Need help with this one!!!

Mathematics

1 answer:

DaniilM [7]4 years ago

8 0

$\bf (2x^2)^4(5x^6)\impliedby \textit{distributing the exponent}
\\\\\\
2^4x^{2\cdot 4}5x^6\implies 2^4\cdot 5x^{2\cdot 4}x^6\implies 16\cdot 5x^8x^6\implies 80x^{8+6}
\\\\\\
80x^{14}$

You might be interested in

Natalie draws an acute triangle.

Nesterboy [21]

Polygon, 3 straight sides

7 0

3 years ago

Shelia's measured glucose level one hour after a sugary drink varies according to the normal distribution with μ = 117 mg/dl and

TiliK225 [7]

Answer:

The level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L is L = 127.1 mg/dl.

Step-by-step explanation:

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

Normal probability distribution

Problems of normally distributed samples are 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 $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$