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3 years ago

8

Jet works in the bookstore he has 300 books to stock on the shows so far he has stopped 120 books today what percent of the book

s does he have left to stock

1 answer:

mafiozo [28]3 years ago

7 0

Stokked 120
how many left?
300-120=180
180/300=x%
percent means parts out of 100
180/300=60/100=60%

60% left

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To the nearest hundredth of a yard, determine the circumference of the given circle. Approximate as 3.14. r=6

Bond [772]

Answer:

37.68

Step-by-step explanation:

c = 2 x 3.14 x 6

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3 years ago

In Problems 1-8, determine the level of measurement of each variable.

lbvjy [14]

The level of measurement of each given variable are:

1. Ordinal

2. Nominal

3. Ratio

4. Interval

5. Ordinal

6. Nominal

7. Ratio

8. Interval

Level of measurement is used in assigning measurement to variables depending on their attributes.

There are basically four (4) levels of measurement (see image in the attachment):

1. <u>Nominal:</u> Here, values are assigned to variables just for naming and identification sake. It is also used for categorization.

Examples of variables that fall under the measurement are: Favorite movie, Eye Color.

<u>2. Ordinal:</u> This level of measurement show difference between variables and the direction of the difference. In order words, it shows magnitude or rank among variables.

Examples of such variables that fall under this are: highest degree conferred, birth order among siblings in a family.

<u>3. Interval Scale:</u> this third level of measurement shows magnitude, a known equal difference between variables can be ascertain. However, this type of measurement has <em>no true zero</em> point.

Examples of the variables that fall here include: Monthly temperatures, year of birth of college students

4. Ratio Scale: This scale of measurement has a "true zero". It also has every property of the interval scale.

Examples are: ages of children, volume of water used.

Therefore, the level of measurement of each given variable are:

1. Ordinal

2. Nominal

3. Ratio

4. Interval

5. Ordinal

6. Nominal

7. Ratio

8. Interval

Learn more about level of measurement here:

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2 years ago

(x +y)(x-y), ifx= -3, y=-5

Rufina [12.5K]

Answer:

-16

Step-by-step explanation:

(-3 + (-5))(-3 - (-5))

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(-8)*2

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3 years ago

Read 2 more answers

8 +2/3 (6x + 30) = 80

amid [387]

Answer:

X=13

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

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3 years ago

2. In how many ways can 3 different novels, 2 different mathematics books and 5 different chemistry books be arranged on a books

insens350 [35]

The number of ways of the books can be arranged are illustrations of permutations.

When the books are arranged in any order, the number of arrangements is 3628800
When the mathematics book must not be together, the number of arrangements is 2903040
When the novels must be together, and the chemistry books must be together, the number of arrangements is 17280
When the mathematics books must be together, and the novels must not be together, the number of arrangements is 302400

The given parameters are:

$\mathbf{Novels = 3}$

$\mathbf{Mathematics = 2}$

$\mathbf{Chemistry = 5}$

<u />

<u>(a) The books in any order</u>

First, we calculate the total number of books

$\mathbf{n = Novels + Mathematics + Chemistry}$

$\mathbf{n = 3 + 2 + 5}$

$\mathbf{n = 10}$

The number of arrangement is n!:

So, we have:

$\mathbf{n! = 10!}$

$\mathbf{n! = 3628800}$

<u>(b) The mathematics book, not together</u>

There are 2 mathematics books.

If the mathematics books, must be together

The number of arrangements is:

$\mathbf{Maths\ together = 2 \times 9!}$

Using the complement rule, we have:

$\mathbf{Maths\ not\ together = Total - Maths\ together}$

This gives

$\mathbf{Maths\ not\ together = 3628800 - 2 \times 9!}$

$\mathbf{Maths\ not\ together = 2903040}$

<u>(c) The novels must be together and the chemistry books, together</u>

We have:

$\mathbf{Novels = 3}$

$\mathbf{Chemistry = 5}$

First, arrange the novels in:

$\mathbf{Novels = 3!\ ways}$

Next, arrange the chemistry books in:

$\mathbf{Chemistry = 5!\ ways}$

Now, the 5 chemistry books will be taken as 1; the novels will also be taken as 1.

Literally, the number of books now is:

$\mathbf{n =Mathematics + 1 + 1}$

$\mathbf{n =2 + 1 + 1}$

$\mathbf{n =4}$

So, the number of arrangements is:

$\mathbf{Arrangements = n! \times 3! \times 5!}$

$\mathbf{Arrangements = 4! \times 3! \times 5!}$

$\mathbf{Arrangements = 17280}$

<u>(d) The mathematics must be together and the chemistry books, not together</u>

We have:

$\mathbf{Mathematics = 2}$

$\mathbf{Novels = 3}$

$\mathbf{Chemistry = 5}$

First, arrange the mathematics in:

$\mathbf{Mathematics = 2!}$

Literally, the number of chemistry and mathematics now is:

$\mathbf{n =Chemistry + 1}$

$\mathbf{n =5 + 1}$

$\mathbf{n =6}$

So, the number of arrangements of these books is:

$\mathbf{Arrangements = n! \times 2!}$

$\mathbf{Arrangements = 6! \times 2!}$

Now, there are 7 spaces between the chemistry and mathematics books.

For the 3 novels not to be together, the number of arrangement is:

$\mathbf{Arrangements = ^7P_3}$

So, the total arrangement is:

$\mathbf{Total = 6! \times 2!\times ^7P_3}$

$\mathbf{Total = 6! \times 2!\times 210}$

$\mathbf{Total = 302400}$

Read more about permutations at:

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2 years ago