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never [62]
2 years ago
11

What will be the sales price of the car if it has a regular price of $16 000 and is on sale for 20% off? Determine the cost of t

he car including 13% tax.
Mathematics
1 answer:
jok3333 [9.3K]2 years ago
6 0

Answer:

14,464

Step-by-step explanation:

16,000 - 20% = 12,800

12,800 + 13% = 14,464

I hoope this is correct. Have a wonderful day and dream a good dream of good people. :)

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Des Moines - 40F
Austin Carson City - 42F
Jackson - 73F
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3 years ago
kate spent 3/5 of her money on a manicure and had $45 left. how much money did she have at first?(PLEASE SHOW STEPS!)
prisoha [69]
Lets say she has x money

Now she spent 3/5of x means 2/5 is left
2/5*x=45
x=112.5
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7 0
3 years ago
A recipe called for the ratio of sugar to flour to be 7 : 2. If you used 63 ounce of sugar, how many ounces
lora16 [44]

Answer:

18 ounces of flour

Step-by-step explanation:

63 ÷ 7 = 9

So one share is 9

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4 0
3 years ago
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Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.7
Zinaida [17]

Answer:

(a) The probability that the sample average sediment density is at most 3.00 is 0.092.

    The probability that the sample average sediment density is between 2.70 and 3.00 is 0.477.

(b) The sample size must be at least 77.

Step-by-step explanation:

The random variable <em>X</em> ca be defined as the sediment density (g/cm) of a specimen from a certain region.

The random variable <em>X</em> is normally distributed with mean, <em>μ</em> = 2.7 and standard deviation, <em>σ</em> = 0.75.

(a)

A random sample of <em>n</em> = 25 specimens is selected.

Compute the probability that the sample average sediment density is at most 3.00 as follows:

Apply continuity correction:

P(\bar X\leq 3.00)=P(\bar X

                    =P(\bar X

                    =P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}

                    =P(Z

Thus, the probability that the sample average sediment density is at most 3.00 is 0.092.

Compute the probability that the sample average sediment density is between 2.70 and 3.00 as follows:

P(2.70

                                =P(0

*Use a <em>z</em>-table.

Thus, the probability that the sample average sediment density is between 2.70 and 3.00 is 0.477.

(b)

It is provided that:

P(\bar X\leq 3.00)\geq 0.99

P(\bar X

P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}

P(Z

The value of <em>z</em> for the probability above is

<em>z</em> ≥ 2.33

Compute the value of <em>n</em> as follows:

\frac{|2.50-2.70|}{0.75/\sqrt{n}}\geq 2.33

\frac{|-0.20|}{2.33}\geq \frac{0.75}{\sqrt{n}}

\sqrt{n}\geq \frac{0.75\times 2.33}{|-0.20|}\\\sqrt{n}\geq 8.7375\\n\geq 76.3439\\\approx n\geq 77

Thus, the sample size must be at least 77.

8 0
3 years ago
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