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Mathematics

1 answer:

sergey [27]2 years ago

5 0

Answer:

${(x + 5)}^{2} - {x}^{2} \\ {x }^{2} + 10x + 25 - {x}^{2} \\ 10x + 25 \\ 5(2x + 5)$

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Solve the following quadratic function

earnstyle [38]

Answer:

x = 10/7 or x = -10/7

Step-by-step explanation:

By using the completing the squares method, we have to divide the coefficient of the x term by 2, square it and add to both sides

From what we have here, the coefficient of the x term is 0

Thus, we have it that;

0 = 100 - 49x^2

We can proceed to use the difference of two squares here;

Whereby;

a^2 - b^2 = (a-b)(a + b)

0 = (10-7x)(10+7x)

10 - 7x = 0 or 10 + 7x = 0

7x = 10 or 7x = -10

x = 10/7 or x = -10/7

4 0

2 years ago

For a particular diamond mine, 81% of the diamonds fail to qualify as "gemstone grade". A random sample of 92 diamonds is analyz

Fed [463]

Answer:

68.79% probability that more than 79% of the sample diamonds fail to qualify as gemstone grade

Step-by-step explanation:

We use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

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.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .