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3 years ago

What are the two ways you can find the distance between points (X1,Y1) and (x2,y2) in the coordinate

Mathematics

1 answer:

igomit [66]3 years ago

4 0

The first (and most typical) way to find distance of two points is by using the distance formula.

$distance=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2 }$

One alternative is the Manhattan metric, also called the taxicab metric. This option is much more complicated, and rarely used in high school math. d(x,y)=∑i|xi-yi|

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The manufacturer of hardness testing equipment uses steel-ball indenters to penetrate metal that is being tested. However, the

kumpel [21]

Answer:

Check the explanation

Step-by-step explanation:

Let X denotes steel ball and Y denotes diamond

$\bar{x_1}$ = 1/9( 50+57+......+51+53)

=530/9

=58.89

$\bar{x_2}$ = 1/9( 52+ 56+....+ 51+ 56)

=543/9

=60.33

difference = d =(60.33- 58.89)

=1.44

$s^2=1/n\sum xi^2 - n/(n-1)\bar{x}^2$

s12 = 1/9( 502+572+......+512+532) -9/8 (58.89)2

=31686/8 - 9/8( 3468.03)

=3960.75 - 3901.53

=59.22

s1 = 7.69

s22 = 1/9( 522+ 562+....+ 512+ 562) -9/8 (60.33)2

=33295/8 - 9/8 (3640.11)

=4161.875 - 4095.12

=66.75

s2 =8.17

sample standard deviation for difference is

s= $\sqrt{[(n1-1)s_1^2+ (n2-1)s_2^2]/(n1+n2-2)}$

= $\sqrt{[(9-1)*59.22+ (9-1)*66.75]/(9+9-2)}$

= $\sqrt{1007.76/16}$

=7.93

sd = $s*\sqrt{(1/n1)+(1/n2)}$

= $7.93*\sqrt{(1/9)+(1/9)}$

=7.93* 0.47

=3.74

For 95% confidence level $Z (\alpha /2)$ =1.96

confidence interval is

$d\pm Z(\alpha /2)*s_d$

=(1.44 - 1.96* 3.75 , 1.44+1.96* 3.75)

=(1.44 - 7.35 , 1.44 + 7.35)

=(-2.31, 8.79)

There is sufficient evidence to conclude that the two indenters produce different hardness readings.

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Reptile [31]

Answer:

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Expression of "three fifths of a number b subtracted from thrice the same number" into mathematical symbol

liubo4ka [24]

Answer:

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Step-by-step explanation:

Let the number be "b"

So the expression would be:

3b - $\frac{3}{5}$ b

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