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ale4655 [162]
3 years ago
12

Find the x-intercept and y-intercept

Mathematics
1 answer:
Aneli [31]3 years ago
8 0

Answer:

x - intercept:  (-22, 0)

y - intercept: (0, 11)

Step-by-step explanation:

Given the linear equation

-3x + 11y = 66

<em><u>Determining the x-intercept</u></em>

We know that the x-intercept can be computed by setting the value y=0 and solving for x.

substitute y = 0

-3x + 11y = 66

-3x + 11(0) = 66

-3x + 0 = 66

-3x = 66

Divide both sides by -3

-3x/-3 = 66/-3

x = -22

Thus, the ordered pair representing the x-intercept : (-22, 0)

<em><u>Determining the y-intercept</u></em>

We know that the y-intercept can be computed by setting the value x=0 and solving for y.

substitute x = 0

-3x + 11y = 66

-3(0) + 11y = 66

0 + 11y = 66

11y = 66

Divide both sides by 11

11y/11 = 66/6

y = 11

Thus, the ordered pair representing the y-intercept : (0, 11)

Therefore, we conclude that:

x - intercept:  (-22, 0)

y - intercept: (0, 11)

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Consider the probability that greater than 96 out of 153 DVDs will work correctly. Assume the probability that a given DVD will
ICE Princess25 [194]

Answer:

0.3594 = 35.94% probability that greater than 96 out of 153 DVDs will work correctly.

Step-by-step explanation:

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)}.

In this problem, we have that:

n = 153, p = 0.62

So

\mu = E(X) = np = 153*0.62 = 94.86

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{153*0.62*0.38} = 6

Probability that greater than 96 out of 153 DVDs will work correctly.

97 or more DVDs, which is 1 subtracted by the pvalue of Z when X = 97. So

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

Z = \frac{97 - 94.86}{6}

Z = 0.36

Z = 0.36 has a pvalue of 0.6406

1 - 0.6406 = 0.3594

0.3594 = 35.94% probability that greater than 96 out of 153 DVDs will work correctly.

3 0
3 years ago
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