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Kaylis [27]
3 years ago
9

1. 1/3 x 2/3 2. 2/3 x 4/5

Mathematics
1 answer:
Olin [163]3 years ago
3 0
The answer to #1 is 2/9
the answer to #2 is 8/15
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Postal regulations specify that a parcel sent by priority mail may have a combined length and girth of no more than 144 in. Find
mr Goodwill [35]

Answer:

The volume in such a package is 27648 in³

Step-by-step explanation:

Consider the provided information.

Postal regulations specify that a parcel sent by priority mail may have a combined length and girth of no more than 144 in.

Let the dimension are x by x by y.

Where x is the variable for the square base package and y is the variable for length.

Thus l=x, b=x and h=y

Then the volume of the box is: V(x)=x^2y (∵V=lbh)

The maximum combined length and girth is 144.

Therefore, 4x+y=144

y=144-4x

Substitute the value of y in volume of the box.

V(x)=x^2(144-4x)

V(x)=144x^2-4x^3

V'(x)=288x-12x^2

Substitute V'(x)=0.

0=288x-12x^2

12x(24-x)=0

x=0\ or\ x=24

Now apply second derivative test.

V''(x)=288-24x

V''(0)=288-24(0)>0 (Min)

V''(24)=288-24(24) (Max)

Hence, the maximum volume is at x=24

If x=24 then y=144-4(24)=48

Substituting value of x = 24 and y = 48 V(x)=x^2y gives 27648.

Hence, the volume in such a package is 27648 in³

5 0
3 years ago
A researcher wants to determine whether the number of minutes adults spend online per day is related to gender. A random sample
suter [353]

Answer:

The expected frequency for the cell E2,2 is = 33

Step-by-step explanation:

The given data is

Gender               | 0-30| 30-60| 60-90| 90+       Totals

Male                    | 23    |    35   |   76    |  46        180

<u> Female               |   31   |    42   |   46    |  16         135    </u>

<u>Totals                    54         77        122    62          315  </u>

<u />

The expected frequency for the cell E2,2 is :

Expected for the (30-60 box for females)   =  Total of (30-6)/ (total )* females

                                                                      = ( 77/315)135= 33

Here p= 77/315 and n= 135

therefor X= pn = 33

χ²=  (33-42)²/33= 2.455  ( for the single value of E2,2=33)

Expected for the (30-60 box for males)   =  Total of (30-6)/ (total )* males

                                                                      = ( 77/315)180= 44

χ²=  (44-42)²/44= 0.09

The critical region is  χ² (0.05) 3 ≥  χ²= 11.34

Let the null and alternate hypothesis be

H0:the number of minutes spent online per day is not related to gender

against

Ha: the number of minutes spent online per day is related to gender

The single value of χ² for E2,2 = 2.45 is less than the critical value of 11.34.  

6 0
3 years ago
Compare <br> 1.41___1.397<br><br> &lt;<br> &gt;<br> =
Snowcat [4.5K]
1.41 is greater than 1.39
>
4 0
3 years ago
How to simplify -3 (4×-5y-5×)
VMariaS [17]
Distribute -3 to the rest of the numbers by multiplying them together. That’s all you need to do you don’t have to solve
6 0
4 years ago
Read 2 more answers
How do you find an absolute value equation from a table? ​
Rina8888 [55]
X. 0,2,6,4,8,10
Y. 2,0,2,4,6,8

It’s how far the number is from zero
7 0
3 years ago
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