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LUCKY_DIMON [66]

3 years ago

7

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

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

1 answer:

snow_lady [41]3 years ago

8 0

Answer:

<em>The z score for a wide receiver who dropped 13 footballs over the course of a season</em>

<em> Z = - 1.5</em>

Step-by-step explanation:

<u><em>Explanation</em></u>:-

Given Population mean ' μ'= 16

Standard deviation of Population 'σ' = 2

Let 'x' be the Random Variable of Normal distribution

Given x = 13 foot balls

<em>The z score for a wide receiver who dropped 13 footballs over the course of a season</em>

<em> </em> $Z = \frac{x-mean}{S.D}$ <em></em>

<em> </em> $Z = \frac{13 -16}{2} = \frac{-3}{2} = -1.5$ <em></em>

<u><em>Final answer</em></u><em>:-</em>

<em>The z score for a wide receiver who dropped 13 footballs over the course of a season</em>

<em> Z = - 1.5</em>

<em></em>

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What are the coordinates of the point on the directed line segment from (-10, -2) (10,−10) that partitions the segment into a ra

sveta [45]

Answer:

The answer to your question is (-5/2, -5)

Step-by-step explanation:

Data

A (-10, -2)

B (10, -10)

ratio 3/5

Formula

X = (x₁ + rx₂)/(1 + r)

Y = (y₁ + ry₂)/(1 + r)

Process

1.- Calculate X

x₁ = -10 x₂ = 10

X = (-10 + 3/5(10))/ (1 + 3/5)

X = (-10 + 30/5) / 8/5

X = -4/8/5

X = -5/2

2.- Calculate Y

y₁ = -2 y₂ = -10

Y = (-2 + (3/5)(-10))/(1 + 3/5)

Y = (-2 - 30/5) / 8/5

Y = -8/8/5

Y = -5

3.- The coordinates of the point are (-5/2, -5)

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4 years ago

Consider an object moving along a line with the given velocity v. Assume t is time measured in seconds and velocities have units

Lady_Fox [76]

Answer:

a. the motion is positive in the time intervals: $[0,2)U(6,\infty)$

The motion is negative in the time interval: (2,6)

b. S=7 m

c. distance=71m

Step-by-step explanation:

a. In order to solve part a. of this problem, we must start by determining when the velocity will be positive and when it will be negative. We can do so by setting the velocity equation equal to zero and then testing it for the possible intervals:

$3t^{2}-24t+36=0$

so let's solve this for t:

$3(t^{2}-8t+12)=0$

$t^{2}+8t+12=0$

and now we factor it again:

(t-6)(t-2)=0

so we get the following answers:

t=6 and t=2

so now we can build our possible intervals:

$[0,2) (2,6) (6,\infty)$

and now we test each of the intervals on the given velocity equation, we do this by finding test values we can use to see how the velocity behaves in the whole interval:

[0,2) test value t=1

so:

$v(1)=3(1)^{2}-24(1)+36$

v(1)=15 m/s

we got a positive value so the object moves in the positive direction.

(2,6) test value t=3

so:

$v(1)=3(3)^{2}-24(3)+36$

v(3)=-9 m/s

we got a negative value so the object moves in the negative direction.

$(6,\infty)$ test value t=7

so:

$v(1)=3(7)^{2}-24(7)+36$

v(1)=15 m/s

we got a positive value so the object moves in the positive direction.

the motion is positive in the time intervals: $[0,2)U(6,\infty)$

The motion is negative in the time interval: (2,6)

b) in order to solve part b, we need to take the integral of the velocity function in the given interval, so we get:

$s(t)=\int\limits^7_0 {(3t^{2}-24t+36)} \, dt$

so we get:

$s(t)=[\frac{3t^{3}}{3}-\frac{24t^{2}}{2}+36]^{7}_{0}$

which simplifies to:

$s(t)=[t^{3}-12t^{2}+36t]^{7}_{0}$

so now we evaluate the integral:

$s=7^{3}-12(7)^{2}+36(7)-(0^{3}-12(0)^{2}+36(0))$

s=7 m

for part c, we need to evaluate the integral for each of the given intervals and add their magnitudes:

[0,2)

s(t)=\int\limits^2_0 {(3t^{2}-24t+36)} \, dt

so we get:

$s(t)=[\frac{3t^{3}}{3}-\frac{24t^{2}}{2}+36]^{2}_{0}$

which simplifies to:

$s(t)=[t^{3}-12t^{2}+36t]^{2}_{0}$

so now we evaluate the integral:

$s=2^{3}-12(2)^{2}+36(2)-(0^{3}-12(0)^{2}+36(0))$

s=32 m

(2,6)

s(t)=\int\limits^6_2 {(3t^{2}-24t+36)} \, dt

so we get:

$s(t)=[\frac{3t^{3}}{3}-\frac{24t^{2}}{2}+36]^{6}_{2}$

which simplifies to:

$s(t)=[t^{3}-12t^{2}+36t]^{6}_{2}$

so now we evaluate the integral:

$s=6^{3}-12(6)^{2}+36(6)-(2^{3}-12(2)^{2}+36(2))$

s=-32 m

(6,7)

s(t)=\int\limits^7_6 {(3t^{2}-24t+36)} \, dt

so we get:

$s(t)=[\frac{3t^{3}}{3}-\frac{24t^{2}}{2}+36]^{7}_{6}$

which simplifies to:

$s(t)=[t^{3}-12t^{2}+36t]^{7}_{6}$

so now we evaluate the integral:

$s=7^{3}-12(7)^{2}+36(7)-(6^{3}-12(6)^{2}+36(6))$

s=7 m

and we now add all the magnitudes:

Distance=32+32+7=71m

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3 years ago

Is each line parallel, perpendicular, or neither parallel nor perpendicular to a line whose slope is −3/4?

djverab [1.8K]

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Line n is perpendicular because it has the opposite reciprocal.
Line p is neither because it does not have the opposite reciprocal nor same slope.
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