NCERT Solutions for Class 10 Maths Chapter 14 Statistics Exercise 14.4

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NCERT Solutions for Class 10 Maths Chapter 14 Statistics Exercise 14.4 Solutions are based on representation of cumulative frequency distribution in graphs as a Cumulative Frequency Curve.

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Check out NCERT Solutions for Class 10 Maths Chapter 14 Statistics Exercise 14.4 Solutions

Exercise Solutions of Class 10 Maths Chapter 14 Statistics

Also check other Exercise Solutions of Class 10 Maths Chapter 14 Statistics

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Chapter Related Topics
Mean of Grouped Data	Statistics Important Question	Statistics Revision Notes
Mean Formula	Confidence Interval	Confidence Interval Formula
Step Deviation Method	Differences between Correlation and Regression	Mode of Grouped Data
Median	Class Mark	Statistics for Class 10

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CBSE X Related Questions

1.
Assertion (A) : H.C.F. \((36 m^{2}, 18 m) = 18 m\), where \(m\) is a prime number.
Reason (R) : H.C.F. of two numbers is always less than or equal to the smaller number.
- Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
- Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
- Assertion (A) is true, but Reason (R) is false.
- Assertion (A) is false, but Reason (R) is true.
View Solution
2.
If the zeroes of the polynomial \(p(x) = 2x^2 - 7x + 6\) are \(\alpha\) and \(\beta\), then the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\) is
- \(\frac{7}{2}\)
- \(\frac{6}{7}\)
- \(\frac{-7}{6}\)
- \(\frac{7}{6}\)
View Solution
3.
For an acute angle \(\theta\), if \(\sin \theta = \frac{1}{9}\), then value of \(\frac{9 \csc \theta + 1}{9 \csc \theta - 1}\) is
- \(0\)
- \(\frac{80}{81}\)
- \(1\)
- \(\frac{82}{80}\)
View Solution
4.
Case Study - 1 : A model of Leafy Ball Fountain is made to be kept on the tabletop. Water gently cascades down the ball into a decorative cylindrical pool where it is recycled. The diameter of spherical ball is 21 cm. Cylindrical pool - Outer diameter is 50 cm and inner diameter is 40 cm. Height of solid base is 14 cm. Height of water filled is 7 cm. Observe the figure and answer the following questions : (i) Determine the total height of the fountain. (ii) Find the volume of the ball. (iii) (a) If one-third of the ball is submerged in the water, find the volume of the water filled in the pool. OR (iii) (b) Find the sum of the outer curved surface area of the cylindrical part and surface area of the ball.
View Solution
5.
If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is
- \(4 \text{ cm}\)
- \(4\sqrt{2} \text{ cm}\)
- \(8 \text{ cm}\)
- \(2\sqrt{2} \text{ cm}\)
View Solution
6.
Mean and Median of a frequency distribution are 43 and 40 respectively. The value of mode is
- 34
- 43
- 38.5
- 41.5
View Solution

View All

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NCERT Solutions for Class 10 Maths Chapter 14 Statistics Exercise 14.4

Exercise Solutions of Class 10 Maths Chapter 14 Statistics

CBSE X Related Questions

1.
Assertion (A) : H.C.F. \((36 m^{2}, 18 m) = 18 m\), where \(m\) is a prime number.
Reason (R) : H.C.F. of two numbers is always less than or equal to the smaller number.

2.
If the zeroes of the polynomial \(p(x) = 2x^2 - 7x + 6\) are \(\alpha\) and \(\beta\), then the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\) is

3.
For an acute angle \(\theta\), if \(\sin \theta = \frac{1}{9}\), then value of \(\frac{9 \csc \theta + 1}{9 \csc \theta - 1}\) is

5.
If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is

6.
Mean and Median of a frequency distribution are 43 and 40 respectively. The value of mode is

Similar Mathematics Concepts

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NCERT Solutions for Class 10 Maths Chapter 14 Statistics Exercise 14.4

Exercise Solutions of Class 10 Maths Chapter 14 Statistics

CBSE X Related Questions

1.Assertion (A) : H.C.F. \((36 m^{2}, 18 m) = 18 m\), where \(m\) is a prime number. Reason (R) : H.C.F. of two numbers is always less than or equal to the smaller number.

2.If the zeroes of the polynomial \(p(x) = 2x^2 - 7x + 6\) are \(\alpha\) and \(\beta\), then the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\) is

3.For an acute angle \(\theta\), if \(\sin \theta = \frac{1}{9}\), then value of \(\frac{9 \csc \theta + 1}{9 \csc \theta - 1}\) is

5.If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is

6.Mean and Median of a frequency distribution are 43 and 40 respectively. The value of mode is

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Assertion (A) : H.C.F. \((36 m^{2}, 18 m) = 18 m\), where \(m\) is a prime number.
Reason (R) : H.C.F. of two numbers is always less than or equal to the smaller number.

2.
If the zeroes of the polynomial \(p(x) = 2x^2 - 7x + 6\) are \(\alpha\) and \(\beta\), then the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\) is

3.
For an acute angle \(\theta\), if \(\sin \theta = \frac{1}{9}\), then value of \(\frac{9 \csc \theta + 1}{9 \csc \theta - 1}\) is

5.
If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is

6.
Mean and Median of a frequency distribution are 43 and 40 respectively. The value of mode is