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IrinaK [193]
3 years ago
8

Victoria is shopping for thank-you notes at Sweet Greetings. She is trying to decide between the three different packages of not

es shown below. Package 1 6 notes for $3.00 Package 2 8 notes for $6.00 Package 3 10 notes for $4.00 If Victoria is looking for the best deal, which package of thank-you notes should she buy?
Mathematics
1 answer:
ivann1987 [24]3 years ago
6 0
Your answer would be 10 notes for $4. If you need your answer explained just let me know. :)
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Factor the following polynomial 14x^2 - 35x
KonstantinChe [14]

Answer:

The binomial 14x^{2} - 35x can be factored as 7x(2x+5)

Step-by-step explanation:

When factoring, you want to factor each constant or variable by their greatest common factor (or GCF). Since 14x^{2} and 35x has a greatest common factor of 7x, factor each term by 7x and you'll get 7x(2x+5).

4 0
3 years ago
Bailey spent $12.25 at the arcade on Saturday and $9.50 at the arcade on Sunday. How much money did Bailey spend anogether?
galben [10]

Answer:

21.75

Step-by-step explanation:

i added 12.25 by 9.50

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3 years ago
How is the decimal 2.4 written as a mixed number in its simplest form
My name is Ann [436]
2 4/10
2 2/5
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3 years ago
NewPop produces their brand of soda drinks in a factory where they claim that the
Citrus2011 [14]

Answer:

a) 14960 bottles

b) 502 bottles

Step-by-step explanation:

Given that:

Mean (μ) = 24 ounces, standard deviation (σ) = 0.14 ounces

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Therefore 68% fall within 0.14 ounces of the mean

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b) To solve this we are going to use the z score equation given as:

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z=\frac{x-\mu}{\sigma}=\frac{23.72-24}{0.14} =-2

From the normal probability distribution table: P(X < 23.72) = P (Z < -2) = 0.0228

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bija089 [108]
We can use the binomial theorem to find the probability that 0 out of the 15 samples will be defective, given that 20% are defective.
P(0/15) = (15C0) (0.2)^0 (1 - 0.2)^15 = (1)(1)(0.8)^15 = 0.0352
Then the probability that at least 1 is defective is equal to 1 - 0.0352 = 0.9648. This means there is a 96.48% chance that at least 1 of the 15 samples will be found defective. This is probably sufficient, though it depends on her significance level. If the usual 95% is used, then this is enough.
7 0
3 years ago
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