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Marizza181 [45]
3 years ago
9

Of 20 >

Mathematics
1 answer:
attashe74 [19]3 years ago
8 0

Answer: 1.5%

Step-by-step explanation:

The Pure Interest Rate refers to the rate in a theoretical market where there is no risk as economic conditions can be predicted with certainty.

Sadly, uncertainty will always exist but there are rates that are close enough to the Pure interest rate such as US Government Security rates.

The United States has never defaulted on a debt payment in the modern era and as such is known to offer the least risky rates in the world. The rate on US Securities can therefore be considered as close as possible to the pure interest rate.

Given the rates in the question, the Pure Interest rate would be the 20-year Treasury bond rate of 1.5%.

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Answer:

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Step-by-step explanation:

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a prototype rocket burns 159 gallons of liquid hydrogen fuel each second during a regular launch. Make a table that shows how ma
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Answer:

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Answer:

(x, y)  -->   (x + 14, y + 8)

Step-by-step explanation:

Look at 1 original point and its corresponding translated point.

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Then you need to go up in y 8 units.

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(x, y)  -->   (x + 14, y + 8)

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P(x < 21 | μ = 23 and σ = 3) enter the probability of fewer than 21 outcomes if the mean is 23 and the standard deviation is
Shkiper50 [21]

Answer:

(a) The value of P (X < 21 | <em>μ </em> = 23 and <em>σ</em> = 3) is 0.2514.

(b) The value of P (X ≥ 66 | <em>μ </em> = 50 and <em>σ</em> = 9) is 0.0427.

(c) The value of P (X > 47 | <em>μ </em> = 50 and <em>σ</em> = 5) is 0.7258.

(d) The value of P (17 < X < 24 | <em>μ </em> = 21 and <em>σ</em> = 3) is 0.7495.

(e) The value of P (X ≥ 95 | <em>μ </em> = 80 and <em>σ</em> = 1.82) is 0.

Step-by-step explanation:

The random variable <em>X</em> is Normally distributed.

(a)

The mean and standard deviation are:

\mu=23\\\sigma=3

Compute the value of P (X < 21) as follows:

P(X

                  =P(Z

Thus, the value of P (X < 21 | <em>μ </em> = 23 and <em>σ</em> = 3) is 0.2514.

(b)

The mean and standard deviation are:

\mu=50\\\sigma=9

Compute the value of P (X ≥ 66) as follows:

Use continuity correction.

P (X ≥ 66) = P (X > 66 - 0.5)

                = P (X > 65.5)

                =P(\frac{X-\mu}{\sigma}>\frac{65.5-50}{9})

                =P(Z>1.72)\\=1-P(Z

Thus, the value of P (X ≥ 66 | <em>μ </em> = 50 and <em>σ</em> = 9) is 0.0427.

(c)

The mean and standard deviation are:

\mu=50\\\sigma=5

Compute the value of P (X > 47) as follows:

P(X>47)=P(\frac{X-\mu}{\sigma}>\frac{47-50}{5})

                 =P(Z>-0.60)\\=P(Z

Thus, the value of P (X > 47 | <em>μ </em> = 50 and <em>σ</em> = 5) is 0.7258.

(d)

The mean and standard deviation are:

\mu=21\\\sigma=3

Compute the value of P (17 < X < 24) as follows:

P(17

                          =P(-1.33

Thus, the value of P (17 < X < 24 | <em>μ </em> = 21 and <em>σ</em> = 3) is 0.7495.

(e)

The mean and standard deviation are:

\mu=80\\\sigma=1.82

Compute the value of P (X ≥ 95) as follows:

Use continuity correction:

P (X ≥ 95) = P (X > 95 - 0.5)

                = P (X > 94.5)

                =P(\frac{X-\mu}{\sigma}>\frac{94.5-80}{1.82})

                =P(Z>7.97)\\=1-P(Z

Thus, the value of P (X ≥ 95 | <em>μ </em> = 80 and <em>σ</em> = 1.82) is 0.

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