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3 years ago

Solve the equation. 6 = 2(x + 8) - 5x A. 2/3 B. 3 1/3 C. - 2/3 D. -3 1/3

2 answers:

7 0

$6 = 2(x + 8) - 5x \\ \\ 6 = 2x + 16 - 5x \\ \\ 6 = -3x + 16 \\ \\ 6 - 16 = -3x \\ \\ -10 -3x \\ \\ \frac{-10}{-3} = x \\ \\ \frac{10}{3} = x \\ \\ x = \frac{10}{3} \\ \\ Answer: \fbox {x = 10/3} \ or \ \fbox {x = 3 1/3}$

Zinaida [17]3 years ago

4 0

6 = 2(x+8) - 5x
⇒ 2x+ 2*8 -5x= 6
⇒ (2x -5x)+ 16= 6
⇒ -3x = 6 -16
⇒ -3x = -10
⇒ x= -10/(-3)
⇒ x= 10/3
⇒ x= 3 1/3

The correct answer is B. 3 1/3~

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Write a quadratic equation in factored form to model the given problem. Be sure to write the entire equation.

Veronika [31]

Equation :
[x(x+4)]/2=6 (as (b*h)/2=A)
x*x + 4*x -12 = 0

$x = -b+- \sqrt{ b^{2}-4ac }/2a$

so, x = 2 units(as -6 is negative)

3 0

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

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

<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>

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

5 0

3 years ago

Help me please, I’m stuck!

Veseljchak [2.6K]

Answer:

5. (x, y) ⇒ (x +5, y +1)

6. (x, y) ⇒ (x +4, y -1)

Step-by-step explanation:

A translation rule will generally have the form ...

(x, y) ⇒ (x +h, y +k)

where (h, k) is the horizontal and vertical distance to the right and up that the figure is being translated.d

5) The image is 5 units right and 1 unit up from the original, so in the above formula, (h, k) = (5, 1) and the translation rule is ...

(x, y) ⇒ (x +5, y +1)

6) The image is 4 units right and 1 unit down from the original, so (h, k) = (4, -1) and the transformation rule is ...

(x, y) ⇒ (x +4, y -1)

4 0

3 years ago

Calls to a customer service center last on average 2.8 minutes with a standard deviation of 1.4 minutes. An operator in the call

Natali5045456 [20]

Answer:

The expected total amount of time the operator will spend on the calls each day is of 210 minutes.

Step-by-step explanation:

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Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

n-values of normal variable:

Suppose we have n values from a normally distributed variable. The mean of the sum of all the instances is $M = n\mu$ and the standard deviation is $s = \sigma\sqrt{n}$