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Elina [12.6K]
3 years ago
7

Identifying a Valid Sample Natasha wants to find out if the neighborhood supports lowering the speed limit on the street in fron

t of her school. Which is a valid sample for the survey? students in Natasha’s social studies class parents who live in a town 15 miles away from Natasha’s school the first 5 drivers who drive past Natasha’s school one morning thirty residentsIdentifying a Valid Sample Natasha wants to find out if the neighborhood supports lowering the speed limit on the street in front of her school. Which is a valid sample for the survey?
A. students in Natasha’s social studies class parents who live in a town 15 miles
B.away from Natasha’s school the first 5 drivers who drive past Natasha’s C.school one morning thirty residents who live within a 2-miles radius of D.Natasha’s school who live within a 2-miles radius of Natasha’s school
Mathematics
2 answers:
nekit [7.7K]3 years ago
7 0

D - Thirty Residents Who Live Within a 2-Miles Radius Of Natasha’s School

blsea [12.9K]3 years ago
3 0

Answer:

thirty residents who live within a 2-miles radius of Natasha’s school

Step-by-step explanation:

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The answer is $1200.
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Find the first four terms of the sequence an = 4n-6
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Answer:

The first four terms of the sequence are-6, -2, 2 and 6

Step-by-step explanation:

The given sequence is a_n=4n-6

In order to find the first four term of the sequence we put n =0, 1, 2 and 3.

For n =0

a_0=4(0)-6=-6

For n =1

a_1=4(1)-6=-2

For n =2

a_2=4(2)-6=2

For n =3

a_3=4(3)-6=6

Therefore, the first four terms of the sequence are

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5 0
3 years ago
A road perpendicular to a highway leads to a farmhouse located d miles away. An automobile traveling on this highway passes thro
pshichka [43]

Answer:

\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+900}}

Step-by-step explanation:

A road is perpendicular to a highway leading to a farmhouse d miles away.

An automobile passes through the point of intersection with a constant speed \frac{dx}{dt} = r mph

Let x be the distance of automobile from the point of intersection and distance between the automobile and farmhouse is 'h' miles.

Then by Pythagoras theorem,

h² = d² + x²

By taking derivative on both the sides of the equation,

(2h)\frac{dh}{dt}=(2x)\frac{dx}{dt}

(h)\frac{dh}{dt}=(x)\frac{dx}{dt}

(h)\frac{dh}{dt}=rx

\frac{dh}{dt}=\frac{rx}{h}

When automobile is 30 miles past the intersection,

For x = 30

\frac{dh}{dt}=\frac{30r}{h}

Since h=\sqrt{d^{2}+(30)^{2}}

Therefore,

\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+(30)^{2}}}

\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+900}}

3 0
3 years ago
Two cars leave towns 360 kilometers apart at the same time and travel toward each other. One car's rate is 12 kilometers per hou
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Let x = speed of the first car

the second car is 12 km slower so the seconds car speed is x-12

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2nd car = 2(x-12)
 
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2x+2(x-12) = 360

2x + 2x-24 = 360

4x-24 = 360

4x = 384
x = 384/4
x = 96

1st car drove at 96 km per hour

slower car drove at 96-12 = 84 km per hour



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