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Dmitry_Shevchenko [17]
4 years ago
5

Which is M=9pm solved for p

Mathematics
1 answer:
zaharov [31]4 years ago
6 0
<h2>Answer:</h2>

<u>The value of </u><u>p :M/9m</u>

<h2>Explanation:</h2>

As we have been given value of M

Value of M=9pm

So we can find value of p from here by simple mathematics rule

M=9pm

divide both side with 9m so

M/9m=9pm/9m

we get value of p so M/9m=p.

we can write it as p=M/9m

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Someone please help me??
lapo4ka [179]
The standard equation of a circle with centre (xc,yc) and radius R is given by

(x-x_c)^2+(y-y_c)^2=R^2

Substituting
centre (xc,yc) = (4,-3)
R=2.5

The equation is therefore

(x-x_c)^2+(y-y_c)^2=R^2
(x-4)^2+(y-(-3))^2=2.5^2
(x-4)^2+(y+3)^2=2.5^2  or
(x-4)^2+(y+3)^2=6.25
7 0
3 years ago
Suppose each of the following data sets is a simple random sample from some population. For each dataset, make a normal QQ plot.
adell [148]

Answer:

a) For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

b) For this case the data is skewed to the left and we can't assume that we have the normality assumption.

c) This last case the histogram is not symmetrical and the data seems to be skewed.

Step-by-step explanation:

For this case we have the following data:

(a)data = c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

We can use the following R code to get the histogram

> x1<-c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

> hist(x1,main="Histogram a)")

The result is on the first figure attached.

For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

(b)data = c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> x2<- c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> hist(x2,main="Histogram b)")

The result is on the first figure attached.

For this case the data is skewed to the left and we can't assume that we have the normality assumption.

(c)data = c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> x3<-c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> hist(x3,main="Histogram c)")

The result is on the first figure attached.

This last case the histogram is not symmetrical and the data seems to be skewed.

7 0
3 years ago
What is the value of the expression 24/x when x = 8?
erma4kov [3.2K]

Answer:

Itd be 3 becaus 24/8 is 3

8 0
3 years ago
How to solve y for this problem:
True [87]
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6 0
3 years ago
Is 4/5 a solution of x&lt;2?
krok68 [10]
Yea it is.4\5 is greater than 2
5 0
4 years ago
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