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denis23 [38]
3 years ago
15

The paint used to make lines on roads must reflect enough light to be clearly visible at night. Let μ denote the true average re

flectometer reading for a new type of paint under consideration. A test of H0: μ = 20 versus Ha: μ > 20 will be based on a random sample of size n from a normal population distribution. What conclusion is appropriate in each of the following situations?
(a) n = 14, t = 3.1, α = 0.05

(b) n = 10, t = 1.8, α = 0.01
Mathematics
1 answer:
Sidana [21]3 years ago
4 0
A.) t calc = 3.1
t-table = 1.77
Since the t-calc > t-table, we reject H0 and accep HA.
We conclude that the true average refrectometer reading is greater than 20.

b.) t-calc = 1.8
t-table = 2.82
Since t-calc < t-table, we fail to reject H0.
We conclude that the true average refrectometer reading may be 20.

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I need help with these three problems please​
Liono4ka [1.6K]

Answer:

Q7. 11.3 inches (3 s.f.)

Q8. 96.2 ft

Q9. 36.4cm

Step-by-step explanation:

Q7. Please see attached picture for full solution.

Q8. Let the length of a side of the square be x ft.

Applying Pythagoras' Theorem,

34^{2}  =  {x}^{2}  +  {x}^{2}  \\ 2 {x}^{2}  = 1156 \\  {x}^{2}  = 1156 \div 2 \\  {x}^{2}  = 578 \\ x =  \sqrt{578}  \\

Thus, the perimeter of the square is

= 4( \sqrt{578} ) \\  = 96.2 ft\:  \:  \: (3 \: s.f.)

Q9. Equilateral triangles have 3 equal sides and each interior angle is 60°.

Since the perimeter of the equilateral triangle is 126cm,

length of each side= 126÷3 = 42 cm

The green line drawn in picture 3 is the altitude of the triangle.

Let the altitude of the triangle be x cm.

sin 60°= \frac{x}{42}

\frac{ \sqrt{3} }{2}  =  \frac{x}{42}  \\ x =  \frac{ \sqrt{3} }{2}  \times 42 \\ x = 21 \sqrt{3}  \\ x = 36.4

(to 3 s.f.)

Therefore, the length of the altitude of the triangle is 36.4cm.

4 0
3 years ago
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There are some nickels, dimes, and quarters in a large piggy bank. For every 2 nickels there are 3 dimes. For every 2 dimes ther
Novay_Z [31]

Answer:

thats a lot of work....

Step-by-step explanation:

5 0
3 years ago
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A Roper survey reported that 65 out of 500 women ages 18-29 said that they had the most say when purchasing a computer; a sample
8090 [49]

Answer:

Step-by-step explanation:

<u><em>Step(i):-</em></u>

<em>Given first random sample size n₁ = 500</em>

Given  Roper survey reported that 65 out of 500 women ages 18-29 said that they had the most say when purchasing a computer.

<em>First sample proportion </em>

<em>                              </em>p^{-} _{1} = \frac{65}{500} = 0.13

<em>Given second sample size n₂ = 700</em>

<em>Given a sample of 700 men (unrelated to the women) ages 18-29 found that 133 men said that they had the most say when purchasing a computer.</em>

<em>second sample proportion </em>

<em>                              </em>p^{-} _{2} = \frac{133}{700} = 0.19

<em>Level of significance = α = 0.05</em>

<em>critical value = 1.96</em>

<u><em>Step(ii)</em></u><em>:-</em>

<em>Null hypothesis : H₀: There  is no significance difference between these proportions</em>

<em>Alternative Hypothesis :H₁: There  is significance difference between these proportions</em>

<em>Test statistic </em>

<em></em>Z = \frac{p_{1} ^{-}-p^{-} _{2}  }{\sqrt{PQ(\frac{1}{n_{1} } +\frac{1}{n_{2} } )} }<em></em>

<em>where </em>

<em>         </em>P = \frac{n_{1} p^{-} _{1}+n_{2} p^{-} _{2}  }{n_{1}+ n_{2}  } = \frac{500 X 0.13+700 X0.19  }{500 + 700 } = 0.165<em></em>

<em>        Q = 1 - P = 1 - 0.165 = 0.835</em>

<em></em>Z = \frac{0.13-0.19  }{\sqrt{0.165 X0.835(\frac{1}{500 } +\frac{1}{700 } )} }<em></em>

<em>Z =  -2.76</em>

<em>|Z| = |-2.76| = 2.76 > 1.96 at 0.05 level of significance</em>

<em>Null hypothesis is rejected at 0.05 level of significance</em>

<em>Alternative hypothesis is accepted at 0.05 level of significance</em>

<u><em>Conclusion:</em></u><em>-</em>

<em>There is there is a difference between these proportions at α = 0.05</em>

3 0
3 years ago
Answer two questions about Equations A and B:
Lyrx [107]

Answer:

we can get the value by our IQ... i.e if we take x is equals to 2 then it must be 5 is equal to 3

4 0
3 years ago
Noah’s dog weighs 15 1⁄3 pounds (lbs). Aiden’s dog weighs three times as much as Noah’s. How many pounds does Aiden’s dog weigh?
SpyIntel [72]
<h3>Answer:  46 pounds</h3>

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

Work Shown:

N = weight of Noah's dog = 15 & 1/3 pounds

A = weight of Aiden's dog = unknown

A = 3*N

A = 3*(15 & 1/3)

A = 3*(15 + 1/3)

A = 3*15 + 3*(1/3) .... distribute

A = 45 + 1

A = 46

Aiden's dog weighs 46 pounds.

4 0
3 years ago
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