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3 years ago

A train of length 180 m approaches a tunnel of length 620 m.

1 answer:

7 0

After coming out of the tunnel it has to travel it's own length to pass through the tunnel completely. So the train of length 180 metres takes 0.889 seconds to pass through a tunnel of length 680 metres completely.

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Ramon has applied to MBA programs at both Harvard and Stanford. He thinks the probability that Harvard will admit him is 0.5, th

ivolga24 [154]

Answer:

(a) The Venn diagram is attached below.

(b) The probability that neither university admits Ramon is 0.20.

Step-by-step explanation:

Let's denote the events as follows:

H = Ramon gets admitted to Harvard

S = Ramon gets admitted to Stanford.

The information provided is:

P (H) = 0.50

P (S) = 0.50

P (H ∩ S) = 0.20

(a)

The Venn diagram is attached below.

(b)

Compute the probability that Ramon gets admitted in neither Harvard nor Stanford as follows:

$P(H^{c}\cup S^{c})=1-P(H\cup S) \\= 1 - [P(H)+P(S)-P(H\cap S)]\\=1-[0.50+0.50-0.20]\\=1-0.80\\=0.20$

Thus, the probability that neither university admits Ramon is 0.20.

(c)

Compute the probability that Ramon gets admitted in Stanford but not in Harvard as follows:

$P(H^{c}\cup S)=P(S)-P(H\cap S) \\= 0.50-0.20\\=0.30$

Thus, the probability that Ramon gets into Stanford but not Harvard 0.30.

7 0

3 years ago

Help please Rewrite the equation −4.9t^2+6.1t+2.2=3 in standard form.

Mnenie [13.5K]

COuld you send a picture

4 0

4 years ago

Quinn mailed four postcards that each cost $0.25, five letters that each cost $0.44, two large envelopes that each cost $0.80, a

PilotLPTM [1.2K]

Answer:

the cost of the small package

Step-by-step explanation:

I hope this is correct. I'm sorry if it's not...

7 0

3 years ago

Read 2 more answers

A tour bus normally leaves for its destination at 5:00 p.m. for a 200 mile trip. This week however, the bus leaves at 5:40 p.m.

iren2701 [21]

Answer:

The usually speed of the bus is 50 miles/h.

Step-by-step explanation:

Let the usual speed of the bus be x mile/hour.

We know that

$speed=\frac{distance}{time}$

$\Rightarrow time = \frac{distance}{speed}$

The bus travels 200 miles.

To reach its destination it takes time $=\frac{200}{x}$ h

This week however the bus leaves at 5:40.

The bus late 40 minutes $=\frac{40}{60} h$ $= \frac{2}{3}h$

Now the speed of the bus is = (x+10) miles/h

The new time to reach the destination is $=\frac{200}{x+10}$ h

According to the problem,

$\frac{200}{x}-\frac{200}{x+10}=\frac{2}{3}$

$\Rightarrow 200[\frac{x+10-x}{x(x+10)}]=\frac{2}{3}$

$\Rightarrow 200[\frac{10}{x^2+10x}]=\frac{2}{3}$

⇒2(x²+10x)=200×10×3

⇒x²+10x = 3000

⇒x²+10x -3000=0

⇒x²+60x-50x-3000=0

⇒x(x+60)-50(x+60)=0

⇒(x+60)(x-50)=0

⇒x= -60,50

∴x=50 [since speed does not negative]

The usually speed of the bus is 50 miles/h.

3 0

4 years ago

X=8x=8 and y=3y=3? 8x-9y 8x−9y

cricket20 [7]

Answer:

x h

h d

Step-by-step explanation:

3 0

3 years ago