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dexar [7]
3 years ago
8

What is the 5-digit code?

Mathematics
1 answer:
kvv77 [185]3 years ago
3 0

9514 1404 393

Answer:

  4 2 6 1 7 or 1 2 4 3 7

Step-by-step explanation:

4 are correct of 1,3,4,6,7

3 are correct of 0,2,3,6,7

1 is correct of 0,3,6,8,7 -- so not both 3 and 6, meaning 1, 4, 7 are correct

Correct digits so far are 1, 4, 7 and one of {3, 6}. Line 2 tells another is 0 or 2; line 3 tells it is not 0. This adds 2 to the list of correct digits.

This makes the correct digits 1, 2, {3, 6}, 4, 7. We don't know whether the last digit is 3 or 6.

__

Looking at digits correctly placed, line 2 tells us the code is either ...

  _ 2 6 _ 7  or  _ 2 _ 3 7

Then line 1 fills in those blanks as ...

  4 2 6 1 7  or  1 2 4 3 7

Either of these codes agrees with all of the clues.

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A marketing research consultant hired by Coca-Cola is interested in determining if the proportion of customers who prefer Coke t
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Answer:

A one-tailed hypothesis will be used to perform the test.

Step-by-step explanation:

The purpose of the marketing research consultant hired by Coca-Cola is to determine whether the the proportion of customers who prefer Coke to other brands is over 50%.

The marketing research consultant selected a random sample of <em>n</em> = 200 customers. The sample proportion of people who favored Coca-Cola over other brands was 55%.

The marketing research consultant can perform a single proportion hypothesis test to determine whether greater than 50% of customers prefer Coca-Cola to other brands.

Since we need to determine whether the population percentage is greater than a null value, the hypothesis is not two-tailed.

The hypothesis can be defined as:

<em>H₀</em>: The proportion of people who favor Coca-Cola over other brands was 55%, i.e. <em>p</em> = 0.50.

<em>Hₐ</em>: The proportion of people who favor Coca-Cola over other brands was more than 55%, i.e. <em>p</em> > 0.50.

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6 0
3 years ago
(2a^x + b^x)^2 i need help ASAP
Finger [1]

Answer:

4a^{2x} + 4a^{x}b^{x} + b^{2x}

Step-by-step explanation:

Using the rule of exponents

a^{m} × a^{n} ⇔ a^{(m+n)}

Given

(2a^{x} + b^{x} ) = (2a^{x} + b^{x} )(2a^{x} + b^{x} )

Each term in the second factor is multiplied by each term in the first factor, that is

2a^{x} (2a^{x} + b^{x} ) + b^{x} (2a^{x} + b^{x} ) ← distribute both parenthesis

= 4a^{2x} +2 a^{x}b^{x} + 2a^{x}b^{x} + b^{2x} ← collect like terms

= 4a^{2x} + 4a^{x}b^{x} + b^{2x}

6 0
3 years ago
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