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3 years ago

50 points

Mathematics

2 answers:

Vesna [10]3 years ago

7 0

I agree with The person that answered before me

xxMikexx [17]3 years ago

6 0

Answer:

Step-by-step explanation:

root the 216 as it is a cube

:edge *edge *edge =volume

Since all edges are the same, just root it

I hope this helped if it did please mark me Brainliest

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Whats the difference between range and interquartile range?

miv72 [106K]

Answer:

Step-by-step explanation:

The range is the difference between the highest and the lowest values in a set of data. The interquartile range, the IQR, is what's "inside" the box in a box plot, which consists of the difference between the central measures, which are the first and the third quartiles.

3 0

3 years ago

A certain test preparation course is designed to help students improve their scores on the LSAT exam. A mock exam is given at th

OlgaM077 [116]

Answer:

Interval [16.34 , 21.43]

Step-by-step explanation:

First step. <u>Calculate the mean</u>

$\bar X=\frac{(23+18+23+12+13+23)}{6}=18.666$

Second step. <u>Calculate the standard deviation</u>

$\sigma =\sqrt{\frac{(23-18.666)^2+(18-18.666)^2+(23-18.666)^2+(12-18.666)^2+(13-18.666)^2+(23-18.666)^2}{6}}$

$\sigma=\sqrt\frac{18.783+0.443+18.783+44.435+5.666+18.783}{6}$

$\sigma=\sqrt{17.815}=4.22$

As the number of data is less than 30, we must use the t-table to find the interval of confidence.

We have 6 observations, our level of confidence DF is then 6-1=5 and we want our area A to be 80% (0.08).

We must then choose t = 1.476 (see attachment)

Now, we use the formula that gives us the end points of the required interval

$\bar X \pm t\frac{\sigma}{\sqrt n}$

where n is the number of observations.

The extremes of the interval are then, rounded to the nearest hundreth, 16.34 and 21.43

6 0

3 years ago

A certain radioactive isotope has leaked into a small stream. Three hundred days after the leak, 3% of the original amount of t

sesenic [268]

If the half-life is t, then every t days, the amount of the radioactive isotope will be cut in half.
(1/2)^(number of half-lives) = 3%
number of half-lives = ln(0.03) / ln(0.5)
This gives the number of half-lives as 5.06.
Then 300 days = (5.06)(length of 1 half-life)
length of 1 half-life = 300 / 5.06 = 59.29 days

6 0

4 years ago

A large serving of soup from a take-out restaurant is 4/5 liter. The restaurant has 3 liters of soup. The diagram below models t

Airida [17]

Answer:

The quantity of total serving soup does restaurant has = T = 2 $\dfrac{2}{5}$ liters .