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alexandr402 [8]
3 years ago
12

Please answer asap!

Mathematics
1 answer:
soldi70 [24.7K]3 years ago
6 0
I think it’s 76 degrees not 100% sure though but you can try it lol
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Solve for x.
Ivahew [28]
ANSWER

x=\frac{2-\sqrt{10}} {3}

or

x=\frac{\sqrt{10}+2} {3}

We have

3x^2-4x-2=0

Since we cannot factor easily, we complete the square.

Adding 2 to both sides give,

3x^2-4x=2

Dividing through by 3 gives

x^2-\frac{4}{3}x= \frac{2}{3}

Adding (-\frac{2}{3})^2 to both sides gives

x^2-\frac{4}{3}x+(-\frac{2}{3})^2= \frac{2}{3}+(-\frac{2}{3})^2

The expression on the Left Hand side is a perfect square.

(x-\frac{2}{3})^2= \frac{2}{3}+\frac{4}{9}

\Rightarrow (x-\frac{2}{3})^2= \frac{10}{9}

\Rightarrow (x-\frac{2}{3})=\pm \sqrt{\frac{10}{9}}

\Rightarrow (x)=\frac{2}{3} \pm {\frac{\sqrt{10}}{3}

Splitting the plus or minus sign gives

x=\frac{2- \sqrt{10}} {3}

or

x=\frac{\sqrt{10}+2} {3}
3 0
3 years ago
Read 2 more answers
A patient was ordered to receive 2 pints of blood via a transfusion. How much blood is to be transfused in milliliters?
vagabundo [1.1K]

Answer:

2 pints of blood =946mililiters

Step-by-step explanation:

1 Pint =473mililters

so either 473+473=

or 473x2

2 pints of blood =946mililiters

6 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
Find the complete factored form of the polynomial<br> 28a^4b^6 + 21a^3b^6
n200080 [17]
Yes the fierce rbajeei ironwood he
3 0
3 years ago
Gradient of line AB (3,7) and (5,11)​
larisa [96]

Answer:

Gradient is change in y over change in x.

gradient = y1-y2 ÷ x1-x2

=7-11 ÷ 3-5

=-4 ÷ -2

=2

4 0
3 years ago
Read 2 more answers
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