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3 years ago

(-8y^2+6y-1)+(-8y^2+8y+1)-(-3y^2+9y-9)

Mathematics

2 answers:

taurus [48]3 years ago

3 0

The answer of the given equation is $-13y^{2} -11y+7$

Step-by-step explanation:

The given equation can be written as,

$(-8y^{2}+6y-1) +(-8y^{2}+8y+1)-(-3y^{2}+9y-9)$

= $-8y^{2} +6y-1-8y^{2}-8y -1+3y^{2}-9y+9$

[Taking same powered variables together we get,]

= $(-8y^{2}-8y^{2}+3y^{2}) +(6y-8y-9y)-1-1+9$

= $-13y^{2} -11y+7$

Here the answer.

iogann1982 [59]3 years ago

3 0

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A television station plans to ask a random sample of 400 city residents if they can name the news anchor on the evening news at

hodyreva [135]

Answer:

There is an 8.38% probability that the anchor will be fired.

Step-by-step explanation:

For each resident, there is only two possible outcomes. Either they can name the news anchor on the evening news at their station, or they cannot.

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However, we are working with samples that are considerably big. So i am going to aaproximate this binomial distribution to the normal.

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:

400 city residents are going to be asked. So $n = 400$ .

Suppose that in fact 12% of city residents could name the anchor if asked. This means that $p = 0.12$ .

So,

$\mu = E(X) = 400*0.12 = 48$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{400*0.12*0.88} = 6.5$ .

They plan to fire the news anchor if fewer than 10% of the residents in the sample can do so.

What is the approximate probability that the anchor will be fired?

10% of the residents is 0.10*400 = 40 residents.

Fewer than 10% is 39 residents. So the probability that the anchor will be fired is the pvalue of Z when $X = 39$ .

So:

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

$Z = \frac{39 - 48}{6.5}$

$Z = -1.38$

$Z = -1.38$ has a pvalue of 0.0838. This means that there is an 8.38% probability that the anchor will be fired.

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3 years ago