Question 5: There are 5 lemon and 4 strawberry sweets in a bag. Hailey takes out a sweet at random, writes down its flavour and puts it back into the bag. Then Hailey takes out a second sweet, at random, and writes down its flavour. (a) Copy and complete the tree diagram 1st Sweet 2nd Sweet Lémon (b) Find the probability of two lemon sweets Lemon (c) Find the probability of one of each flavour

Q: 24. Find d²4 zP 2. xp 4=3x² +34-8

A: Differentiate two times

Q: 24 30

A:

Q: 1/x

A:

Q: C dx 1+ C +2 X 3

A:

Q: Grafthe fallawing ustion yinhd ラ2エ55 2 1T

A: we use Trigonometry table to find out values of y = sin(2x)

Q: いZ ch-1)2 いe1 2.

A:

Q: Find the H.C.F of 120,150 and 210

A:

Q: 26 ÷ 22

A: Solve by squarring both side

Q: 63 45 56 48

A:

Q: b. 14C11

A: nCr=n!r!×(n−r)!r!=r(r-1)(r-2)×....×3×2×1

Q: 5+45(669=

A: The given expression is 5+45669. Compute the value as follows.

Q: Evaluate 1,C5 and P,

A: For 12C5, The formula of combination is,   Substitute the values in the above equation.   Thus,…

Q: 6.5+ 4

A: To simplify given

Q: 3.546 is of 10

A:

Q: 13 dと 2) xp 7-x2

A: Here we have to find the integral value.

Q: There are two more questions pleade

A: I have to answer for part-d and part-e, so i use the previous result .

Q: 41 21°

A: As we know that  sin(θ)=opposite hypotanuse

Q: s): olicities: Behavior at x-axis: (through, bounce, flatten) 2. Divide. 4x-10x+5x-7x+1 -x+2

A:

Q: 4 11  ✕  5 9

A: Consider the given expression as 411×59. 411×59=4×511×9=2099

Q: only

A: Solution of the problem as follows

Q: 3,41 140

A:

Q: 1 Evaluate. 5P3 • 6C4 900 300 150

A: Use the formula and evaluate.

Q: 11C6

A: we know that,Cr=n!(n-r)!×r!nand r!=r(r-1)(r-2).....×2×1

Q: 190+45-23=?

A: 190+45-23=?

Q: 2 questions in pictures

A:

Q: 28 47

A:

Q: 2dt O 20 O 8 O 16 0 4 0 4

A:

Q: Evaluate Lt

A:

Q: 5 23 4 48 1 5 75

A: Given       5  2 3 − 4 4 8¯      1  5 7 5

Q: y² 49 1 y? /-

A:

Q: I indy ralue

A: using two formula ln(y²) =2lny (x-y)²=x²-2xy+y²

Q: Evaluate  P43 and C96

A: Solution

Q: 60 90

A:

Q: BCD???

A: Given problems:- b) ∫6w3+7wdwc) ∫y+14ydy d)∫123z-24dz

Q: Xx well l defned

A:

Q: -2 2. d=r Sinp.sin0

A:

Q: 2dt 12 18 O 14 20

A: We have to solve the integration.

Addition Rule of Probability

It simply refers to the likelihood of an event taking place whenever the occurrence of an event is uncertain. The probability of a single event can be calculated by dividing the number of successful trials of that event by the total number of trials.

Expected Value

When a large number of trials are performed for any random variable ‘X’, the predicted result is most likely the mean of all the outcomes for the random variable and it is known as expected value also known as expectation. The expected value, also known as the expectation, is denoted by: E(X).

Probability Distributions

Understanding probability is necessary to know the probability distributions. In statistics, probability is how the uncertainty of an event is measured. This event can be anything. The most common examples include tossing a coin, rolling a die, or choosing a card. Each of these events has multiple possibilities. Every such possibility is measured with the help of probability. To be more precise, the probability is used for calculating the occurrence of events that may or may not happen. Probability does not give sure results. Unless the probability of any event is 1, the different outcomes may or may not happen in real life, regardless of how less or how more their probability is.

Basic Probability

The simple definition of probability it is a chance of the occurrence of an event. It is defined in numerical form and the probability value is between 0 to 1. The probability value 0 indicates that there is no chance of that event occurring and the probability value 1 indicates that the event will occur. Sum of the probability value must be 1. The probability value is never a negative number. If it happens, then recheck the calculation.