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Arisa [49]
3 years ago
13

A shipping company sells large boxes and small boxes. The dimensions of the large box are 3 times the dimensions of the small bo

x.
How does the volume of the large box compare to the volume of the small box?
a) the volume of the large box is 3 times the volume of the small box
b) 6 times
c) 9 times
d) 27 times
Mathematics
1 answer:
VladimirAG [237]3 years ago
7 0

Answer:

d) 27 times

Step-by-step explanation:

Let l,w, and h represent the length, width and height of the small box.

Therefore, 3l,3w, and 3h are the dimensions of the large box.

The volume of a rectangular prism, such as a box, is the product of its length, width and height

Volume of small box = l*w*h = lwh

Volume of large  box = 3l*3w*3h = 27lwh

So we can see that the volume of the large box is 27 times larger.

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Graph this line using the slope and y-intercept: y = 1/10x + 3​
Mila [183]

Answer:

Slope: 1/10

y-intercept: (0,3)

Step-by-step explanation:

Slope intercept form: y = mx + b

Slope: 1/10

y-intercept: (0,3)

8 0
3 years ago
Farmers know that driving heavy equipment on wet soil compresses the soil and injures future crops. Here are data on the "penetr
Greeley [361]

Answer:

Step-by-step explanation:

Hello!

To see if driving heavy equipment on wet soil compresses it causing harm to future crops, the penetrability of two types of soil were measured:

Sample 1: Compressed soil

X₁: penetrability of a plot with compressed soil.

n₁= 20 plots

X[bar]₁= 2.90

S₁= 0.14

Sample 2: Intermediate soil

X₂: penetrability of a plot with intermediate soil.

n₂= 20 (with outlier) n₂= 19 plots (without outlier)

X[bar]₂= 3.34 (with outlier) X[bar]₂= 2.29 (without outlier)

S₂= 0.32 (with outlier) S₂= 0.24 (without outlier)

Outlier: 4.26

Assuming all conditions are met and ignoring the outlier in the second sample, you have to construct a 99% CI for the difference between the average penetration in the compressed soil and the intermediate soil. To do so, you have to use a t-statistic for two independent samples:

Parámeter of interest: μ₁-μ₂

Interval:

[(X[bar]₁-X[bar]₂)±t_{n_1+n_2-2;1-\alpha/2}*Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} }]

t_{n_1+n_2-2;1-\alpha/2}= t_{20+19-2;1-(0.01/2)}= t_{37; 0.995}= 2.715

Sa= \sqrt{\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} } = \sqrt{\frac{19*0.0196+18*0.0576}{20+19-2} }= 0.195= 0.20

[(2.90-2.29)±2.715*0.20\sqrt{\frac{1}{20} +\frac{1}{19} }]

[0.436; 0.784]

I hope this helps!

5 0
3 years ago
Simplify 3(6 – 4)2. i am now adding words to make my question 20 figures so i can write it.
sergiy2304 [10]
LOL, that question. But the answer is 12.
4 0
3 years ago
Read 2 more answers
HURRY PLZ!!! IM TAKING THE TEST RIGHT NOW!!!
Vesnalui [34]

Answer :   YOU HAVE TO SHOW THE WHOLE PIC PLS SO I CAN HELP U

Step-by-step explanation:

7 0
3 years ago
A submarine travels 8.3 km due North from its base and then turns and travels due West for 13.9 km.
olga_2 [115]

Answer:

He is 5.6 km away

Step-by-step explanation:

he starts traveling away from base and is 8.3 km away to start with

then he turns west and travels 13.9 km away

what this mean is that you have to subtract the distance he turned from the distance he started off with

ex: 13.9 - 8.3 = 5.9

to get your answer of how far away he is from the base

5 0
2 years ago
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