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2 years ago

32% of 300 is what number?

Mathematics

1 answer:

Wewaii [24]2 years ago

8 0

Answer:

Four options to solve:

1. 32% of 300 = 96

OR

2. 32% × 300 = 96

OR

3. 0.32 × 300 = 96

OR

4. $\frac{32}{100}$ × 300 = 96

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3 years ago

How to factor 4x^2+24x+27

fomenos

Answer:

(2 x + 3) (2 x + 9)

Step-by-step explanation:

Factor the following:

4 x^2 + 24 x + 27

Factor the quadratic 4 x^2 + 24 x + 27. The coefficient of x^2 is 4 and the constant term is 27. The product of 4 and 27 is 108. The factors of 108 which sum to 24 are 6 and 18. So 4 x^2 + 24 x + 27 = 4 x^2 + 18 x + 6 x + 27 = 9 (2 x + 3) + 2 x (2 x + 3):

9 (2 x + 3) + 2 x (2 x + 3)

Factor 2 x + 3 from 9 (2 x + 3) + 2 x (2 x + 3):

Answer: (2 x + 3) (2 x + 9)

3 0

3 years ago

Solve the problem. Use the Central Limit Theorem.The annual precipitation amounts in a certain mountain range are normally distr

bazaltina [42]

Answer:

0.8944 = 89.44% probability that the mean annual precipitation during 25 randomly picked years will be less than 112 inches.

Step-by-step explanation:

To solve this question, we use the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

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