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Dmitriy789 [7]
3 years ago
10

Which transformation is not a rigid motion?

Mathematics
1 answer:
Bumek [7]3 years ago
8 0
D. Dilate by a factor of 2 is Not a rigid transformation
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Chaleah deposited $900 in a new account that earns 6% simple interest. After 2 years, how much interest will she have earned?
nikklg [1K]

Answer:

$108

Step-by-step explanation:

<u>Use the equation i = prt </u><u>(interest = principle*rate*time)</u><u>:</u>

Principle: $900

Rate: 0.06 (6%)

Time: 2 years

<u>Put it all in an equation:</u>

i = 900*0.06*2

i = $108

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What is the relationship between the two quantities in the table?
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The relationship is 9x
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Step-by-step explanation:

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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
Is 10/9 closest to 0,1/2 or 1
Juliette [100K]
Since 10/9 is 1.1 repeating, it's closer to 1
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4 years ago
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