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mr Goodwill [35]
3 years ago
11

A distribution of numbers has the following five-number summary: 13.7, 24.5, 38.9, 50.0, 73.2 True or False? These numbers can b

e used to calculate the standard deviation of the distribution. A. True B. False
Mathematics
2 answers:
Oxana [17]3 years ago
6 0

Answer:

False

Step-by-step explanation:

Given that a distribution of numbers has the following five-number summary: 13.7, 24.5, 38.9, 50.0, 73.2

These five numbers are known as minimum, I quartile, Median, III quartile and max respectively.

This gives an idea about the scatter of data on either side of median, the interquartile range, and also help to find outliers if any.

But std deviation is the square root of variance. Variance is the average of sum of squares of deviations from the mean.

Hence without knowing mean and each entry just with five number summary we may not be able to find out the standard deviation.

DIA [1.3K]3 years ago
3 0

Answer:

the answer is false

Step-by-step explanation:

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The submarine is 217 feet below sea level. Identify the integer
Sever21 [200]

Answer:

okay

Step-by-step explanation:

6 0
2 years ago
0.3,1/5,1/2,0.706,7/8,57.1%<br> Place following in ascending order
harina [27]

Answer:

the answer is 0.31/ 5.1/8.57.1% /207067

Step-by-step explanation:

ascending order from the smallest to the largest

5 0
3 years ago
-2,-1,0,1 whichave pair of numbers has a sum of zero
Verizon [17]
-1+1=0. So your answer is -1 and 1
3 0
3 years ago
Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by
11111nata11111 [884]

Answer:

The dimension of the open rectangular box is 8.216\times 4.216\times 1.392.

The volume of the box is 8.217 cubic inches.

Step-by-step explanation:

Given : The open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by cutting congruent squares from the corners and folding up the sides.

To find : The dimensions and the volume of the open rectangular box ?

Solution :

Let the height be 'x'.

The length of the box is '11-2x'.

The breadth of the box is '7-2x'.

The volume of the box is V=l\times b\times h

V=(11-2x)\times (7-2x)\times x

V=4x^3-36x^2+77x

Derivate w.r.t x,

V'(x)=4(3x^2)-2(36x)+77

V'(x)=12x^2-72x+77

The critical point when V'(x)=0

12x^2-72x+77=0

Solve by quadratic formula,

x=\frac{18+\sqrt{93}}{6},\frac{18-\sqrt{93}}{6}

x=4.607,1.392

Derivate again w.r.t x,

V''(x)=24x-72

Now, V''(4.607)=24(4.607)-72=38.568>0 (+ve)

V''(1.392)=24(1.392)-72=-38.592 (-ve)

So, there is maximum at x=1.392.

The length of the box is l=11-2x

l=11-2(1.392)=8.216

The breadth of the box is b=7-2x

b=7-2(1.392)=4.216

The height of the box is h=1.392.

The dimension of the open rectangular box is 8.216\times 4.216\times 1.392.

The volume of the box is V=l\times b\times h

V=8.216\times 4.216\times 1.392

V=48.217\ in.^3

The volume of the box is 8.217 cubic inches.

5 0
3 years ago
NEED HELP ASAP WILL MARK BRANLIEST!!!!!!
Blababa [14]
The heck, what math are you in?
7 0
3 years ago
Read 2 more answers
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