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3 years ago

Please help me solve this, and show your work

1 answer:

6 0

$6r+5=11r\ \ \ \ |\text{subtract 6r from both sides}\\\\5=5r\ \ \ \ |\text{divide both sides by 5}\\\\\boxed{r=1}\\-------------------------\\\dfrac{2x-4}{2}=6x+3\\\\\dfrac{2x}{2}-\dfrac{4}{2}=6x+3\\\\x-2=6x+3\ \ \ \ \ |\text{add 2 to both sides}\\\\x=6x+5\ \ \ \ |\text{subtract 6x from both sides}\\\\-5x=5\ \ \ \ \ |\text{divide both sides by -5}\\\\\boxed{x=-1}$

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The probability that your call to a service line is answered in less than 30 seconds is 0.85. Assume that your calls are indepen

aev [14]

Answer:

a) 0.1720

b) 0.8298

c) 19

Step-by-step explanation:

For each call, there are only two possible outcomes. Either they are answered in less than 30 seconds. Or they are not. The probabilities for each call are independent. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

$p = 0.85$

(a) If you call 12 times, what is the probability that exactly 9 of your calls are answered within 30 seconds? Round your answer to four decimal places (e.g. 98.7654).

This is $P(X = 9)$ when $n = 12$ .

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 9) = C_{12,9}.(0.85)^{9}.(0.15)^{3} = 0.1720$

(b) If you call 20 times, what is the probability that at least 16 calls are answered in less than 30 seconds? Round your answer to four decimal places (e.g. 98.7654).

This is $P(X \geq 16)$ when $n = 20$

$P(X \geq 16) = P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 16) = C_{20,16}.(0.85)^{16}.(0.15)^{4} = 0.1821$

$P(X = 17) = C_{20,17}.(0.85)^{17}.(0.15)^{3} = 0.2428$

$P(X = 18) = C_{20,18}.(0.85)^{18}.(0.15)^{2} = 0.2293$

$P(X = 19) = C_{20,19}.(0.85)^{19}.(0.15)^{1} = 0.1368$

$P(X = 20) = C_{20,20}.(0.85)^{20}.(0.15)^{0} = 0.0388$

$P(X \geq 16) = P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.1821 + 0.2428 + 0.2293 + 0.1368 + 0.0388 = 0.8298$

(c) If you call 22 times, what is the mean number of calls that are answered in less than 30 seconds? Round your answer to the nearest integer.

The expected value of the binomial distribution is:

$E(X) = np$

In this question, we have $n = 22$

$E(X) = 22*0.85 = 18.7$

The nearest integer to 18.7 is 19.

7 0

3 years ago

A college requires applicants to have an ACT score in the top 12% of all test scores. The ACT scores are normally distributed, w

DochEvi [55]

Answer:

a) The lowest test score that a student could get and still meet the colleges requirement is 27.0225.

b) 156 would be expected to have a test score that would meet the colleges requirement

c) The lowest score that would meet the colleges requirement would be decreased to 26.388.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 21.5, \sigma = 4.7$