2m^2 ( -3m + 4m^3 ) - (3m)^3 simplify completely

Mathematics

2 answers:

patriot [66]3 years ago

6 0

Answer:

8m^5-33m^3

Step-by-step explanation:

Combine like terms Expand and simplify

erastova [34]3 years ago

6 0

Answer:

8m^5-33m^3

Step-by-step explanation:

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What is the side length of the largest square that can fit into a circle with a radius of 5 units?

Mariana [72]

So 1st consider that it's a square! That's very important. So for a square, all 4 sides are equal.

And now considering that the given information is the diameter. So any angle made at the circle extended from the 2 points of diameter gives an angle of 90°

Now consider one triangle. So we already know that 2 sides of the triangle are equal (because they are 2 sides of a square) , has a side of 10 (diameter) and and angle of 90°. So remaining 2 angles are 45°

Now solve it by applying

$\sin(45) \: \: \: \: = x \div 10 \\ (1 \div \sqrt{2} ) = (x \div 10) \\ 10 \div \sqrt{2} \: = x$

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Schach [20]

Answer:

97.72% probability that their mean body temperature is greater than 98.4degrees° F.

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$