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I have taught as a Lecturer in different Universities and designed engineering courses, resource materials and papers. This is lecture 4 of Instrumentation & Measurement.

Engineering

 Instrumentation & Measurement (Lecture 4)

I have taught as a Lecturer in different Universities and designed engineering courses, resource materials and papers. This is lecture 3 of Instrumentation & Measurement.

Instrumentation & Measurement (Lecture 3)

I have taught as a Lecturer in different Universities and designed engineering courses, resource materials and papers. This is lecture 2 of Instrumentation & Measurement.

Instrumentation & Measurement (Lecture 2)

I have taught as a Lecturer in different Universities and designed engineering courses, resource materials and papers. This is lecture 1 of Instrumentation & Measurement.

Instrumentation & Measurement (Lecture 1)

Answer for the given expression will be 4x+6(y^3)

Maths

2x+x+x+2y×3y×y

Average for the exams scores 97, 73, 88 and 80 can be calculated as,

Average after 4th exam = (97+73+88+80)/4 = 84.5

Ahmad earned scores of 97, 73 and 88 respectively in her first three examinations. If she scored 80 in the fourth examination, then her average score will be *

Amount after n years = P{(1+i/100)^n}

Where P is the principle amount
i is the annual interest rate
n is the number of compounding periods in years

Thus,
Amount after 2 years = 2000{(1+3/100)^2}= £2121.8 

F £2000 is placed into a bank account that pays 3% compound per year, how much will be in the account after 2 years

(5/9) divide by 5 is same as,
(5/9)*(1/5) = 1/9

5/9 divide by 5

Notice that the difference between consecutive numbers of the sequence is increasing by 1.

5 (next term) - 1 (previous term) = 4
10 (next term) - 5 (previous term) = 5
16 (next term) - 10 (previous term) = 6
23 (next term) - 16 (previous term) = 7
X (next term) - 23 (previous term) = 8
So, X = 8+23 = 31 (6th term)
Similarly, Y(next term) - 31(previous term) = 9
So, Y = 9+31 = 40 (7th term)
Therefore, next two terms of the given sequence are 31 and 40 😊

The first 5 terms of the sequence are:1,5,10,16,23 find the next two terms of the sequence

Range can be calculated by taking the difference between the highest and the lowest value in the given sequence.

Here highest value is 8
And lowest value is 1
Range = highest value - lowest value = 8 - 1 = 7

What is the range of 3 7 2 4 7 5 7 1 8 8

Put n=3 in given equation,
5n-2 = 5(3)-2 = 15-2 =13

5n-2 find the 3rd term

5h:33min can be written in fraction as 5.55 hours.

What is 5h:33min to fraction

Here I am assuming that n on the left hand side of the equation is in subscript.
Put n=10 in the given equation, we get
S(10th)=(5(10)^2+5)/2 = 252.5

Sn= ( 5n^2 +5)/ (2), find 10^th term

Given:
8 black pens
3 red pens
11 total pens
Solution:
2 pens can be drawn out of 11 pens in 11C2 = n(S) = 55 ways
Let A = Both are black then n(A) = 8C2 = 28 ways
P(A) = n(A)/n(S) = 28/55 = 0.509

Let B = Both are red then n(B) = 3C2 = 3 ways
P(B) = n(B)/n(S) = 3/55 = 0.0545

Probability that both are of the same colors will be now,
P(AUB) = P(A) + P(B)                [Independent Events]
P(AUB) = 0.509 + 0.0545 = 0.5635 (Answer)


There are 11 pens in a box 8 are black and 3 are red two pens are taken at random without replacement workout out probability that the colour is thr same

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Muhammad Hassan