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VashaNatasha [74]
2 years ago
6

The two-way table below gives the thousands of commuters in Massachusetts in 2015 by transportation method and one-way length of

commute.
Less than
15
1515 minutes
15
1515-
29
2929 minutes
30
3030-
44
4444 minutes
45
4545-
59
5959 minutes
60
6060 or more minutes Total
Private vehicle
636
636636
908
908908
590
590590
257
257257
256
256256
2647
26472647
Public Transportation
9
99
54
5454
96
9696
62
6262
108
108108
329
329329
Other
115
115115
70
7070
23
2323
7
77
7
77
222
222222
Total
760
760760
1032
10321032
709
709709
326
326326
371
371371
3198
31983198
Given that the commuter used public transportation, find the probability that the commuter had a commute of
60
6060 or more minutes.
P
(
60
+
minutes
∣
public transportation
)
=
P(60+ minutes ∣ public transportation)=
Physics
2 answers:
Ugo [173]2 years ago
8 0

Answer:

36

Explanation:

Step review

forsale [732]2 years ago
7 0

Answer:

wow uh what? hdfgdfgf

Explanation:?

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The Electromagnetic spectrum.

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3 years ago
Una furgoneta circula por una carretera a 55km/h. Diez km atrás , un coche circula en el mismo sentido a 85km/h ¿ En cuanto tiem
statuscvo [17]

Answer:

t = 0.33h = 1200s

x = 18.33 km

Explanation:

If the origin of coordinates is at the second car, you can write the following equations for both cars:

car 1:

x=x_o+v_1t    (1)

xo = 10 km

v1 = 55km/h

car 2:

x'=v_2t    (2)

v2 = 85km/h

For a specific value of time t the positions of both cars are equal, that is, x=x'. You equal equations (1) and (2) and solve for t:

x=x'\\\\x_o+v_1t=v_2t\\\\(v_2-v_1)t=x_o\\\\t=\frac{x_o}{v_2-v_1}

t=\frac{10km}{85km/h-55km/h}=0.33h*\frac{3600s}{1h}=1200s

The position in which both cars coincides is:

x=(55km/h)(0.33h)=18.33km

6 0
3 years ago
A jet liner must reach a speed of 82 m/s for takeoff. If the
SIZIF [17.4K]

Answer:

The acceleration that the jet liner that must have is 2.241 meters per square second.

Explanation:

Let suppose that the jet liner accelerates uniformly. From statement we know the initial (v_{o}) and final speeds (v_{f}), measured in meters per second, of the aircraft and likewise the runway length (d), measured in meters. The following kinematic equation is used to calculate the minimum acceleration needed (a), measured in meters per square second:

a = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot d}

If we know that v_{o} = 0\,\frac{m}{s}, v_{f} = 82\,\frac{m}{s} and d = 1500\,m, then the acceleration that the jet must have is:

a = \frac{\left(82\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot (1500\,m)}

a = 2.241\,\frac{m}{s^{2}}

The acceleration that the jet liner that must have is 2.241 meters per square second.

3 0
2 years ago
A straight, cylindrical wire lying along the x axis has a length L and a diameter d . It is made of a material described by Ohm'
Zepler [3.9K]

The magnitude and direction of the electric field in the wire are mathematically given as

L &=[(v / L) v / m] \hat{i}

<h3>What is the magnitude and direction of the electric field in the wire?</h3>

Generally, the equation for is  mathematically given as

A cylindrical wire that is straight and parallel to the x-axis has the following dimensions: length L, diameter d, resistivity p, diameter d, potential v, and z length. combining elements from both sides  

E d x=\int d v.

\begin{aligned}&-E \int_0^L d x=\int_v^0 d v \\\therefore E \cdot L &=v \\L &=[(v / L) v / m] \hat{i}\end{aligned}

In conclusion, the magnitude and direction of the electric field in the wire are given as

L &=[(v / L) v / m]

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Brainliest if correct Question 1 of 10
Art [367]

Answer:

B

Explanation:

hope this helps you

4 0
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