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oksano4ka [1.4K]
2 years ago
8

The traffic-control monitor on the freeway shows 200 vehicles per minute passing the camera in 5 minutes. Of those vehicles, on

average, 125 have one passenger, 60 have four or fewer passengers, and 15 have more than four passengers. What is P(vehicles with more than four passengers)?
Mathematics
1 answer:
oksian1 [2.3K]2 years ago
4 0

Answer: maybe 7.5

Step-by-step explanation: I don't really know the question isn't real clear

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Identify the property used in the first four steps to solve the equation 4(x-7)=2x-6
myrzilka [38]

Answer:

x = 11

distributive law

additive law

multiplication property

Step-by-step explanation:

Given in the question an equation

4(x-7) = 2x-6

Distributive law

a(b+c) = ab +ac

4x - 4(7) = 2x - 6

4x -28 = 2x - 6

Use the addition property of equality to reduce the 2x on the right.

4x - 2x - 28 = 2x - 2x -6

4x -2x -28 = -6

Use the addition property of equality to reduce -28 to zero on the left.

4x - 2x- 28 + 28 = -6 + 28

2x = 22

Use the multiplication property of equality to reduce 2x to just x

2x2 = 22/22

x = 22/2

x = 11

4 0
3 years ago
Given that €1 =£0.72 a) how much is €410 b) what is the £ to € exchange rate ?
nekit [7.7K]

Answer:

As per the given statement: €1 = £0.72Find how much is €410 in £.then;€410 =  = £295.2Hence, £295.2 much is €410.to find, the exchange rate of £ to €:€1 = £0.72Divide both sides by 0.72 we get;£1 = €1.38

7 0
3 years ago
What is the slope of the line represented by the equation y = x – 3?
lyudmila [28]
<span>What is the slope of the line represented by the equation?
</span>C. 4/5
5 0
3 years ago
Read 2 more answers
a theater is designed with 15 seats in the first row, 19 in the second, 23 in the third, and so on. if this seating pattern cont
marusya05 [52]

Answer:

<u>131 seats</u> are in the 30th row.

Step-by-step explanation:

The theater is designed with the first row there are 15 seats, in second row 19 seats and in the third row there are 23 seats.

Now, to find the number of seats in the 30th row.

So, we get the common difference(d) from the arithmetic sequence first:

19-15=4.

Thus, d=4.

So, the first tem a(1) = 15.

The number of last row (n) = 30.

Now, to get the number of seat in the 30th row we put formula:

a(n) = a(1) + d(n-1)

a(30)=15+4(30-1)

a(30)=15+4\times 29

a(30)=15+116

a(30)=131

Therefore, 131 seats are in the 30th row.

4 0
3 years ago
Find the point(s) on the surface z^2 = xy 1 which are closest to the point (7, 11, 0)
leonid [27]
Let P=(x,y,z) be an arbitrary point on the surface. The distance between P and the given point (7,11,0) is given by the function

d(x,y,z)=\sqrt{(x-7)^2+(y-11)^2+z^2}

Note that f(x) and f(x)^2 attain their extrema, if they have any, at the same values of x. This allows us to consider the modified distance function,

d^*(x,y,z)=(x-7)^2+(y-11)^2+z^2

So now you're minimizing d^*(x,y,z) subject to the constraint z^2=xy. This is a perfect candidate for applying the method of Lagrange multipliers.

The Lagrangian in this case would be

\mathcal L(x,y,z,\lambda)=d^*(x,y,z)+\lambda(z^2-xy)

which has partial derivatives

\begin{cases}\dfrac{\mathrm d\mathcal L}{\mathrm dx}=2(x-7)-\lambda y\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dy}=2(y-11)-\lambda x\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dz}=2z+2\lambda z\\\\\dfrac{\mathrm d\mathcal L}{\mathrm d\lambda}=z^2-xy\end{cases}

Setting all four equation equal to 0, you find from the third equation that either z=0 or \lambda=-1. In the first case, you arrive at a possible critical point of (0,0,0). In the second, plugging \lambda=-1 into the first two equations gives

\begin{cases}2(x-7)+y=0\\2(y-11)+x=0\end{cases}\implies\begin{cases}2x+y=14\\x+2y=22\end{cases}\implies x=2,y=10

and plugging these into the last equation gives

z^2=20\implies z=\pm\sqrt{20}=\pm2\sqrt5

So you have three potential points to check: (0,0,0), (2,10,2\sqrt5), and (2,10,-2\sqrt5). Evaluating either distance function (I use d^*), you find that

d^*(0,0,0)=170
d^*(2,10,2\sqrt5)=46
d^*(2,10,-2\sqrt5)=46

So the two points on the surface z^2=xy closest to the point (7,11,0) are (2,10,\pm2\sqrt5).
5 0
3 years ago
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