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Vikki [24]
3 years ago
10

Which expressions are equivalent to the expression 32 ÷ (8 – 4)? Choose exactly two answers that are correct. A. 18 – 2 ÷ 2 B. 4

8 ÷ (10 – 4) C. 15 – 21 ÷ 3 D. 36 – 15 ÷ 3 + 1
Mathematics
1 answer:
kap26 [50]3 years ago
3 0
32 \div (8-4)=32 \div 4=8 \\ \\
A. \\
18-2 \div 2=18-1=17 \\ \\
B. \\
48 \div (10-4) = 48 \div 6=8 \\ \\
C. \\
15-21 \div 3=15-7=8 \\ \\
D. \\
36 - 15 \div 3 +1=36 - 5+1=31+1=32

B and C are equivalent to the expression.
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Can you guy's help me quick!
nadya68 [22]

Answer:

480,700 sets (first choice)

Step-by-step explanation:

If all 25 questions are different,

When order of selection counts:

P(25,7)

= 25!/(25-7)!

= 15511210043330985984000000/6402373705728000

= 2422728000

When order of selection does not count count:

C(25,7)

=25!/(7!*(25-7)!)

=15511210043330985984000000/(6402373705728000*5040)

= 480,700

Hope this helps, have a nice day.

6 0
3 years ago
A whole number is 6 more than 2 times another number. The sum of the two numbers is less than 50. This can be written in an ineq
alukav5142 [94]

Answer:

13, 14

Step-by-step explanation:

The parameters of the numbers are;

A whole number value = 2 × Another number + 6

The sum of the two numbers is less than 50

Given that the first number is equal to more than twice the second number, we have that the first number is the larger number, while the second number is the smaller number

Where 'x' represents the second number, we get;

x + 2·x + 6 < 50

Simplifying gives;

3·x + 6 < 50

x < (50 - 6)/3 = 14.\overline 6

x < 14.\overline 6

Therefore, the numbers for which the inequality holds true are numbers less than 14.\overline 6. From the given option, the numbers are 13, and 14.

3 0
3 years ago
Evaluate the expression 3 * 5 x when x is 2.
Ivahew [28]

Answer:

3 * 5x

3 * 5(2)

3 * 10

30

5 0
3 years ago
Read 2 more answers
An 8-meter roll of blue ribbon costs $5.12. What is the unit price?<br> per meter<br> Submit
ivann1987 [24]

Answer:

$0.64

Step-by-step explanation:

Unit price is the cost of a single metre of ribbon

5.12/8=0.64

7 0
3 years ago
Suppose <img src="https://tex.z-dn.net/?f=m" id="TexFormula1" title="m" alt="m" align="absmiddle" class="latex-formula"> men and
ollegr [7]

Firstly, we'll fix the postions where the n women will be. We have n! forms to do that. So, we'll obtain a row like:

\underbrace{\underline{~~~}}_{x_2}W_2 \underbrace{\underline{~~~}}_{x_3}W_3 \underbrace{\underline{~~~}}_{x_4}... \underbrace{\underline{~~~}}_{x_n}W_n \underbrace{\underline{~~~}}_{x_{n+1}}

The n+1 spaces represented by the underline positions will receive the men of the row. Then,

x_1+x_2+x_3+...+x_{n-1}+x_n+x_{n+1}=m~~~(i)

Since there is no women sitting together, we must write that x_2,x_3,...,x_{n-1},x_n\ge1. It guarantees that there is at least one man between two consecutive women. We'll do some substitutions:

\begin{cases}x_2=x_2'+1\\x_3=x_3'+1\\...\\x_{n-1}=x_{n-1}'+1\\x_n=x_n'+1\end{cases}

The equation (i) can be rewritten as:

x_1+x_2+x_3+...+x_{n-1}+x_n+x_{n+1}=m\\\\&#10;x_1+(x_2'+1)+(x_3'+1)+...+(x_{n-1}'+1)+x_n+x_{n+1}=m\\\\&#10;x_1+x_2'+x_3'+...+x_{n-1}'+x_n+x_{n+1}=m-(n-1)\\\\&#10;x_1+x_2'+x_3'+...+x_{n-1}'+x_n+x_{n+1}=m-n+1~~~(ii)

We obtained a linear problem of non-negative integer solutions in (ii). The number of solutions to this type of problem are known: \dfrac{[(n)+(m-n+1)]!}{(n)!(m-n+1)!}=\dfrac{(m+1)!}{n!(m-n+1)!}

[I can write the proof if you want]

Now, we just have to calculate the number of forms to permute the men that are dispposed in the row: m!

Multiplying all results:

n!\times\dfrac{(m+1)!}{n!(m-n+1)!}\times m!\\\\&#10;\boxed{\boxed{\dfrac{m!(m+1)!}{(m-n+1)!}}}

4 0
3 years ago
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