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2 years ago

4b) Work out the mean and standard deviation of the following: picture included

Mathematics

1 answer:

Jlenok [28]2 years ago

4 0

Answer:

Yes, your simultaneous equations are correct;

Solve them like so:

170 + 1.6449σ = 180 - 0.8416

2.4865σ = 10

σ = 4.0217...

μ = 170 + 1.6449(4.0217...)

μ = 176.615...

MS is correct, you must've made an error in solving or rearranging

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2 years ago

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3 years ago

Given that 'n' is a natural number. Prove that the equation below is true using mathematical induction.

LenaWriter [7]

<h3>To ProvE :- </h3>

1 + 3 + 5 + ..... + (2n - 1) = n²

Method :-

If P(n) is a statement such that ,

P(n) is true for n = 1
P(n) is true for n = k + 1 , when it's true for n = k ( k is a natural number ) , then the statement is true for all natural numbers .

$\sf\to \textsf{ Let P(n) : 1 + 3 + 5 + $\dots$ +(2n-1) = n$^{\sf 2}$ }$

Step 1 : Put n = 1 :-

$\sf\longrightarrow LHS = \boxed{\sf 1 } \\$

$\sf\longrightarrow RHS = n^2 = 1^2 = \boxed{\sf 1 }$

Step 2 : Assume that P(n) is true for n = k :-

$\sf\longrightarrow 1 + 3 + 5 + \dots + (2k - 1 ) = k^2$

Add (2k +1) to both sides .

$\sf\longrightarrow 1 + 3+5+\dots+(2k-1)+(2k+1)=k^2+(2k+1)$

RHS is in the form of ( a + b)² = a²+b²+2ab .

$\sf\longrightarrow 1 + 3+5+\dots+(2k-1)+(2k+1)= (k +1)^2$

Adding and subtracting 1 to LHS .

$\sf\longrightarrow 1 + 3+5+\dots+(2k-1)+(2k+1) + 1 -1 = (k +1)^2 \\$

$\sf\longrightarrow 1 + 3+5+\dots+(2k-1)+(2k+2) - 1 = (k +1)^2$

Take out 2 as common .

$\sf\longrightarrow 1 + 3+5+\dots+(2k-1)+\{2(k+1)-1\}= (k +1)^2$

P(n) is true for n = k + 1 .

Hence by the principal of Mathematical Induction we can say that P(n) is true for all natural numbers 'n' .

**Edits are welcomed**

8 0

2 years ago

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