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mamaluj [8]
3 years ago
7

On a finance exam, 15 accounting majors had an average grade of 90. On the same exam, 7 marketing majors averaged 85, and 10 fin

ance majors averaged 93. What is the weighted mean for all 32 students taking the exam?
Mathematics
1 answer:
lilavasa [31]3 years ago
6 0

Answer:

89.84375

Step-by-step explanation:

Given that on a finance exam, 15 accounting majors had an average grade of 90. On the same exam, 7 marketing majors averaged 85, and 10 finance majors averaged 93.

To find weighted mean.

Here weights can be considered as number of majors in each subject

Marks x     90        85       93\\Weights     15          7         10\\Product     1350    595     930\\

Weighted total = 1350+595+930 = 2875

Weighted mean = weighted total/total students

=\frac{2875}{32} =89.84375

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19) Suppose that there are 55 Democrats and 45 Republicans in the US Senate. A committee of seven senators is to be formed by se
andre [41]

Answer:

Probability = 0.12676992

Step-by-step explanation:

The probability of selecting committee of seven composed of all Democrats.

Total number = 45+55= 100

Number of Democrats = 55

Number of committee = 7

Probability= (55C7)/(100/7)

Where C represent combination sign

Probability = (202927725/1.60075608*10^10)

Probability = 0.12676992

7 0
2 years ago
Calculus hw, need help asap with steps.
nikdorinn [45]

Answers are in bold

S1 = 1

S2 = 0.5

S3 = 0.6667

S4 = 0.625

S5 = 0.6333

=========================================================

Explanation:

Let f(n) = \frac{(-1)^{n+1}}{n!}

The summation given to us represents the following

\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n!}=\sum_{n=1}^{\infty} f(n)\\\\\\\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n!}=f(1) + f(2)+f(3)+\ldots\\\\

There are infinitely many terms to be added.

-------------------

The partial sums only care about adding a finite amount of terms.

The partial sum S_1 is the sum of the first term and nothing else. Technically it's not really a sum because it doesn't have any other thing to add to. So we simply say S_1 = f(1) = 1

I'm skipping the steps to compute f(1) since you already have done so.

-------------------

The second partial sum is when things get a bit more interesting.

We add the first two terms.

S_2 = f(1)+f(2)\\\\S_2 = 1+(-\frac{1}{2})\\\\S_2 = \frac{1}{2}\\\\S_2 = 0.5\\\\\\

The scratch work for computing f(2) is shown in the diagram below.

-------------------

We do the same type of steps for the third partial sum.

S_3 = f(1)+f(2)+f(3)\\\\S_3 = 1+(-\frac{1}{2})+\frac{1}{6}\\\\S_3 = \frac{2}{3}\\\\S_3 \approx 0.6667\\\\\\

The scratch work for computing f(3) is shown in the diagram below.

-------------------

Now add the first four terms to get the fourth partial sum.

S_4 = f(1)+f(2)+f(3)+f(4)\\\\S_4 = 1+(-\frac{1}{2})+\frac{1}{6}-\frac{1}{24}\\\\S_4 = \frac{5}{8}\\\\S_4 \approx 0.625\\\\\\

As before, the scratch work for f(4) is shown below.

I'm sure you can notice by now, but the partial sums are recursive. Each new partial sum builds upon what is already added up so far.

This means something like S_3 = S_2 + f(3) and S_4 = S_3 + f(4)

In general, S_{n+1} = S_{n} + f(n+1) so you don't have to add up all the first n terms. Simply add the last term to the previous partial sum.

-------------------

Let's use that recursive trick to find S_5

S_5 = [f(1)+f(2)+f(3)+f(4)]+f(5)\\\\S_5 = S_4 + f(5)\\\\S_5 = \frac{5}{8} + \frac{1}{120}\\\\S_5 = \frac{19}{30}\\\\S_5 \approx 0.6333

The scratch work for f(5) is shown below.

7 0
2 years ago
Which expression shows 36 as a product of prime factors?
Lapatulllka [165]

Answer:

This is ur answer

HAVE A NICE DAY

4 0
2 years ago
What is -8 square root of 9 when simplified​
gregori [183]

Answer:

\large\boxed{-8\sqrt9=-24}

Step-by-step explanation:

\sqrt{a}=b\iff b^2=a\\\\\sqrt9=3\ \text{because}\ 3^2=9\\\\-8\sqrt9=-8(3)=-24

7 0
3 years ago
how much money should I deposited today in an account that earns 5% compounded semiannually so that it will accumulate to 12,000
Nookie1986 [14]
800 or 2400 not sure which
6 0
3 years ago
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