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Andrews [41]
3 years ago
13

9u²v and which term are like terms?

Mathematics
2 answers:
madam [21]3 years ago
8 0
Answer: Since you didn't complete the problem, I can only give you some advice on how to pick the correct answer.

For terms to be like, they must contain the same variables. Also, the variables must be raised to the same power.

Look for another term that includes the exact variables: u^2v
NARA [144]3 years ago
8 0
I attached a diagram with the complete expression.

Answer:
5vu² is like 9u²v

Explanation:
Like terms are terms having the same degree for all variable it contains

The given term has two variables:
u² and v
This means that the term that is considered to be like this one should also have the variables u² and v.

Taking a look at the given terms, we would find that the only term satisfying this condition is 5vu².

Hope this helps :)

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260/500 = .52
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It takes Mike 18 minutes to finish reading 4 pages of a book. How long does it take for him to finish reading 30 pages
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Answer:

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3 years ago
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

7 0
2 years ago
Your part time job pays you $14/h. You work approximately 12 hours/week.
kirill115 [55]

Answer:

  1. <u>BOX 1 </u>=  672.00

(14 * 12 = 168

168 * 4 = $ 672 )

     2. <u>BOX 2</u> = 282.95

(Cell phone= $45 A MONTH

Eating outside = $50 A MONTH

12 .50 * 4   = 50   (4 WEEKS A MONTH )

Transportation =  131.2 A MONTH

32.80 * 4 = 131.2  (4 WEEKS A MONTH )

Movie =  $16.75 A MONTH

Charity = $ 40 A MONTH

10 * 4  (4 WEEKS A MONTH )

<u> MONTHLY EXPENSE </u> = 45 + 50 + 131.2 + 16.75 + 40

<u> MONTHLY EXPENSE </u> = 282.95 )

      3. <u>BOX 3</u> = 389.05

Money that can be saved = Monthly income - Monthly expense

Money that can be saved = 672 -  282.95

                                          = 389.05

<h2><em><u>please give brainliest </u></em></h2>
4 0
3 years ago
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