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3 years ago

4x-1=3(x-1) Please help!

Mathematics

2 answers:

Vsevolod [243]3 years ago

7 0

X = -2

Step-by-step explanation:

Solve for x by simplifying both sides of the equation, then isolating the variable.

Sveta_85 [38]3 years ago

5 0

Answer:

x=-2

Step-by-step explanation:

4x-1=3x-3

x=-2

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Find the unit price for each option shown below. Round to the nearest cent when necessary. Option I: 10 candy bars for $6.75 Opt

natita [175]

Option I: 10 candy bars for $6.75

6.75/10 = 0.68 

Option II: 12 candy bars for $7.25 

7.25 / 12 = 0.60

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2 years ago

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The winner of a basketball contest is the individual who scores the highest percentage of free throws. The top three finishers a

Kamila [148]

Answer:

Step-by-step explanation:

Convert all numbers to decimals

85% --> 0.85
0.88 --> 0.88

21/25 --> 0.84

Since these are the top three, the rest of the competitors are less than 0.84.

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3 years ago

The life of a red bulb used in a traffic signal can be modeled using an exponential distribution with an average life of 24 mont

BartSMP [9]

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

$\bf f(x)=\frac{1}{24}e^{-x/24}\;(x\geq 0)$

and its cumulative distribution function (CDF) is

$\bf P(X\leq t)=\int_{0}^{t} f(x)dx=1-e^{-t/24}$

• 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

$\bf P(X\leq 12)=1-e^{-12/24}=1-e^{-0.5}=0.39347$

• 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

$\bf P(A | B) = P(B)P(B) = (P(B))^2$

We have

$\bf P(B)=P(X>18)=1-P(X\leq 18)=1-(1-e^{-18/24})=e^{-3/4}=0.47237$

hence,

$\bf P(A | B)=(P(B))^2=(0.47237)^2=0.22313$

• 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

$\bf \binom{6}{k}(0.39347)^k(1-0.39347)^{6-k}$

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.

This means that,

Probability that at least one of them needs replacement at the first inspection =

$\bf 1-\binom{6}{0}(0.39347)^0(1-0.39347)^{6}=1-(0.60653)^6=0.95021$