<img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%20%E2%88%923x2%E2%88%9211x%2B7" id="TexFormula1" title="f(x) = −3x2&minus

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Vilka [71]

3 years ago

;11x+7" alt="f(x) = −3x2−11x+7" align="absmiddle" class="latex-formula">

Mathematics

1 answer:

bixtya [17]3 years ago

4 0

Answer:

$x= \frac{ - 11 + \sqrt{205} }{6} \: \: \: \: \: or \: \: \: \: \\x = \frac{ - 11 - \sqrt{205} }{6}$

Step-by-step explanation:

$- 3x {}^{2} - 11x + 7 = 0 \\ 3x {}^{2} + 11x - 7 = 0$

here,

$a = 3 \\ b = 11 \\ c = - 7$

now,

$D = b {}^{2} - 4ac \\ = (11) {}^{2} - 4(3)( - 7) \\ = 121 + 84 \\ = 205 > 0 \\$

so,

$x = \frac{ - b + - \sqrt{b {}^{2} - 4ac} }{2a} \\ \\ = \frac{ - 11 + - \sqrt{205} }{2(3)} \\ = \frac{ - 11 + - \sqrt{205} }{6} \\ \\ = \frac{ - 11 + \sqrt{205} }{6} \: \: \: \: \: or \: \: \: \: = \frac{ - 11 - \sqrt{205} }{6}$

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Suppose the weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams. The weights

user100 [1]

Answer:

a) 0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.

b) The weight that 80% of the apples exceed is of 78.28g.

Step-by-step explanation:

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.

Weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams.

This means that $\mu = 85, \sigma = 8$