If the probability of observing at least one car on a highway during any 20-minute time interval is 609/625, then the probability of observing at least one car during any 5-minute time interval is 609/2500
Given The probability of observing at least one car on a highway during any 20 minute time interval is 609/625.
We have to find the probability of observing at least one car during any 5 minute time interval.
Probability is the likeliness of happening an event among all the events possible. It is calculated as number/ total number. Its value lies between 0 and 1.
Probability during 20 minutes interval=609/625
Probability during 1 minute interval=609/625*20
=609/12500
Probability during 5 minute interval=(609/12500)*5
=609/2500
Hence the probability of observing at least one car during any 5 minute time interval is 609/2500.
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Answer:
32+d
Step-by-step explanation:
because it's asking for 32 added on to whatever d is
example
if d=5 it would be asking what's 32 +5
Answer:
75.391
Step-by-step explanation:
193x100/256=75.391
Step-by-step explanation:
step 1. let's call the amount of money A, the initial amount P, the yearly rate r, the number of compounds per year n.
step 2. A = P(1 + r/n)^(nt)
step 3. A = 1600(1 + .03/12)^((12)(5)
step 4. A = 1600(1.0025)^(60)
step 5. A = $1858.59
Answer:
Equation → 5y = 3y + 6
Value of ST = 15
Step-by-step explanation:
From the picture attached,
In right triangles ΔVST and ΔVUT,
Acute angles ∠SVT ≅ ∠UVT [Given]
TV ≅ TV [By reflexive property of congruence]
ΔVST ≅ ΔVUT [Hypotenuse angle congruence of right triangles]
Therefore, corresponding parts of the congruent triangles are congruent.
Therefore, ST ≅ TU
5y = 3y + 6
5y - 3y = 6
2y = 6
y = 3
Therefore, ST = 5y
ST = 5(3)
= 15