Police investigating traffic accidents often estimate the speed of a vehicle by measuring

Find all relative extrema of the function. use the second derivative test where applicable. (if an answer does not exist, enter

Alinara [238K]

The given quadratic describes a parabola that opens upward. Its one absolute extreme is a minimum that is found at x = -3/2. The value of the function there is
(-3/2 +3)(-3/2) -1 = -13/4

The one relative extreme is a minimum at (-1.5, -3.25).

_____
For the parabola described by ax² +bx +c, the vertex (extreme) is found where
x = -b/(2a)
Here, that is x=-3/(2·1) = -3/2.

7 0

3 years ago

a survey amony freshman at a certain university revealed that the number of hours spent studying the week before final exams was

Marat540 [252]

Answer:

Probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

Step-by-step explanation:

We are given that the number of hours spent studying the week before final exams was normally distributed with mean 25 and standard deviation 15.

A sample of 36 students was selected.

Let $\bar X$ = sample average time spent studying

The z-score probability distribution for sample mean is given by;

Z = $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ ~ N(0,1)

where, $\mu$ = population mean hours spent studying = 25 hours

$\sigma$ = standard deviation = 15 hours

n = sample of students = 36

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

Now, Probability that the average time spent studying for the sample was between 29 and 30 hours studying is given by = P(29 hours < $\bar X$ < 30 hours)

P(29 hours < $\bar X$ < 30 hours) = P( $\bar X$ < 30 hours) - P( $\bar X$ $\leq$ 29 hours)

P( $\bar X$ < 30 hours) = P( $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ < $\frac{ 30-25}{\frac{15}{\sqrt{36} } }} }$ ) = P(Z < 2) = 0.97725

P( $\bar X$ $\leq$ 29 hours) = P( $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ $\leq$ $\frac{ 29-25}{\frac{15}{\sqrt{36} } }} }$ ) = P(Z $\leq$ 1.60) = 0.94520

So, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2 and x = 1.60 in the z table which has an area of 0.97725 and 0.94520 respectively.

Therefore, P(29 hours < $\bar X$ < 30 hours) = 0.97725 - 0.94520 = 0.0321

Hence, the probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

7 0

3 years ago

Y-(5-(y-x)) x = 4 y = 6 I need the steps please, this is a little bit confusing..

KATRIN_1 [288]

Answer:

3

Step-by-step explanation:

substitute x = 4 , y = 6 into the expression

y - (5 - (x - y))

= 6 - (5 - (6 - 4))

= 6 - (5 - 2))

= 6 - 3

= 3

3 0

2 years ago

Simplify. Assume all variables are non-zero. HELP ASAP!

Alexandra [31]

Answer:

D

Step-by-step explanation:

((p^4*q)/p^8)^2.

p^4/p^8=p^(4-8)=p^-4=1/p^4

(q/p^4)^2=(q^2/p^8)

4 0

3 years ago

Power Up Test Practice

omeli [17]

Answer: 26

Step-by-step explanation: Add all the booths together and you will get 300. You will get 300 because the t-shirts booth is $26. Now divide 300 by how many booths there are, which is six. And your mean will be 50.

6 0

2 years ago