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Reil [10]
3 years ago
15

list some things I could spend a million dollars on.. doing a math worksheet and I can’t think of things.. help!

Mathematics
2 answers:
Lisa [10]3 years ago
6 0
A car, a plane, a boat, a house.
svet-max [94.6K]3 years ago
4 0

Answer:

A car, college, and house, bills. and Airplane? would those work?

Step-by-step explanation:

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An apple has an unknown cost of a

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I will have to pay 6a for the apples
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Area of circle
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Answer:

1. area

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Step-by-step explanation:

4 0
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Find the length of each arc. Round to the nearest tenth.
allochka39001 [22]

Answer:

  5.5 units

Step-by-step explanation:

Arc length is given by ...

  s = rθ

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6 0
3 years ago
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
David is building wooden picture frames. Each frame requires 3/4 feet of wood. If David started with 12 feet of wood and now has
ZanzabumX [31]

Answer: 8

Step-by-step explanation:

6 0
4 years ago
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