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3 years ago

Seven is not an example of? my choices are a variable, an expression, a constant or a term

2 answers:

4 0

Seven is not an example of a variable since a Variable is unknown but seven is known so that why it cannot be an example of a variable

Akimi4 [234]3 years ago

3 0

An expression because it can not be expressive

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Custumers of a phone company can choose between two service plans for long distance calls. The first plan has a $9 monthly fee a

saul85 [17]

The answer is 350.

m= minutes

Plan 1: 9+0.13m
Plan 2: 23+0.09m

9+0.13m = 23+0.09m
-9 -9
0.13m = 14+0.09m
-0.09m -0.09m
0.04m= 14
14÷0.04= 350
m= 350

3 0

3 years ago

Read 2 more answers

What is the y intercept of this line? y=x-2 PLS SOLVE ASAP!!!

OleMash [197]

-2 is the y intercept

6 0

3 years ago

One year Perry had the lowest ERA (earned-run average, mean number of runs yielded per nine innings pitched) of any male pitch

Blizzard [7]

Answer:

z score Perry $z=-1.402$

z score Alice $z=-1.722$

Alice had better year in comparison with Perry.

Step-by-step explanation:

Consider the provided information.

One year Perry had the lowest ERA of any male pitcher at his school, with an ERA of 3.02. For the males, the mean ERA was 4.206 and the standard deviation was 0.846.

To find z score use the formula.

$z=\frac{x-\mu}{\sigma}$

Here μ=4.206 and σ=0.846

$z=\frac{3.02-4.206}{0.846}$

$z=\frac{-1.186}{0.846}$

$z=-1.402$

Alice had the lowest ERA of any female pitcher at the school with an ERA of 3.16. For the females, the mean ERA was 4.519 and the standard deviation was 0.789.

Find the z score

where μ=4.519 and σ=0.789

$z=\frac{3.16-4.519 }{0.789}$

$z=\frac{-1.359}{0.789}$

$z=-1.722$

The Perry had an ERA with a z-score is –1.402. The Alice had an ERA with a z-score is –1.722.

It is clear that the z-score value for Perry is greater than the z-score value for Alice. This indicates that Alice had better year in comparison with Perry.

7 0

3 years ago

Can u solve this problem in show your work

aksik [14]

Hey there!

We can solve this problem by doing PEMDAS, which lists:

$Parentheses \\ Exponents \\ Multiplication \\ Division \\ Addition \\ Subtraction$

We have in this equation

$Parentheses \\ Exponents \\ Dvision$

Let's sp;ve this equation now!

Parentheses ↓

$(6 -3) = 3$

Exponents ↓

$3^2 = 9$

Division ↓

$\frac{36}{9} = 9$

Okay now , I show you the steps so it can be easier to solve!

$\frac{36}{(6-3)^2} \\ \\ 6-3 = 3 \\ \\ \frac{36}{3^2} \\ \\ 3^2 = 9 \\ \\ \frac{36}{9} \\ \\ \\ \\ Answer: 4$

Good luck on your assignment and enjoy your day!

~ $LoveYourselfFirst:)$

5 0

3 years ago

Read 2 more answers

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:

$\mathbf{Novels = 3}$

$\mathbf{Mathematics = 2}$

$\mathbf{Chemistry = 5}$

(a) The books in any order

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:

$\mathbf{n! = 10!}$

$\mathbf{n! = 3628800}$

(b) The mathematics book, not together

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

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

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

(c) The novels must be together and the chemistry books, together

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:

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

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

$\mathbf{Arrangements = 17280}$

(d) The mathematics must be together and the chemistry books, not together

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:

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

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

$\mathbf{Total = 302400}$