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Dmitry [639]
3 years ago
13

The starting salaries of individuals with an mba degree are normally distributed with a mean of $40,000 and a standard deviation

of $5,000. refer to exhibit 6-4. what is the probability that a randomly selected individual with an mba degree will get a starting salary of at least $30,000
Mathematics
1 answer:
horsena [70]3 years ago
6 0
<span>With the mean of 40k and standard deviation of 5k, we need to find P(x>=30k). P((30k-40k)/5k) = P(z<-2) = 1-0.0228 which is equal to 0.9772. There is a 97.72% chances that the starting salary will be at least 30k.</span>
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Answer:

Maria's book was 8 days late

Step-by-step explanation:

According to the information given, you can write an equation that states that 15¢ multiply for the number of days that the book was late plus $2 is equal to $3.20:

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\large\boxed{\dfrac{n^3}{8}\sqrt{15n}}

Step-by-step explanation:

\sqrt{\dfrac{60n^{11}}{256n^4}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\ \text{and}\ \sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}\\\\=\sqrt{\dfrac{60}{256}n^{11-4}}=\dfrac{\sqrt{60n^7}}{\sqrt{256}}=\dfrac{\sqrt{4n^{6+1}\cdot15}}{16}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=\dfrac{\sqrt{4n^6\cdot15n}}{16}\qquad\text{use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\=\dfrac{\sqrt{4n^{3\cdot2}}\cdot\sqrt{15n}}{16}\qquad\text{use}\ (a^n)^m=a^{nm}

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