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2 years ago

I really don’t know what to do someone help ASAP

Mathematics

2 answers:

Archy [21]2 years ago

5 0

Answer:

0.001

Step-by-step explanation:

-BARSIC- [3]2 years ago

3 0

Answer:

5.5

Step-by-step explanation:

The volume of a sphere is represented by the formula 4/3πr^3. So, if we set 4/3πr^3 equal to 2048/3, we can start solving. Start by multiplying both sides by 3/4, the reciprocal of 4/3 to get rid of the fraction. This gives us πr^3 = 512. Divide 512 by π to get 163.1 (when rounded to the nearest tenth). This leaves us with r^3 = 163.1; find the cubic root of 163.1, which is 5.5, to get r

(a more exact answer is 5.4631958645870141851784948846314‬)

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Ksenya-84 [330]

The answer is 3X + 12 because in he distributive property you must multiply both the 3X and the 4 by 3

3x1=3 hence the 3X

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8 0

3 years ago

Let a population consist of the values cigarettes, cigarettes, and cigarettes smoked in a day. Show that when samples of size

Katen [24]

The "sample" with "mean absolute deviation" indicate about a sample mean absolute deviation is being used as an estimator of the mean absolute deviation of a population

Mean of the sample MAD=3.3
Population MAD =6.4

<h3>What does this indicate about a sample mean absolute deviation used as an estimator of the mean absolute deviation of a population?</h3>

Generally, The MAD measures the average dispersion around the mean of a given data collection.

$1/n \sum_i-1^{n} |x_i -m(X)|$

In conclusion, for the corresponding same to mean

the sample mean absolute deviation

7,7 ↔ 0

7,21 ↔ 7

7,22 ↔ 7.5

21,7 ↔ 7

21,21 ↔ 0

21,22 ↔ 0.5

22,7 ↔ 7.5

Therefore

Mean of the sample MAD=3.3
Population MAD =6.4

Read more about mean absolute deviation

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5 0

2 years ago

A teacher was interested in knowing the amount of physical activity that his students were engaged in daily. He randomly sampled

klasskru [66]

Answer:

The standard error of the mean is 4.5.

Step-by-step explanation:

As we don't know the standard deviation of the population, we can estimate the standard error of the mean from the standard deviation of the sample as:

$\sigma_{\bar{x}}\approx\frac{s}{\sqrt{n}}$

The sample is [30mins, 40 mins, 60 mins, 80 mins, 20 mins, 85 mins]. The size of the sample is n=6.

The mean of the sample is:

$\bar{x}=\frac{1}n} \sum x_i =\frac{30+40+60+80+20+85}{6}=52.5$

The standard deviation of the sample is calculated as:

$s=\sqrt{\frac{1}{n-1}\sum (x_i-\bar x)^2} \\\\ s=\sqrt{\frac{1}{5}\cdot ((30-52.5)^2+(40-52.5)^2+(60-52.5)^2+(80-52.5)^2+(20-52.5)^2+(85-52.5)^2}\\\\s=\sqrt{\frac{1}{5} *3587.5}=\sqrt{717.5}=26.8$