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1 year ago

Sequence: 1,5,13,25 Find nth term (Equation/formula thingy) Need answer ASAP

Mathematics

1 answer:

masha68 [24]1 year ago

4 0

Answer:

Formula for a linear sequence is

Tn=a+(n-1)d

where Tn is the nth term

a is the first term

n is the number of term

d is common difference

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x1= -1-√17/4

x2= -1+√17/4

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2 years ago

The accompanying data on x = current density (mA/cm2) and y = rate of deposition (m/min)μ appeared in a recent study.

gtnhenbr [62]

Answer:

a) $r=\frac{4(333)-(200)(5.37)}{\sqrt{[4(12000) -(200)^2][4(9.3501) -(5.37)^2]}}=0.9857$

The correlation coefficient for this case is very near to 1 so then we can ensure that we have linear correlation between the two variables

b) $m=\frac{64.5}{2000}=0.03225$

Now we can find the means for x and y like this:

$\bar x= \frac{\sum x_i}{n}=\frac{200}{4}=50$

$\bar y= \frac{\sum y_i}{n}=\frac{5.37}{4}=1.3425$

$b=\bar y -m \bar x=1.3425-(0.03225*50)=-0.27$

So the line would be given by:

$y=0.3225 x -0.27$

Step-by-step explanation:

Part a

The correlation coeffcient is given by this formula:

$r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}$

For our case we have this:

n=4 $\sum x = 200, \sum y = 5.37, \sum xy = 333, \sum x^2 =12000, \sum y^2 =9.3501$

$r=\frac{4(333)-(200)(5.37)}{\sqrt{[4(12000) -(200)^2][4(9.3501) -(5.37)^2]}}=0.9857$

The correlation coefficient for this case is very near to 1 so then we can ensure that we have linear correlation between the two variables

Part b

$m=\frac{S_{xy}}{S_{xx}}$

Where:

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}$

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}$

With these we can find the sums:

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=12000-\frac{200^2}{4}=2000$

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i){n}}=333-\frac{200*5.37}{4}=64.5$

And the slope would be:

$m=\frac{64.5}{2000}=0.03225$

Now we can find the means for x and y like this:

$\bar x= \frac{\sum x_i}{n}=\frac{200}{4}=50$

$\bar y= \frac{\sum y_i}{n}=\frac{5.37}{4}=1.3425$

And we can find the intercept using this:

$b=\bar y -m \bar x=1.3425-(0.03225*50)=-0.27$

So the line would be given by:

$y=0.3225 x -0.27$

4 0

2 years ago

A cheetah's speed was timed over a set distance. The cheetah was clocked running 60 miles per hour. Which equation is the relati

Troyanec [42]

Answer:

D. y=70x

Step-by-step explanation:

The cheetah was clocked running 70 miles per hour.

i.e. in 1 hour cheetah covered 70 miles

in 2 hours cheetah covered 140 miles

in 3 hours cheetah covered 210 miles

in general, in n hour cheetah covered 70n miles

If y denote the distance covered

and x denotes the time cheetah runs

Then, y=70x

Hence, the equation which shows the relationship between the distance, y, and time, x, the cheetah runs is:

D y=70x

8 0

2 years ago