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MArishka [77]
3 years ago
10

The large rectangle below represents one whole.What percent is represented by the shaded are​

Mathematics
2 answers:
kompoz [17]3 years ago
7 0
30 percent. 6/20 is equal to 30/100
spin [16.1K]3 years ago
3 0

Answer:

3 over 10or 6 over 20

Step-by-step explanation: there are twenty squares so 20 will be your denominator and six are shaded so 6 will be your numerator so it will be 6/20 or you could simplify the fraction to 3/10 because you divided both your numerator and denominator by 2 which is allowed in math problems

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Find an equation of the sphere with center (2, −10, 3) and radius 5. $$25=(x−2)2+(y−(−10))2+(z−3)2 use an equation to describe i
lana66690 [7]

In the x,y plane, we have z=0 everywhere. So in the equation of the sphere, we have

25=(x-2)^2+(y+10)^2+(-3)^2\implies(x-2)^2+(y+10)^2=16=4^2

which is a circle centered at (2, -10, 0) of radius 4.

In the x,z plane, we have y=0, which gives

25=(x-2)^2+10^2+(z-3)^2\implies(x-2)^2+(z-3)^2=-75

But any squared real quantity is positive, so there is no intersection between the sphere and this plane.

In the y,z plane, x=0, so

25=(-2)^2+(y+10)^2+(z-3)^2\implies(y+10)^2+(z-3)^2=21=(\sqrt{21})^2

which is a circle centered at (0, -10, 3) of radius \sqrt{21}.

3 0
4 years ago
GIVING BRAINLIEST AND 30 POINTS TO BEST ANSWER!! REMEMBER, FOLLOW DIRECTIONS. YOU CAN DO IT! :D
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Answer:

2 3 4 5 6 7 8 9 10

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3 0
2 years ago
Listed below are the number of years it took for a random sample of college students to earn bachelor's degrees (based on data f
schepotkina [342]

Answer:

a) Mean = 6.5, sample standard deviation = 3.50

b) Standard error = 0.7826

c) Point estimate = 6.5

d) Confidence interval:  (5.1469 ,7.8531)

Step-by-step explanation:

We are given the following data set for students to earn bachelor's degrees.

4, 4, 4, 4, 4, 4, 4.5, 4.5, 4.5, 4.5, 4.5, 4.5, 6, 6, 8, 9, 9, 13, 13, 15

a) Formula:

\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}  

where x_i are data points, \bar{x} is the mean and n is the number of observations.  

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{130}{20} = 6.5

Sum of squares of differences = 6.25 + 6.25 + 6.25 + 6.25 + 6.25 + 6.25 + 4 + 4 + 4 + 4 + 4 + 4 + 0.25 + 0.25 + 2.25 + 6.25 + 6.25 + 42.25 + 42.25 + 72.25 = 233.5

S.D = \sqrt{\frac{233.5}{19}} = 3.50

b) Standard Error

= \displaystyle\frac{s}{\sqrt{n}} = \frac{3.50}{\sqrt{20}} = 0.7826

c) Point estimate for the mean time required for all college is given by the sample mean.

\bar{x} = 6.5

d) 90% Confidence interval:  

\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}  

Putting the values, we get,  

t_{critical}\text{ at degree of freedom 19 and}~\alpha_{0.10} = \pm 1.729  

6.5 \pm 1.729(\frac{3.50}{\sqrt{20}} ) = 6.5 \pm 1.3531 = (5.1469 ,7.8531)

e) No, the confidence interval does not contain the value of 4 years. Thus, confidence interval is not a good estimator as most of the value in the sample is of 4 years. Most of the sample does not lie in the given confidence interval.

5 0
4 years ago
the perimeter of a square is 80-64y units. which expression can be used to show the side lenght of one side of the square?
leonid [27]
80x by 80x up and down and 64y by 64y left and right 
5 0
3 years ago
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