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2 years ago

Q2) for both of the following, calculate A5 and A100. (i) An= 6n-3

Mathematics

2 answers:

dlinn [17]2 years ago

4 0

Answer:

27 and 597

Step-by-step explanation:

To evaluate A5 and A100, substitute n = 5, n = 100 into An , that is

A5 = 6(5) - 3 = 30 - 3 = 27

A100 = 6(100) - 3 = 600 - 3 = 597

Alexxandr [17]2 years ago

4 0

Answer:

5n=6n-3

C.L.T

6n-5n=3

n=3

ii.100n=6n-3

C.L.T

100n-6n= -3

94n/94=-3/94

n=-0.032

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3 years ago

A drawer contains 3 white shirts, 2 blue shirts, and 5 gray shirts. A shirt is randomly

shutvik [7]

Answer:

Probability that first shirt is white and second shirt is gray if first shirt selected is set aside = $\frac{1}{4}$

Step-by-step explanation:

Given that

3 white, 2 blue and 5 gray shirts are there.

To find:

Probability that first shirt is white and second shirt is gray if first shirt selected is set aside = ?

Solution:

Here, total number of shirts = 3+2+5 = 10

First of all, let us learn about the formula of an event E:

$P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}$

$P(First\ White) = \dfrac{\text{Number of white shirts}}{\text {Total number of shirts left}}$

$P(First\ White) = \dfrac{3}{10}$

Now, this shirt is set aside.

So, total number of shirts left are 9 now.

$P(First\ White\ and\ second\ gray) = P(First White) \times P(Second\ Gray)\\\Rightarrow P(First\ White\ and\ second\ gray) = P(First White) \times \dfrac{\text{Number of gray shirts}}{\text{Total number of shirts left}}\\\\\Rightarrow P(First\ White\ and\ second\ gray) = \dfrac{3}{10} \times \dfrac{5}{9}\\\Rightarrow P(First\ White\ and\ second\ gray) = \dfrac{1}{2} \times \dfrac{1}{2}\\\Rightarrow P(First\ White\ and\ second\ gray) = \bold{\dfrac{1}{4} }$

So, the answer is:

Probability that first shirt is white and second shirt is gray if first shirt selected is set aside = $\frac{1}{4}$

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3 years ago