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3 years ago

If 12 is 10% of a value, what is that value

Mathematics

2 answers:

Vedmedyk [2.9K]3 years ago

7 0

B.120 !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

irina1246 [14]3 years ago

7 0

The answer is 120 because if you take ten and multiply it by 12, you get 120. Or if you multiply 120 by .1 (which is 10% converted to a decimal) you still get twelve

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A certain bookstore chain has two stores, one in San Francisco and one in Los Angeles. It stocks three kinds of books: hardcover

riadik2000 [5.3K]

Answer:

<h2>See the explanation.</h2>

Step-by-step explanation:

(a)

$\left[\begin{array}{cccc}T&H&S&P\\S&600&1300&2000\\L&400&300&400\end{array}\right]$ = A.

In the above matrix A, the columns refers the three type of books and the rows refers the from which stores the books are been sold.

The numbers represents the corresponding sales in the month of January.

The sale is same for the 6 months.

Hence, 6A = $\left[\begin{array}{cccc}T&H&S&P\\S&3600&7800&12000\\L&2400&1800&2400\end{array}\right]$ . This matrix 6A represents the total sales over the 6 months.

(b)

If we denote the books in stock at the starting of January by B, then

B = $\left[\begin{array}{cccc}T&H&S&P\\S&1000&3000&6000\\L&1000&6000&3000\end{array}\right]$ .

Each month, the chain restocked the stores from its warehouse by shipping 500 hardcover, 1,400 softcover, and 1,400 plastic books to San Francisco and 500 hardcover, 500 softcover, and 500 plastic books to Los Angeles.

If we represent the amount restocked books at the end of each month by another matrix C, then

C = $\left[\begin{array}{cccc}T&H&S&P\\S&500&1400&1400\\L&500&500&500\end{array}\right]$ .

This restocking will be done for 5 times before the end of June.

If there would be no sale, then the stock would be

B + 5C = $\left[\begin{array}{cccc}T&H&S&P\\S&1000+2500&3000+7000&6000+7000\\L&1000+2500&6000+2500&3000+2500\end{array}\right] \\= \left[\begin{array}{cccc}T&H&S&P\\S&3500&10000&13000\\L&3500&8500&5500\end{array}\right]$ .

Since, the total sale is given by 6A, at the end of June, the inventory in each store can be shown as following,

B+5C-6A $\left[\begin{array}{cccc}T&H&S&P\\S&3500&10000&13000\\L&3500&8500&5500\end{array}\right] - \left[\begin{array}{cccc}T&H&S&P\\S&3600&7800&12000\\L&2400&1800&2400\end{array}\right] \\= \left[\begin{array}{cccc}T&H&S&P\\S&-100&2200&1000\\L&1100&6700&3100\end{array}\right]$

6 0

3 years ago

What is the slope and rate change?

jekas [21]

Slope is 3/2 rate of change is

7 0

3 years ago

2 less than the product of 5 and n

Deffense [45]

The product of 5 and n is written as $5n$ .

Two less than this product is $5n-2$

7 0

3 years ago

Solve for a in the figure shown.

aleksandrvk [35]

Answer: 58 is the answer

hope this help and here my steps use a mark or a pin

and line it with the line if it's wrong sorry for the wrong

answer

7 0

3 years ago

Read 2 more answers

Suppose that you had the following data set. 500 200 250 275 300 Suppose that the value 500 was a typo, and it was suppose to be

hodyreva [135]

Answer:

$\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305$

$s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108$

$\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105$

$s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221$

The absolute difference is:

$Abs = |340.221-115.108|= 225.113$

If we find the % of change respect the before case we have this:

$\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%$

So then is a big change.

Step-by-step explanation:

The subindex B is for the before case and the subindex A is for the after case

Before case (with 500)

For this case we have the following dataset:

500 200 250 275 300

We can calculate the mean with the following formula:

$\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305$

And the sample deviation with the following formula:

$s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108$

After case (With -500 instead of 500)

For this case we have the following dataset:

-500 200 250 275 300

We can calculate the mean with the following formula:

$\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105$

And the sample deviation with the following formula:

$s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221$

And as we can see we have a significant change between the two values for the two cases.

The absolute difference is:

$Abs = |340.221-115.108|= 225.113$

If we find the % of change respect the before case we have this:

$\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%$

So then is a big change.

8 0

3 years ago