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Alina [70]
3 years ago
13

Am I correct? Will give brainliest!!

Mathematics
2 answers:
nevsk [136]3 years ago
7 0

Yes that would be the correct answer

Katarina [22]3 years ago
6 0
That is the correct answer
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PLEASE HELP 100 POINTS TO WHO HELPS ME
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Answer:

1. 13.65, 136.5, 1365    

2. 57.14, 5.714, .5714

Step-by-step explanation:

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A particular shade of orange paint has 2 cups of yellow paint for every 3 cups of red paint.According to the double number line,
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Answer:9 cups of red paint

Step-by-step explanation:1 batch = 3 cups of paint and you need to find the number of cups of paint for 3 batches so you multiply 3 x 3 and get 9. Hope this helped!!!:)

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Adding and Subtracting Functions (20 POINTS and MARK BRAINLIEST)
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Hi I wanna is a good day to you 568
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A random sample of 500 registered voters in Phoenix is asked if they favor the use of oxygenated fuels year-round to reduce air
Stells [14]

Answer:

a) 0.0853

b) 0.0000

Step-by-step explanation:

Parameters given stated that;

H₀ : <em>p = </em>0.6

H₁ : <em>p  = </em>0.6, this explains the acceptance region as;

p° ≤ \frac{315}{500}=0.63 and the region region as p°>0.63 (where p° is known as the sample proportion)

a).

the probability of type I error if exactly 60% is calculated as :

∝ = P (Reject H₀ | H₀ is true)

   = P (p°>0.63 | p=0.6)

where p° is represented as <em>pI</em><em> </em>in the subsequent calculated steps below

   

    = P  [\frac{p°-p}{\sqrt{\frac{p(1-p)}{n}}} >\frac{0.63-p}{\sqrt{\frac{p(1-p)}{n}}} |p=0.6]

    = P  [\frac{p°-0.6}{\sqrt{\frac{0.6(1-0.6)}{500}}} >\frac{0.63-0.6}{\sqrt{\frac{0.6(1-0.6)}{500}}} ]

    = P   [Z>\frac{0.63-0.6}{\sqrt{\frac{0.6(1-0.6)}{500} } } ]

    = P   [Z > 1.37]

    = 1 - P   [Z ≤ 1.37]

    = 1 - Ф (1.37)

    = 1 - 0.914657 ( from Cumulative Standard Normal Distribution Table)

    ≅ 0.0853

b)

The probability of Type II error β is stated as:

β = P (Accept H₀ | H₁ is true)

  = P [p° ≤ 0.63 | p = 0.75]

where p° is represented as <em>pI</em><em> </em>in the subsequent calculated steps below

  = P [\frac{p°-p} \sqrt{\frac{p(1-p)}{n} } }\leq \frac{0.63-p}{\sqrt{\frac{p(1-p)}{n} } } | p=0.75]

  = P [\frac{p°-0.6} \sqrt{\frac{0.75(1-0.75)}{500} } }\leq \frac{0.63-0.75}{\sqrt{\frac{0.75(1-0.75)}{500} } } ]

  = P[Z\leq\frac{0.63-0.75}{\sqrt{\frac{0.75(1-0.75)}{500} } } ]

  = P [Z ≤ -6.20]

  = Ф (-6.20)

  ≅ 0.0000 (from Cumulative Standard Normal Distribution Table).

6 0
3 years ago
Adult male heights are normally distributed with a mean of 70 inches and a standard deviation of 3 inches. The average basketbal
Marizza181 [45]
Z = (x - mean) / SD = (79 - 70) / 3 = 3 
<span>P (Z > 3)? = 1 - F (z) = 1 - F (3) = 0.00135</span>
5 0
3 years ago
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