Whats 28,456 +92 / 2 * 8 + 3?

A tech company bought a number of computers for its employees: 66 costing \$1600$1600 each, 44 costing \$1200$1200 each, and yy

daser333 [38]

Answer:

5

Step-by-step explanation:

A tech company bought a number of computers for its employees: 6 costing $1600 each, 4 costing $1200 each, and y costing $900 each, where y is a positive odd integer. what is the greatest possible value of y

Solution:

The median of a group of numbers is the middle number when the group of numbers have been arranged either in ascending of descending other. The median of an even group of numbers is the average of the two most middle most numbers while the median of an odd group of numbers is the middle number.

Since there are 6 computers costing $1600, 4 computers costing $1200 each and y computers costing $900 each. Since y is odd, hence y + 6 + 4 would be odd and the median would be the middle number.

Given that the median price = $1200, this means that the middle of the group of numbers is $1200

900, 900, ..., 1200, 1200, 1200, 1200, 1600, 1600, 1600, 1600, 1600, 1600

Since y is odd, for the median price to be $1200, y has to be 3 or 5.

If y = 3:

900, 900, 900, 1200, 1200, 1200, 1200, 1600, 1600, 1600, 1600, 1600, 1600

median = $1200

If y = 5:

900, 900, 900, 900, 900, 1200, 1200, 1200, 1200, 1600, 1600, 1600, 1600, 1600, 1600

Median = $1200

Therefore the greatest possible value of y is 5

5 0

3 years ago

Evaluate each finite series for the specified number of terms. 1+2+4+...;n=5

zaharov [31]

Answer:

31

Step-by-step explanation:

The series are given as geometric series because these terms have common ratio and not common difference.

Our common ratio is 2 because:

1*2 = 2

2*2 = 4

The summation formula for geometric series (r ≠ 1) is:

$\displaystyle \large{S_n=\frac{a_1(r^n-1)}{r-1}}$ or $\displaystyle \large{S_n=\frac{a_1(1-r^n)}{1-r}}$

You may use either one of these formulas but I’ll use the first formula.

We are also given that n = 5, meaning we are adding up 5 terms in the series, substitute n = 5 in along with r = 2 and first term = 1.

$\displaystyle \large{S_5=\frac{1(2^5-1)}{2-1}}\\\displaystyle \large{S_5=\frac{2^5-1}{1}}\\\displaystyle \large{S_5=2^5-1}\\\displaystyle \large{S_5=32-1}\\\displaystyle \large{S_5=31}$

Therefore, the solution is 31.

__________________________________________________________

Summary

If the sequence has common ratio then the sequence or series is classified as geometric sequence/series.

Common Ratio can be found by either multiplying terms with common ratio to get the exact next sequence which can be expressed as $\displaystyle \large{a_{n-1} \cdot r = a_n}$ meaning “previous term times ratio = next term” or you can also get the next term to divide with previous term which can be expressed as:

$\displaystyle \large{r=\frac{a_{n+1}}{a_n}}$

Once knowing which sequence or series is it, apply an appropriate formula for the series. For geometric series, apply the following three formulas:

$\displaystyle \large{S_n=\frac{a_1(r^n-1)}{r-1}}\\\displaystyle \large{S_n=\frac{a_1(1-r^n)}{1-r}}$

Above should be applied for series that have common ratio not equal to 1.

$\displaystyle \large{S_n=a_1 \cdot n}$

Above should be applied for series that have common ratio exactly equal to 1.

__________________________________________________________

Topics

Sequence & Series — Geometric Series

__________________________________________________________

Others

Let me know if you have any doubts about my answer, explanation or this question through comment!

__________________________________________________________

7 0

3 years ago

RSM HOMEWORK HELP!!!

Gnesinka [82]

Answer:

Step-by-step explanation:

∠CAB = 33 {alternate interior angles}

5x = 33 + 2x { exterior angle equals sum of opposite interior angles}

5x - 2x = 33

3x = 33

x = 33/3

x = 11

∠B = 2x = 2* 11 = 22

∠ECB = 5x = 5*11 = 55

8 0

4 years ago

Read 2 more answers

What eigen value for this matix <br> (1 -2)<br> (-2 0)

natali 33 [55]

You find the eigenvalues of a matrix A by following these steps:

Compute the matrix $A' = A-\lambda I$ , where I is the identity matrix (1s on the diagonal, 0s elsewhere)
Compute the determinant of A'
Set the determinant of A' equal to zero and solve for lambda.

So, in this case, we have

$A = \left[\begin{array}{cc}1&-2\\-2&0\end{array}\right] \implies A'=\left[\begin{array}{cc}1&-2\\-2&0\end{array}\right]-\left[\begin{array}{cc}\lambda&0\\0&\lambda\end{array}\right]=\left[\begin{array}{cc}1-\lambda&-2\\-2&-\lambda\end{array}\right]$