Ask question

Ask question

All categories

2 years ago

14

Can someone help me please

1 answer:

Kitty [74]2 years ago

7 0

Answer:

I guess its c.

i am not sure

You might be interested in

there are 2,300 licensed dogs in clarkson. A random sample of 50 of the dogs in Clarkson shows that 8 have ID microchips implant

algol [13]

Set up a ratio comparing the sample portion to the whole portion.

x= # of all dogs likely to have chip

Sample: 8 out of 50
Whole: x out of 2,300

8/50= x/2300
cross multiply

(50*x)= (8*2300)
50x= 18400

divide both sides by 50
x= 368

ANSWER: 368 dogs in Clarkson are likely to have a microchip.

Hope this helps! :)

7 0

2 years ago

Read 2 more answers

A card is drawn from a well-shuffled deck of 52 cards. find p(drawing an ace or a 9). 8

Maru [420]

P=8/52
p=0.15
p=15 percent

7 0

2 years ago

E=I×R and P=E×I what is the formula that will solve for R when only E and I are known

blagie [28]

You wrote it there yourself, at the beginning of your question.

You wrote E = I x R

Divide each side by ' I ' : E / I = R

That's what you're asking for. ^

4 0

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

Sorry, but I also need help on this

Fudgin [204]

Answer:

I think it's A

Step-by-step explanation:

┓┏┓┏┓┃

┛┗┛┗┛┃＼○／

┓┏┓┏┓┃ /

┛┗┛┗┛┃ノ)

┓┏┓┏┓┃

┛┗┛┗┛┃

┓┏┓┏┓┃

┛┗┛┗┛┃

┓┏┓┏┓┃

┃┃┃┃┃┃

┻┻┻┻┻┻

6 0

2 years ago

Read 2 more answers