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3 years ago

11

Suppose the number of dropped footballs for a wide receiver, over the course of a season, are normally distributed with a mean o

f 16 and a standard deviation of 2.
What is the z-score for a wide receiver who dropped 13 footballs over the course of a season?

−3

−1.5

1.5

3

1 answer:

inessss [21]3 years ago

8 0

The z-score is the number of standard deviations away from the mean. We first subtract the actual score from the mean, then divide the difference by the standard deviation. In this case, our actual score is 13, so we subtract from the mean of 16, then divide by the SD of 2
(13 - 16) / 2 = -1.5

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Solve for h.<br> A=3h <br> Can anyone help me?

IgorC [24]

Answer:

h = 3/A

Step-by-step explanation:

A = 3h

Divide both sides by 3 to isolate h:

A/3 = 3h/3

Simplify and get

A/3 = h or h = A/3

Hope this helps!

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2 years ago

The population of a certain town was 10,000 in 1990. The rate of change of the population, measured in people per year, is model

Katena32 [7]

The question is incomplete. The complete question is :

The population of a certain town was 10,000 in 1990. The rate of change of a population, measured in hundreds of people per year, is modeled by P prime of t equals two-hundred times e to the 0.02t power, where t is measured in years since 1990. Discuss the meaning of the integral from zero to twenty of P prime of t, d t. Calculate the change in population between 1995 and 2000. Do we have enough information to calculate the population in 2020? If so, what is the population in 2020?

Solution :

According to the question,

The rate of change of population is given as :

$\frac{dP(t)}{dt}=200e^{0.02t}$ in 1990.

Now integrating,

$\int_0^{20}\frac{dP(t)}{dt}dt=\int_0^{20}200e^{0.02t} \ dt$

$=\frac{200}{0.02}\left[e^{0.02(20)}-1\right]$

$=10,000[e^{0.4}-1]$

$=10,000[0.49]$

=4900

$\frac{dP(t)}{dt}=200e^{0.02t}$

$\int1.dP(t)=200e^{0.02t}dt$

$P=\frac{200}{0.02}e^{0.02t}$

$P=10,000e^{0.02t}$

$P=P_0e^{kt}$

This is initial population.

k is change in population.

So in 1995,

$P=P_0e^{kt}$

$=10,000e^{0.02(5)}$

$=11051$

In 2000,

$P=10,000e^{0.02(10)}$

$=12,214$

Therefore, the change in the population between 1995 and 2000 = 1,163.

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3 years ago

Find the z score that corresponds to P30 (the 30th percentile).

Leokris [45]

Answer:

Step-by-step explanation:

From the standard normal distribution tables it is -0.524.

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3 years ago

A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 30 ft/s. Its height

Crank

Answer:

a) h = 0.1: $\bar v = -11\,\frac{ft}{s}$ , h = 0.01: $\bar v = -10.1\,\frac{ft}{s}$ , h = 0.001: $\bar v = -10\,\frac{ft}{s}$ , b) The instantaneous velocity of the ball when $t = 2\,s$ is $-10$ feet per second.

Step-by-step explanation:

a) We know that $y = 30\cdot t -10\cdot t^{2}$ describes the position of the ball, measured in feet, in time, measured in seconds, and the average velocity ( $\bar v$ ), measured in feet per second, can be done by means of the following definition:

$\bar v = \frac{y(2+h)-y(2)}{h}$

Where:

$y(2)$ - Position of the ball evaluated at $t = 2\,s$ , measured in feet.

$y(2+h)$ - Position of the ball evaluated at $t =(2+h)\,s$ , measured in feet.

$h$ - Change interval, measured in seconds.

Now, we obtained different average velocities by means of different change intervals:

$h = 0.1\,s$

$y(2) = 30\cdot (2) - 10\cdot (2)^{2}$

$y (2) = 20\,ft$

$y(2.1) = 30\cdot (2.1)-10\cdot (2.1)^{2}$

$y(2.1) = 18.9\,ft$

$\bar v = \frac{18.9\,ft-20\,ft}{0.1\,s}$

$\bar v = -11\,\frac{ft}{s}$

$h = 0.01\,s$

$y(2) = 30\cdot (2) - 10\cdot (2)^{2}$

$y (2) = 20\,ft$

$y(2.01) = 30\cdot (2.01)-10\cdot (2.01)^{2}$

$y(2.01) = 19.899\,ft$

$\bar v = \frac{19.899\,ft-20\,ft}{0.01\,s}$

$\bar v = -10.1\,\frac{ft}{s}$

$h = 0.001\,s$

$y(2) = 30\cdot (2) - 10\cdot (2)^{2}$

$y (2) = 20\,ft$

$y(2.001) = 30\cdot (2.001)-10\cdot (2.001)^{2}$

$y(2.001) = 19.99\,ft$

$\bar v = \frac{19.99\,ft-20\,ft}{0.001\,s}$

$\bar v = -10\,\frac{ft}{s}$

b) The instantaneous velocity when $t = 2\,s$ can be obtained by using the following limit:

$v(t) = \lim_{h \to 0} \frac{x(t+h)-x(t)}{h}$

$v(t) = \lim_{h \to 0} \frac{30\cdot (t+h)-10\cdot (t+h)^{2}-30\cdot t +10\cdot t^{2}}{h}$

$v(t) = \lim_{h \to 0} \frac{30\cdot t +30\cdot h -10\cdot (t^{2}+2\cdot t\cdot h +h^{2})-30\cdot t +10\cdot t^{2}}{h}$

$v(t) = \lim_{h \to 0} \frac{30\cdot t +30\cdot h-10\cdot t^{2}-20\cdot t \cdot h-10\cdot h^{2}-30\cdot t +10\cdot t^{2}}{h}$

$v(t) = \lim_{h \to 0} \frac{30\cdot h-20\cdot t\cdot h-10\cdot h^{2}}{h}$

$v(t) = \lim_{h \to 0} 30-20\cdot t-10\cdot h$

$v(t) = 30\cdot \lim_{h \to 0} 1 - 20\cdot t \cdot \lim_{h \to 0} 1 - 10\cdot \lim_{h \to 0} h$

$v(t) = 30-20\cdot t$

And we finally evaluate the instantaneous velocity at $t = 2\,s$ :

$v(2) = 30-20\cdot (2)$

$v(2) = -10\,\frac{ft}{s}$

The instantaneous velocity of the ball when $t = 2\,s$ is $-10$ feet per second.

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3 years ago

I need help finding the answers someone plz help

Harlamova29_29 [7]

Happy New Year from MrBillDoesMath!

Answer:

Proof by ASA congruence postulate. See below

Discussion:

Fact 1 : angle A = angle T (given)

Fact 2: The angles on both sides of point X are equal as vertical angles

are equal.

From these facts it follows that angle M = angle H (as all plane triangles have 180 degrees). Also AM = TH (given) so

In the left triangle In the right triangle

(angle M, side AM, angle A) = (angle H, side TH, angle T)

Hence the triangles have two congruent angles, and congruent sides included between the angles, so they are congruent by ASA.

Thank you,

MrB

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3 years ago