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ziro4ka [17]
3 years ago
6

Example:The missing value is x than Sin 7 / 18

Mathematics
2 answers:
Snezhnost [94]3 years ago
4 0

Answer:

44.9

Step-by-step explanation:

12/17 = PQ/PR

(sin^-1)(12/17) = 44.9

Darya [45]3 years ago
3 0

Answer:

44.9

Step-by-step explanation:

12/17 = PQ/PR

(sin^-1)(12/17) = 44.9

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Joey is buying plants for his garden. He wants to have at least twice as many flowering plants as nonflowering plants and a mini
vodomira [7]
(24, 12) and (36, 0).  The least amount of flowering plants occurs when x=2y, and the largest amount occurs when y=0.  These two points satisfy both conditions and both sum to 36.
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Read 2 more answers
Find the value of h(x) = -3x – 12 when x = -2? Show work
7nadin3 [17]

Answer:

-6

Step-by-step explanation:

h(x) = -3x – 12 becomes h(-2) =  -3(-2) – 12 = -6

7 0
2 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
Let V be the volume of the solid obtained by rotating about the y-axis the region bounded y = 4x and y = x2/4 . Find V by slicin
aliya0001 [1]

Answer:

Step-by-step explanation:

Consider the graphs of the y = 4x  and  y = \frac{x^{2} }{4}.

By equating the expressions, the intersection points of the graphs can be found and in this way delimit the area that will rotate around the Y axis.

4x = \frac{x^{2} }{4} \\   x^{2}  = 16x \\ x^{2}  - 16x = 0 \\   x(x-16) = 0 then x=0  o  x=16. Therefore the integration limits are:

y = 4(0) = 0  and  y = 4(16) = 64

The inverse functions are given by:

x = 2 \sqrt{y}  and  x = \frac{y}{4}. Then

The volume of the solid of revolution is given by:

\int\limits^{64}_ {0} \, [2\sqrt{y} - \frac{y}{4}]^{2}  dy = \int\limits^{64}_ {0} \, [4y - y^{3/2} + \frac{y^{2}}{16} ]\  dy = [2y^{2} - \frac{2}{5}y^{5/2} + \frac{y^{3}}{48} ]\limits^{64}_ {0} = 546.133 u^{2}

6 0
3 years ago
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