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Shkiper50 [21]
3 years ago
10

Determine whether carpeted rooms have more bacteria than uncarpeted rooms at the level of significance. Normal probability plots

indicate that the data are approximately normal and boxplots indicate that there are no outliers.
Carpeted Un-Carpeted
15.9 8.1 15 7.2 10.4 6.1
11.2 16 11.2 6.8 11.3 6.8
8.6 10 12.1 6.3

Requried:
State the null and alternative hypotheses. Let population 1 be carpeted rooms and population 2 be uncarpeted rooms.
Mathematics
1 answer:
Mashcka [7]3 years ago
3 0

Answer:

The null (H_{0}) and alternative (H_{1}) hypotheses can be stated as follows:

H_{0}:  μ_{1} = μ_{2}.

H_{1}: μ_{1} > μ_{2}.

Step-by-step explanation:

In the question, the given assertion is that carpeted rooms have more bacteria than uncarpeted rooms, i.e. μ_{1} > μ_{2}.

The assertion made in the question can either be the null hypothesis (H_{0}) or the alternate hypothesis (H_{1}). An equality must be present in the null hypothesis. The alternative hypothesis expresses the exact opposite of the null hypothesis if the claim is the null hypothesis.

Therefore, the null (H_{0}) and alternative (H_{1}) hypotheses can be stated as follows:

H_{0}:  μ_{1} = μ_{2}.

H_{1}: μ_{1} > μ_{2}.

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Answer:

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Step-by-step explanation:

Using the rules of exponents

a^{m} ÷ a^{n} = a^{(m-n)}

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Note 81 = 9² and 729 = 9³

Given

(81)^{-4} ÷ (729)^{2-p}

= (9^{2}) ^{-4} ÷ (9^{3}) ^{2-p}

= 9^{-8} ÷ 9^{6-3p}

= 9^{-8-(6-3p)}

= 9^{-8-6+3p}

= 9^{3p-14}

Thus

9^{3p-14} = 9^{4p}

Since bases on both sides are equal then equate the exponents

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6 0
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36.2 rounding to the nearest tenth​
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This is what I found

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Answer:

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4 years ago
A hot air balloon is flying at a constant speed of 20 mi/h at a bearing of N 36° E. There is a 10mi/h cross wind blowing due eas
Ksju [112]

Answer:

The actual speed = 27.12 mi/h

Direction = 36.5° in NE(north of east)

Step-by-step explanation:

As given , A hot air balloon is flying at a constant speed of 20 mi/h at a bearing of N 36° E.

⇒θ = 36°

Let v₀ be the constant speed, then v₀ = 20

Let vₓ be the speed in East direction

     v_{y} be the speed in North direction

So,

vₓ  = v₀ sin(θ) = 20 sin(36°) + 10 ( As given, There is a 10mi/h cross wind blowing due east.)

⇒vₓ  = 20(0.588) + 10 = 11.76 + 10 = 21.76 mi/h

and   v_{y}  = v₀ cos(θ) = 20 cos(36°) = 20(0.809) = 16.18 mi/h

Now,

the actual speed = √(vₓ)² + (v_{y})²

                            = √(21.76)² + (16.18)²

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⇒The actual speed = 27.12 mi/h

Now,

Direction = θ = tan^{-1}(\frac{v_{y} }{v_{x} } ) = tan^{-1}(\frac{16.18 }{21.76 } ) =  tan^{-1}(0.74) = 36.5

⇒ Direction = 36.5° in NE(north of east)

7 0
3 years ago
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