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DENIUS [597]
2 years ago
8

The cost of printing a catalog of the products that your company distributes is a flat rate of $52 plus $0.20 per copy. Let C re

present the total cost of printing and x represent the number of copies. Write in symbolic form an equation that will relate C and x. Organize the data into a table of values. Begin with 500 copies, increase by increments of 500, and end with 2500 copies.
Answer by selected letter:

A.
B.
C.
D.
​

Mathematics
1 answer:
marusya05 [52]2 years ago
7 0
The answer is d, just find the equation and then plug in the first few numbers of copies and you’ll find the correct answer

mark me brainliest if the answer is correct, thank you in advance :)
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Answer:

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Step-by-step explanation:

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What is the multiplicative rate of change for the exponential function graphed to the left​
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- Draw a line AB 8 cm long. What is the locus of a point C such that the angle ACB is always a right angle?​
bagirrra123 [75]

Answer:

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2 years ago
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
I forget how to use fractions, what's 7 times 1/2
bekas [8.4K]
In exact form it is 7/2

In decimal form it is 3.5

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