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melamori03 [73]

3 years ago

5

Kahlil is recording a beat for a song that he is working on he wants the length of the beat to be more than 17 seconds long. His

friend tells him the beat needs to be 9 seconds longer than that to match the lyrics he has written.
write an inequality to represent the beat's length. Give three possible beat lengths that satisfy the inequality.

a. T < 17 c. T > 8
8, 6, 5 27, 35, 33

b. T > 26 d. T < 26
27, 35, 33 8, 6, 5

1 answer:

Svet_ta [14]3 years ago

5 0

In the problem, the length of the beat is take by variable T.

According to problem,

Kahlil wants, "the length of the beat to be more than 17 seconds long".

We took length of beat by variable T.

Therefore, the length of the beat to be more than 17 seconds long could be rephrase as

" T is more than 17".

And in inequality form we can write it as

T>17.

Now, Kahlil's friend told that "beat needs to be 9 seconds longer than that".

So, 17 longer than 9 is 17+9=26.

But as we can see that the beat should be longer than 17 but less than 26 (because 26 is the maximum length in seconds suggested by friend).

So, the final beat should be less than 26.

And we would setup an inequality as

T < 26.

Therefore, Correct option is option D) T < 26 ; 8, 6, 5.

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Step-by-step explanation:

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3 years ago

Read 2 more answers

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.9 minutes and a standard deviation of 2.9

Eva8 [605]

Answer:

a) 0.2981 = 29.81% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

b) 0.999 = 99.9% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes

c) 0.2971 = 29.71% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

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.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 8.9 minutes and a standard deviation of 2.9 minutes.

This means that $\mu = 8.9, \sigma = 2.9$

Sample of 37:

This means that $n = 37, s = \frac{2.9}{\sqrt{37}}$

(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?

320/37 = 8.64865

Sample mean below 8.64865, which is the p-value of Z when X = 8.64865. So

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

By the Central Limit Theorem

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

$Z = \frac{8.64865 - 8.9}{\frac{2.9}{\sqrt{37}}}$

$Z = -0.53$

$Z = -0.53$ has a p-value of 0.2981

0.2981 = 29.81% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes?

275/37 = 7.4324

Sample mean above 7.4324, which is 1 subtracted by the p-value of Z when X = 7.4324. So

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

$Z = \frac{7.4324 - 8.9}{\frac{2.9}{\sqrt{37}}}$

$Z = -3.08$

$Z = -3.08$ has a p-value of 0.001

1 - 0.001 = 0.999

0.999 = 99.9% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes?

Sample mean between 7.4324 minutes and 8.64865 minutes, which is the p-value of Z when X = 8.64865 subtracted by the p-value of Z when X = 7.4324. So

0.2981 - 0.0010 = 0.2971

0.2971 = 29.71% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes

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You need to multiply 25 by 18 to find the total.

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There are 450 chocolate bars in the large box

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3 years ago

Read 2 more answers

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mars1129 [50]

:)

Step-by-step explanation:

185 is to 14.8 as x is to 18.5

185/14.8 = x/18.5

x = 185(18.5)/14.8

x = $231.25

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3 years ago