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3 years ago

How to write the number 2,937,082 in expanded form

Mathematics

2 answers:

stich3 [128]3 years ago

6 0

2,000,000+ 900,000+ 30,000+ 7,000+ 82

ololo11 [35]3 years ago

6 0

2,000,000
900,000
30,000
7,000
80
2

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Using an appropriately designed and analyzed study, a researcher observes a p-value of 0.010000. Consider the following proposit

mafiozo [28]

Answer: a.

a. PROPOSITION ONE is true and PROPOSITION TWO is true

b. PROPOSITION ONE is true and PROPOSITION TWO is false

c. PROPOSITION ONE is false and PROPOSITION TWO is true

d. PROPOSITION ONE is false and PROPOSITION TWO is false

e. None of the above

7 0

3 years ago

Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 21 in. by 12 in. by

a_sh-v [17]

Answer:

Dimension of the box is $16.1\times 7.1\times 2.45$

The volume of the box is 280.05 in³.

Step-by-step explanation:

Given : The open rectangular box of maximum volume that can be made from a sheet of cardboard 21 in. by 12 in. by cutting congruent squares from the corners and folding up the sides.

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

Solution :

Let h be the height of the box which is the side length of a corner square.

According to question,

A sheet of cardboard 21 in. by 12 in. by cutting congruent squares from the corners and folding up the sides.

The length of the box is $L=21-2h$

The width of the box is $W=12-2h$

The volume of the box is $V=L\times W\times H$

$V=(21-2h)\times (12-2h)\times h$

$V=(21-2h)\times (12h-2h^2)$

$V=252h-42h^2-24h^2+4h^3$

$V=4h^3-66h^2+252h$

To maximize the volume we find derivative of volume and put it to zero.

$V'=12h^2-132h+252$

$0=12h^2-132h+252$

Solving by quadratic formula,

$h=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$h=\frac{-(-132)\pm\sqrt{132^2-4(12)(252)}}{2(12)}$

$h=\frac{132\pm72.99}{24}$

$h=2.45,8.54$

Now, substitute the value of h in the volume,

$V=4h^3-66h^2+252h$

When, h=2.45

$V=4(2.45)^3-66(2.45)^2+252(2.45)$

$V\approx 280.05$

When, h=8.54

$V=4(8.54)^3-66(8.54)^2+252(8.54)$

$V\approx -170.06$

Rejecting the negative volume as it is not possible.

Therefore, The volume of the box is 280.05 in³.

The dimension of the box is

The height of the box is h=2.45

The length of the box is $L=21-2(2.45)=16.1$

The width of the box is $W=12-2(2.45)=7.1$

So, Dimension of the box is $16.1\times 7.1\times 2.45$

6 0

3 years ago

3z-4=6z-17 how do solve

Maru [420]

3z-4=6z-17
-3z-4=-17 (subtracted the 6z from both sides)
-3z=-13 (added the 4 to both sides)
z= $\frac{-13}{-3z}$ (divided both sides by -3z)
z= $\frac{13}{3}$