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1 year ago

The graph shows the linear parent function.Y5-55----Which statement best describes the function?

Mathematics

1 answer:

Vikki [24]1 year ago

6 0

From the picture, we have the graph of the linear parent function f(x) = x.

We have the following statements as descriptions of the function:

A. The function is negative when x < 0.

From the graph, we see that the function takes negative values for x < 0. This statement is true.

B. The function is negative when x < 0 and also when x > 0.

The first part of this statement is true, but the second is not because that we see that the function takes positives values for x > 0. So this statement is false.

C. The function is never negative.

If we see the graph, the function is negative when x < 0. So this statement is false.

D. The function is negative when x > 0.

Again, seeing the graph we note that the function takes positive values for x > 0. So this statement is false.

So the only statement that it is true, is option A.

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The distribution of passenger vehicle speeds traveling on a certain freeway in California is nearly normal with a mean of 71.5 m

Zinaida [17]

Answer:

$G = 0.007523$

The number of cars the highway patrol officer would watch before a car that is seen is $E(X) = 1.6027$

The standard deviation is $s = 0.9829$

Step-by-step explanation:

From the question we are told that

The mean is $\mu = 71.5 \ miles/hour$

The standard deviation is $\sigma = 4.75 \ miles/hour$

The speed limit is $x = 70 \ miles /hour$

Generally the probability of getting a car that is moving with speed greater than the speed limit is mathematically represented as

$p =P(X > x ) = P(X > 70) = P(\frac{X - \mu }{\sigma } > \frac{70 - 71.5 }{4.75})$

=> $p= P(X > 70) = P(\frac{X - \mu }{\sigma } > \frac{70 - 71.5 }{4.75})$

=> $p= P(X > 70) = P(\frac{X - \mu }{\sigma } > -0.31579 )$

Here

$\frac{X - \mu }{\sigma } = Z(The \ standardized \ value \ of X )$

=> $p= P(X > 70) = P(Z > -0.31579 )$

From the z-table

$p = P(Z > -0.31579 ) = 0.62392$

$p = P(X > 70) = 0.62392$

Generally the probability of getting a car that is not moving with speed greater than the speed limit is mathematically represented as

$q = 1 - p$

=> $q = 1 - 0.62392$

=> $q = 0.37608$

Generally the probability of getting 5 cars that are not speeding is mathematically represented as

$G = q^5$

=> $G = (0.37608)^5$

=> $G = 0.007523$

Generally the number of cars that the highway patrol officer is expected to watch until the first car that is speeding is gotten is mathematically represented as

$E(X) = \frac{1}{p}$