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3 years ago

What is the third quartile of this data set 20,21,24,25,28,29,35,37,42

Mathematics

2 answers:

deff fn [24]3 years ago

7 0

36 is the upper or third quartile the range is 22
the mean is 29 the interquartile range is 13.5

stepan [7]3 years ago

3 0

Answer:

Step-by-step explanation:

The middle number 28 is the median.

The upper quartile is 'middle' number of the last 4 numbers.

So its the mean of 35 and 37 = 36 (answer)

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Solve the equation 1/3=6 1/2

Rufina [12.5K]

Answer:

false

Step-by-step explanation:

what exactly is the equation? If it's true or false it's false.

3 0

3 years ago

How can i prove this property to be true for all values of n, using mathematical induction.

chubhunter [2.5K]

Proof -

So, in the first part we'll verify by taking n = 1.

$\implies \: 1 = {1}^{2} = \frac{1(1 + 1)(2 + 1)}{6}$

$\implies{ \frac{1(2)(3)}{6} }$

$\implies{ 1}$

Therefore, it is true for the first part.

In the second part we will assume that,

$\: { {1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} = \frac{k(k + 1)(2k + 1)}{6} }$

and we will prove that,

$\sf{ \: { {1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k + 1)(k + 1 + 1) \{2(k + 1) + 1\}}{6}}}$

$\: {{1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k + 1)(k + 2) (2k + 3)}{6}}$

${1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{k (k + 1) (2k + 1) }{6} + \frac{(k + 1) ^{2} }{6}$

${1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{k(k+1)(2k+1)+6(k+1)^ 2 }{6}$

${1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k+1)\{k(2k+1)+6(k+1)\} }{6}$

${1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k+1)(2k^2 +k+6k+6) }{6}$

${1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k+1)(2k^2+7k+6) }{6}$

${1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k+1)(k+2)(2k+3) }{6}$

Henceforth, by using the principle of mathematical induction 1²+2² +3²+....+n² = n(n+1)(2n+1)/ 6 for all positive integers n.

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8 0

2 years ago

Find the y-intercept of the following equation. Simplify your answer. y = 5x + 7

kramer

The y-intercept is the “+ (number)” so in this case the y-intercept is 7

3 0

3 years ago

Read 2 more answers

Cual es la respuesta de la imagen adjunta

FrozenT [24]

Answer:

Step-by-step explanation:

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3 years ago

I really dont know how to start this Calculus 1 question!

hichkok12 [17]

According to the given plot, $f'(1)=2$ . Then the linear approximation to $f(x)$ at $x=1$ is

$L(x)=f(1)+f'(1)(x-1)=6+2(x-1)=2x+4$

Then

$f(0.95)\approx L(0.95)=5.9$

$f(1.05)\approx L(1.05)=6.1$

On the interval [0, 1], the plot of $f'(x)$ is positive, so $f(x)$ is an increasing function here. But we can see that $f'(x)$ is approaching 0. This means tangent lines to $f(x)$ have a positive slope, but the slopes are approaching 0 and are thus becoming less steep. This in turn means the tangent lines lie above the curve, so the approximations are greater than the actual values of $f(0.95)$ and $f(1.05)$ .

7 0

3 years ago