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romanna [79]
3 years ago
8

The average price of a laptop is $965. Assume laptop prices are approximately normally distributed with a standard

Mathematics
1 answer:
miss Akunina [59]3 years ago
6 0

Answer:

$836.8

Step-by-step explanation:

Average price = mean = $965

Standard deviation, = $100

Given that distribution is approximately normal ;

The least expensive 10% of the laptops :

We Obtain the Zscore that corresponds to P(Z ≤ 0.1) ; this means the least 10% of the laptops ;

From, a normal probability distribution table ;

P(Z ≤ 0.1) = - 1.282

We substitute this into the Zscore formula :

Zscore = (x - mean ) / standard deviation

x = price

-1.282 = (x - 965) / 100

-128.2 = (x - 965)

x = - 128.2 + 965

x = $836.8

Hence, price is $836.8

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42 is what % of 90 as a percent
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<h3>Answer:  g(x) = (-2/3)*x^2</h3>

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

Work Shown:

f(x) = x^2

g(x) = a*f(x) for some constant 'a' since g(x) is a scaled version of f(x).

The value of 'a' vertically stretches f(x) upward if a > 0

If 'a' is negative, then we have a reflection going on as shown in the diagram.

We want (x,y) = (3,-6) to be on the graph of g(x). This means g(3) = -6

If we plugged x = 3 into f(x), we get

f(x) = x^2

f(3) = 3^2

f(3) = 9

So,

g(x) = a*f(x)

g(3) = a*f(3) ... replace x with 3

g(3) = a*9 ... replace f(3) with 9 since f(3) = 9

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A sample of 500 nursing applications included 60 from menThe 95% confidence interval of the true proportion of men who applied t
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Answer:

The correct solution is "(0.092, 0.148)".

Step-by-step explanation:

Given:

Sample size,

n = 500

True proportion,

\hat {p} = \frac{60}{500}

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The standard error will be:

SE=\sqrt{\frac{\hat p(1-\hat p)}{n} }

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The confidence interval is "1.96".

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The required confidence interval will be:

= (\hat p - 1.96 SE, \hat p + 1.96 SE)

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