I would say -5.5., but I might be wrong.
a car travels 157 miles on 10 gallons of gasoline mentally calculate the number of miles that the car travels on one gallon of g
2 answers:
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4sqrt(3)
See the diagram below.
See the attached figure.
========================
AB = 10 , FD = 3
∵ D is the midpoint of AB, and F is the mid point of CB
∴ FD // AC , FD = 0.5 AC
∵ Δ ABC is a right triangle at C
∴ FD ⊥ BC
∴ BD = 0.5 AB = 5
∴ in Δ FDB ⇒⇒ BF² = BD² - FD² = 5² - 3² = 16
∴ BF = √16 = 4
∵ F is the mid point of CB
∴ CF = BF = 4 , and CB = 2 BF = 2*4 = 8
∵ D is the midpoint of AB, and E is the mid point of AC
∴ DE // CB , and DE = 0.5 CB = 0.5 * 8 = 4
∴ T<span>he length of line ED is 4
</span>
========================
AB = 10 , FD = 3
∵ D is the midpoint of AB, and F is the mid point of CB
∴ FD // AC , FD = 0.5 AC
∵ Δ ABC is a right triangle at C
∴ FD ⊥ BC
∴ BD = 0.5 AB = 5
∴ in Δ FDB ⇒⇒ BF² = BD² - FD² = 5² - 3² = 16
∴ BF = √16 = 4
∵ F is the mid point of CB
∴ CF = BF = 4 , and CB = 2 BF = 2*4 = 8
∵ D is the midpoint of AB, and E is the mid point of AC
∴ DE // CB , and DE = 0.5 CB = 0.5 * 8 = 4
∴ T<span>he length of line ED is 4
</span>
Answer:
See steps below
Step-by-step explanation:
Let X be the random variable that measures the lifespan of a bulb.
If the random variable X is exponentially distributed and X has an average value of 24 month, then its probability density function is
and its cumulative distribution function (CDF) is
• What is probability that the red bulb will need to be replaced at the first inspection?
The probability that the bulb fails the first year is
• If the bulb is in good condition at the end of 18 months, what is the probability that the bulb will be in good condition at the end of 24 months?
Let A and B be the events,
A = “The bulb will last at least 24 months”
B = “The bulb will last at least 18 months”
We want to find P(A | B).
By definition P(A | B) = P(A∩B)P(B)
but B⊂A, so A∩B = B and
We have
hence,
• If the signal has six red bulbs, what is the probability that at least one of them needs replacement at the first inspection? Assume distribution of lifetime of each bulb is independent
If the distribution of lifetime of each bulb is independent, then we have here a binomial distribution of six trials with probability of “success” (one bulb needs replacement at the first inspection) p = 0.39347
Now the probability that exactly k bulbs need replacement is
<em>Probability that at least one of them needs replacement at the first inspection = 1- probability that none of them needs replacement at the first inspection. </em>
This means that,
<em>Probability that at least one of them needs replacement at the first inspection = </em>