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3 years ago

Please help with this math question

Mathematics

2 answers:

Ilya [14]3 years ago

8 0

Answer:

(a) The continuously compounding account

(b) $1.39

Step-by-step explanation:

(a) At identical interest rates, continuous compounding will yield the greatest account balance when compared to periodic forms of compounding (i.e., yearly, quarterly, monthly, weekly, etc.).

(b) By compounding interest monthly, Gregory's account balance will be $6,291.61 after 10 years. By compounding interest continuously, his balance will be $6,293.00. Therefore, by choosing the continuously compounding account, he will earn $1.39 more over 10 years.

Hatshy [7]3 years ago

3 0

Answer:

(a) Account B

(b) $1.39,

Step-by-step explanation:

(a) The account with continuous compound interest will earn the most.

(b) Amount with 2.3% interest = 5000(1 + (0,023/12))^120

= $6291.61.

Amount with Continuous compound interest

= 5000e^(0.023*10) = $6293.00.

Greg earns $1.39 more with Account B.

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A professor pays 25 cents for each blackboard error made in lecture to the student who pointsout the error. In a career ofnyears

marta [7]

Answer:

(a) The probability that Y₂₀ exceeds 1000 is 3.91 × 10⁻⁶.

(b) n = 28.09

Step-by-step explanation:

The random variable Yₙ is defined as the total numbers of dollars paid in n years.

It is provided that Yₙ can be approximated by a Gaussian distribution, also known as Normal distribution.

The mean and standard deviation of Yₙ are:

$\mu_{Y_{n}}=40n\\\sigma_{Y_{n}}=\sqrt{100n}$

(a)

For n = 20 the mean and standard deviation of Y₂₀ are:

$\mu_{Y_{n}}=40n=40\times20=800\\\sigma_{Y_{n}}=\sqrt{100n}=\sqrt{100\times20}=44.72\\$

Compute the probability that Y₂₀ exceeds 1000 as follows:

$P(Y_{n}>1000)=P(\frac{Y_{n}-\mu_{Y_{n}}}{\sigma_{Y_{n}}}>\frac{1000-800}{44.72})\\=P(Z> 4.47)\\=1-P(Z$

**Use a z table for probability.

Thus, the probability that Y₂₀ exceeds 1000 is 3.91 × 10⁻⁶.

(b)

It is provided that P (Yₙ > 1000) > 0.99.

$P(Y_{n}>1000)=0.99\\1-P(Y_{n}$

The value of z for which P (Z < z) = 0.01 is 2.33.

Compute the value of n as follows:

$z=\frac{Y_{n}-\mu_{Y_{n}}}{\sigma_{Y_{n}}}\\2.33=\frac{1000-40n}{\sqrt{100n}}\\2.33=\frac{100}{\sqrt{n}}-4\sqrt{n} \\2.33=\frac{100-4n}{\sqrt{n}} \\5.4289=\frac{(100-4n)^{2}}{n}\\5.4289=\frac{10000+16n^{2}-800n}{n}\\5.4289n=10000+16n^{2}-800n\\16n^{2}-805.4289n+10000=0$

The last equation is a quadratic equation.

The roots of a quadratic equation are:

$n=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

a = 16

b = -805.4289

c = 10000

On solving the last equation the value of n = 28.09.

8 0

3 years ago

Use the numbers 6 2 9 5 2 3 to make 36.

VashaNatasha [74]

Answer:

6+2+9+5+2+3+6+3 = 36

Step-by-step explanation:

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3 years ago

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sammy [17]

Answer:

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Step-by-step explanation:

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3 years ago

NEED HELP ASAP PLEASS AND THANK YOU

Ghella [55]

Answer:

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7 0

3 years ago

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mart [117]

Answer:

28 square units

Step-by-step explanation:

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3 years ago