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3 years ago

Number three please I need you’re guys help

Mathematics

2 answers:

Evgesh-ka [11]3 years ago

5 0

Answer:

F. $1.50 per mile

Step-by-step explanation:

In order to solve this, you can first change it to slope-intercept form to find the slope, or rate of change, easier:

y = 3.75 + 1.5(x - 1) (given)

y = 3.75 + 1.5x - 1.5 (multiply 1.5 by each term in the parenthesis)

y = 2.25 + 1.5x (subtract 1.5 from 2.25)

y = 1.5x + 2.25 (order the terms by exponent value)

In slope-intercept form, the m (slope) of this equation is 1.5, so the rate of change is $1.50 per mile (with an initial fee of $2.25).

leva [86]3 years ago

4 0

Answer:

F

Step-by-step explanation:

Simplify then graph it, find the x-intercept.

y = 3.75 + 1.5 (x - 1)

y = 1.5x + 2.25

When graphed, the x-intercept is 1.5

So it is, $1.50 per mile

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3 years ago

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nikdorinn [45]

Answer:

0.3821 = 38.21% probability that no less than 95 out of 152 registered voters will vote in the presidential election.

Step-by-step explanation:

Binomial probability distribution

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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 distributions 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 $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

Assume the probability that a given registered voter will vote in the presidential election is 61%.

This means that $p = 0.61$

152 registed voters:

This means that $n = 152$

Mean and Standard deviation:

$\mu = E(X) = 152*0.61 = 92.72$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{152*0.61*0.39} = 6.01$

Probability that no less than 95 out of 152 registered voters will vote in the presidential election.

This is, using continuity correction, $P(X \geq 95 - 0.5) = P(X \geq 94.5)$ , which is 1 subtracted by the pvalue of Z when X = 94.5. So

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

$Z = \frac{94.5 - 92.72}{6.01}$

$Z = 0.3$

$Z = 0.3$ has a pvalue of 0.6179

1 - 0.6179 = 0.3821

0.3821 = 38.21% probability that no less than 95 out of 152 registered voters will vote in the presidential election.

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