up the pool, we will have to divide as follows:

208,000cm ^{3} / 34000cm^3 =6.118 minutes

Therefore it will take 6.118 minutes to fill up the pool to a depth of 20 cm

4 0
3 years ago
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