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soldier1979 [14.2K]
3 years ago
10

Ignacio is curious about the average age of cars in the commuter lot at Mis large university. He takes a random

Mathematics
1 answer:
spin [16.1K]3 years ago
4 0

Answer:

Between 12.614 years and 16.386 years

Step-by-step explanation:

Given that:

Mean age (μ) = 14.5 years, standard deviation (σ) = 4.6 years, number o sample (n) and the confidence interval (c) = 90% = 0,9

α = 1 -c = 1 -0.9 = 0.1

\frac{\alpha }{2}  = \frac{0.1}{2} = 0.05

The z score of \alpha /2 is the same as the z score of 0.45 (0.5 - 0.05). This can be gotten from the probability distribution table. Therefore:

z_{0.05}=1.64

The margin of error (e) = z_{0.05}\frac{\sigma}{\sqrt{n} } = 1.64*\frac{4.6}{\sqrt{16} }=1.886

The interval = μ ± e = 14.5 ± 1.886 = (12.614 ,16.386)

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Your 71,1482 doesn't make sense. But regardless,

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You know the total spent and the cost of one car. So logically, if you subtract the two you'd get the price of the second car.

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7 0
4 years ago
Solve each equation y-6.72=253
diamong [38]
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4 years ago
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-1+(-3) is blank units from -1 in the blank direction
Thepotemich [5.8K]

-1+(-3) is  3 units from -1 in the left direction

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Which of the following represents a function?
zhuklara [117]
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a standard deck of cards missing the queen of hearts in the 2 of clubs what is the probability of pulling either an ace or a spa
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<h3>Answer:  8/25</h3>

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

Explanation:

In a standard deck, there are 52 cards.

If this deck is missing the queen of hearts and 2 of clubs, then we really have 52-2 = 50 cards in the deck.

There are 4 aces and 13 spades. Those values add to 4+13 = 17, but we need to subtract off 1 to account for the ace of spades counted twice. We have 17-1 = 16 cards that are either an ace, a spade, or both.

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