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your ans.. is 3/4 because TT HH HT TH

41/12 answer..........

a coin tossed twice so probability is 1/2

Let B, C and H be the sets of students that plays Basketball, Cricket and Hockey

n(H) = 23, n(B) = 15, n(C) = 20

n(H intersection B) = 7, n(C intersection B) = 5, n(H intersection C) = 4

n(H union B union C) = 60 – 15 = 45

1. Hence, students that plays all games,

n(H intersection B intersection C) = n(H union B union C) – n(H) – n(B) – n(C) + n(H intersection B) + n(B intersection C) + n(C intersection H)

= 45 – 23 – 15 – 20 + 7 + 5 + 4 = 3

2. Who plays hockey not cricket

n(H) – n(H intersection C) = 23 – 4 = 19

3. Plays cricket and hockey not basketball

n(H intersection C) – n(H intersection C intersection B)

4 -3 = 1

Option B is correct as badminton and football altogether consists of 70 students and Cricket alone consists of 50.

Therefore , 70-50=20

yes he is correct the option b

send the ans of 2 part also

Class interval is the difference between the class marks, which is equal to 5.

905.5+27.197

Petroleumis alsoknown as Black Gold- because when crude oil is extracted from the land it isblackin color. People call itgoldbecause of its oils and value . It is very difficult to find.

question is not clear

yes question is not clear..

yes pls upload clear question

x–2tan—1x+c

1/(x-1) bahupad nahi hai , kyoki uski

degree negitive hai

class interval frequency 120_125 5125_130 6 130_135 6135_140 4140_145 5145_150 4

class interval frequency 120_125 5125_130 6130_135 6135_140 4140_145 5145_150 4

write the following in standard from -72 / 144

write the following in standard from -72/144

class intervals. pupils height120-125. 7126-130. 5131-135. 7136-140. 3141-145. 4146-150. 4

i love you and where your from in india

Please post the question by image for better understanding.

please post clear pic

Solution:Let us assume Jacob’s current age is x and his son’s current age is y.

Five years hence, Jacob’s age = x + 5 and his son’s age = y + 5

As per question;

x + 5 = 3(y + 5)

Or, x + 5 = 3y + 15

Or, x = 3y + 15 – 5 = 3y + 10

Five years ago, Jacob’s age = x – 5 and his son’s age = y – 5

As per question;

x – 5 = 7(y – 5)

Or, x – 5 = 7y – 35

Or, x = 7y – 35 + 5 = 7y – 30

Substituting the value of x from first equation in this equation, we get;

3y + 10 = 7y – 30

Or, 3y + 10 + 30 = 7y

Or, 7y – 3y = 40

Or, 4y = 40

Or, y = 10

Substituting the value of y in first equation, we get;

x = 3y + 10

Or, x = 3 x 10 + 10 = 40

Hence, Jacob’s age = 40 years and son’s age = 10 years

thank you i

thank u

24is the correct answer of this question

Lateral Surface area = height * circumference = 100 * 176 = 17600 sq cm

thanx bahiya

answer is=75,15

Solution:Let us assume Jacob’s current age is x and his son’s current age is y.

Five years hence, Jacob’s age = x + 5 and his son’s age = y + 5

As per question;

x + 5 = 3(y + 5)

Or, x + 5 = 3y + 15

Or, x = 3y + 15 – 5 = 3y + 10

Five years ago, Jacob’s age = x – 5 and his son’s age = y – 5

As per question;

x – 5 = 7(y – 5)

Or, x – 5 = 7y – 35

Or, x = 7y – 35 + 5 = 7y – 30

Substituting the value of x from first equation in this equation, we get;

3y + 10 = 7y – 30

Or, 3y + 10 + 30 = 7y

Or, 7y – 3y = 40

Or, 4y = 40

Or, y = 10

Substituting the value of y in first equation, we get;

x = 3y + 10

Or, x = 3 x 10 + 10 = 40

Hence, Jacob’s age = 40 years and son’s age = 10 years

thanks

AREA = SIDE X SIDE= 1.5* 1.5 x 0.91 x0.91= 22.5* 0.91 *0.91 m ^2

area=side^2 =(1.5)^2 =2.25yards

2.25 is the correct answer of the given question

According to question Let present the age of a man=x years present the age of a man his son=45-× years From question five years ago the product of their ages was four times the man's age.so (x-5)(45-x-5)=4(x-5)or (40-x)=4(x-5)/(x-5)or 40-x=4or 40-4=xso x=36 years Hence present the age of a man=x=36 years and present the age of a man his son =45-x years =45-36 years =9 years