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Sonja [21]
3 years ago
9

Gracie is shipping a cube shaped box. It measures 14 in x 14 in x 14 in. What is the surface area of the box. Please show work)

Mathematics
1 answer:
Aliun [14]3 years ago
6 0
So you do length times width and in your case it would be 14 times 14 = 196
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Direction: True or False
Goshia [24]

Answer:

<u>False </u>Any three points are always coplanar

<u>False </u>Two points are always collinear

<u>False </u>Two planes intersect at a point

Read explanations, as the answer to these questions, is subjective.

Step-by-step explanation:

The definition of coplanar is; figures that exist on the same plane. Any three points might not always be colinear, as some points might exist on one plane, but other points could exist on other planes. To visualize this phenomenon, refer to the attached image. Two points could be on the red plane, but the third could be on the green. Therefore, while there are three points, not all of them exist on the same plane. However, another plane can be constructed to connect these points, bear in mind certain problems might specifically indicate that three points are not coplanar, therefore, the answer is false.

The definition of collinear is existing on the same line. This statement is a little subject, it can be both true and false depending on the way one looks at it. A line can be drawn to connect any two points, one only needs two points to determine a line. So technically two points are always collinear. However, it also depends on the circumstance. Certain geometry problems might specifically indicate that two points are not collinear. Thus, it really depends on the context. Therefore, the answer is technically false.

Two planes usually intersect on a line. To understand this, please refer to the image attached. Tehcnailly, if a corner of one plane intersects another, then one can state that the plane intersects on a point, but typically, planes intersect on lines. Therefore, technically the answer to this problem is false, as two planes usually intersect on a line.

Images credits: Geogebra

8 0
3 years ago
Read 2 more answers
Add. (6x2 – 2x) + (5x – 7)
11Alexandr11 [23.1K]

Answer:

Thats going to be 12x

Step-by-step explanation:

8 0
3 years ago
i bought 6 apples for0.30 dollars each and 5 oranges for 0.40 dollars each. how much did i spend ? write an equation then solve​
katovenus [111]

Answer:

3.8

Step-by-step explanation:

if the apples are 0.30 cents each then you multiply .3 X 6 and you get 1.8 because 1 apple is .30 cents you want to know how much is 6.

Next you pay .40 cents each for 5 oranges so .4 X 5 which is 2. And once again one orange is .40 cents so you multiply that by 5 to see the price of 5 oranges

Now you add the 1.8 to 2 and get 3.8

Now for the equation you just need to plug in what you did so...

(0.3 X 6) + (0.40 X 5) =  3.8

Glad i could help :)

7 0
3 years ago
I need the first and second one, please anyone
tia_tia [17]

Answer:

f(-1) = 0

f(2) = 16

Step-by-step explanation:

f(-1) = 4(-1) + 4 = 0

f(2) = 4(2) + 8 = 16

5 0
3 years ago
Three populations have proportions 0.1, 0.3, and 0.5. We select random samples of the size n from these populations. Only two of
IRINA_888 [86]

Answer:

(1) A Normal approximation to binomial can be applied for population 1, if <em>n</em> = 100.

(2) A Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50 and 40.

(3) A Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50, 40 and 20.

Step-by-step explanation:

Consider a random variable <em>X</em> following a Binomial distribution with parameters <em>n </em>and <em>p</em>.

If the sample selected is too large and the probability of success is close to 0.50 a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

  • np ≥ 10
  • n(1 - p) ≥ 10

The three populations has the following proportions:

p₁ = 0.10

p₂ = 0.30

p₃ = 0.50

(1)

Check the Normal approximation conditions for population 1, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.10=1

Thus, a Normal approximation to binomial can be applied for population 1, if <em>n</em> = 100.

(2)

Check the Normal approximation conditions for population 2, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.30=310\\\\n_{c}p_{1}=50\times 0.30=15>10\\\\n_{d}p_{1}=40\times 0.10=12>10\\\\n_{e}p_{1}=20\times 0.10=6

Thus, a Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50 and 40.

(3)

Check the Normal approximation conditions for population 3, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.50=510\\\\n_{c}p_{1}=50\times 0.50=25>10\\\\n_{d}p_{1}=40\times 0.50=20>10\\\\n_{e}p_{1}=20\times 0.10=10=10

Thus, a Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50, 40 and 20.

8 0
3 years ago
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