Ask question

Ask question

All categories

3 years ago

13

a box of crackers has a volume of 48 cubic inches. what is the volume of a similar box that is smaller by a scale factor of 2/3

1 answer:

earnstyle [38]3 years ago

6 0

For this case what you need to know is that the original volume of the cookie box is:
V = (w) * (l) * (h)
Where,
w: width
l: long
h: height.
We have then:
V = (w) * (l) * (h) = 48 in ^ 3
The volume of a similar box is:
V = (w * (2/3)) * (l * (2/3)) * (h * (2/3))
We rewrite:
V = ((w) * (l) * (h)) * ((2/3) * (2/3) * (2/3))
V = (w) * (l) * (h) * ((2/3) ^ 3)
V = 48 * ((2/3) ^ 3)
V = 14.22222222 in ^ 3
Answer:
the volume of a similar box that is smaller by a scale factor of 2/3 is:
V = 14.22222222 in ^ 3

You might be interested in

The diameter of a circle is 10 ft. Find its area in terms of pi

Alexxandr [17]

A = pi(r) ^2
A = 3.14(5) ^2
A = 15.7 ^2
A = 246.49
Hope this helps.

5 0

3 years ago

Prove ΔPAB is isosceles.

Licemer1 [7]

Answer:

See explanation

Step-by-step explanation:

If $PX\cong PY,$ then triangle PXY is isosceles triangle. Angles adjacent to the base XY of an isosceles triangle PXY are congruent, so

$\angle 1\cong \angle 2$

and

$m\angle 1=m\angle 2$

Angles 1 and 3 are supplementary, so

$m\angle 3=180^{\circ}-m\angle 1$

Angles 2 and 4 are supplementary, so

$m\angle 4=180^{\circ}-m\angle 2$

By substitution property,

$m\angle 4=180^{\circ}-m\angle 2=180^{\circ}-m\angle 1=m\angle 3$

Hence,

$\angle 3\cong \angle 4$

Consider triangles APX and BPY. In these triangles:

$PX\cong PY$ - given;
$\angle 5\cong \angle 6$ - given;
$\angle 3\cong \angle 4$ - proven,

so $\triangle APX\cong \triangle BPY$ by ASA postulate.

Congruent triangles have congruent corresponding sides, then

$AP\cong BP$

Therefore, triangle APB is isosceles triangle (by definition).

6 0

3 years ago

Suppose a box of Cracker Jacks contains one of 5 toy prizes: a small rubber ball, a whistle, a Captain America decoder ring, a r

lakkis [162]

Answer:

11.42 boxes

Step-by-step explanation:

For the first box bought, there is a 100% chance of getting a unique toy (since you still don't have any). E₁ = 1.

After that, there is a 4 in 5 chance of getting a unique toy from the next box, the expected number of boxes required is:

$E_2 = (\frac{4}{5})^{-1} = 1.25$

For the next unique toy, there is now a 3 in 5 chance of getting it:

$E_3 = (\frac{3}{5})^{-1} = 1.67$

Following that logic, there is a 2 in 5 chance of getting the 4th unique toy:

$E_4 = (\frac{2}{5})^{-1} = 2.5$

Finally, there is a 1 in 5 chance to get the last unique toy:

$E_5 = (\frac{1}{5})^{-1} = 5$

The expected number of boxes to obtain a full set is:

$E=E_1+E_2+E_3+E_4+E_5\\E=1+1.25+1.67+2.5+5\\E=11.42\ boxes$

5 0

3 years ago

How many rays intersect at point O <br> 1. 5 <br> 2. 4 <br> 3. 6 <br> 4. 3

Vilka [71]

Answer:

3 is correct ans

ray on point o

is 3

option 4 3 is correct

line intersecting on same point on o

5 0

2 years ago

The probability that it will rain on Thursday is 67%. What is the probability that it won’t rain on Thursday? Express your answe

kondaur [170]

Answer:

33%

Step-by-step explanation:

Since the probability of rain on Thursday is 67%, the probability of no rain on Thursday is 100% - 67% = 33%.

4 0

2 years ago