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torisob [31]
4 years ago
7

What number is midway between 2/5 and 1?

Mathematics
2 answers:
jeka57 [31]4 years ago
5 0
4/10 and 10/10:

7/10.
snow_lady [41]4 years ago
4 0

Answer: \frac{7}{10}

Step-by-step explanation:

To find the mid point 'c' between any two numbers 'a' and 'b' we apply the following rule:-

c=\frac{a+b}{2}

Now, the number is midway between 2/5 and 1 will be given by :-

\c=\frac{\frac{2}{5}+1}{2}\\\\\Rightarrow\ c=\frac{\frac{7}{5}}{2}....\text{[by taking LCM]}\\\\\Rightarrow\ c=\frac{7}{2\times5}\\\\\Rightarrow\ c=\frac{7}{10}

Hence, the number is midway between 2/5 and 1 =\frac{7}{10}

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To furnish a room in a model home, an interior decorator is to select 2 chairs and 2 tables from a collection of chairs and tabl
Free_Kalibri [48]

Answer:

A. 6

Step-by-step explanation:

Let n represent number of tables.

We have been given that to furnish a room in a model home, an interior decorator is to select 2 chairs and 2 tables from a collection of chairs and tables in a warehouse that are all different from each other. There are 5 chairs in the warehouse and if 150 different combinations are possible.

Since 2 chairs are being selected from 5 chairs, so we can choose 2 chairs in 5C2 ways.

There are n tables and we can choose 2 tables from n table in nC2 ways.

We can represent our given information in an equation as:

5C2\times nC2=150

\frac{5!}{2!(5-2)!}\times \frac{n!}{2!(n-2)!}=150

\frac{5*4*3!}{2*1*3!}\times \frac{n!}{2!(n-2)!}=150

10\times \frac{n!}{2!(n-2)!}=150

\frac{n!}{2!(n-2)!}=15

\frac{n!}{2*1*(n-2)!}=15

\frac{n*(n-1)*(n-2)!}{2*(n-2)!}=15

\frac{n(n-1)}{2}=15

n(n-1)=30

n^2-n=30

n^2-n-30=0

n^2-6n+5n-30=0

n(n-6)+5(n-6)=0

(n-6)(n+5)=0

(n-6)=0\text{ (or) }(n+5)=0

n=6\text{ (or) }n=-5

Since tables cannot be negative quantity, therefore, 6 tables are in the warehouse.

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