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german
3 years ago
5

Please help!! I need to go to the next question!

Mathematics
1 answer:
tino4ka555 [31]3 years ago
5 0

Given:

The inequality is:

3t+1\leq 5

To find:

The graph of the given inequality.

Solution:

We have,

3t+1\leq 5

Subtract both sides by 1.

3t\leq 5-1

3t\leq 4

Divide both sides by 3.

t\leq \dfrac{4}{3}

The value of t is less than or equal to \dfrac{4}{3}.

Since t\leq \dfrac{4}{3}, it means \dfrac{4}{3} is included in the solution, therefore there is a  closed circle at t=\dfrac{4}{3} and an arrow approaches to left from t=\dfrac{4}{3}.

Therefore, the correct option is A.

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Which expression has the same value as 7÷9 over 10
Maurinko [17]
This may be wrong but I'm going to go with B. 1/7 x 9/10



Hope this helps! ( Probably didn't help in any way but I tried).
5 0
3 years ago
The Revenue for the week is $2000, and labor cost consists of two workers earning $8 per hour who work 40 hours each, what is la
kkurt [141]

8*40 = 320 * 2 = $640 for the 2 workers

640/2000 = 0.32

labor cost is 32% of the revenue

8 0
3 years ago
Help with this asap!​
zmey [24]

Answer:

26?

Step-by-step explanation

because they are vertical? (i think the word) they are equal so set them equal to each other then solve for x

5 0
3 years ago
The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3
In-s [12.5K]

Answer:

a) There is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

b) There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

c) There is a 68.74% 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:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

Problems of normally distributed samples can be 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:

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3 minutes. This means that \mu = 8.3, \sigma = 3.3.

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

We are working with a sample mean of 37 jets. So we have that:

s = \frac{3.3}{\sqrt{37}} = 0.5425

Total time of 320 minutes for 37 jets, so

X = \frac{320}{37} = 8.65

This probability is the pvalue of Z when X = 8.65. So

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

Z = \frac{8.65 - 8.3}{0.5425}

Z = 0.65

Z = 0.65 has a pvalue of 0.7422. This means that there is a 74.22% 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?

Total time of 275 minutes for 37 jets, so

X = \frac{275}{37} = 7.43

This probability is subtracted by the pvalue of Z when X = 7.43

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

Z = \frac{7.43 - 8.3}{0.5425}

Z = -1.60

Z = -1.60 has a pvalue of 0.0548.

There is a 1-0.0548 = 0.9452 = 94.52% 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?

Total time of 320 minutes for 37 jets, so

X = \frac{320}{37} = 8.65

Total time of 275 minutes for 37 jets, so

X = \frac{275}{37} = 7.43

This probability is the pvalue of Z when X = 8.65 subtracted by the pvalue of Z when X = 7.43.

So:

From a), we have that for X = 8.65, we have Z = 0.65, that has a pvalue of 0.7422.

From b), we have that for X = 7.43, we have Z = -1.60, that has a pvalue of 0.0548.

So there is a 0.7422 - 0.0548 = 0.6874 = 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.

7 0
3 years ago
What is 0.75 over 100 simplest form?
STALIN [3.7K]
.0075 is simplest... or .0075/1 if it needs to be a fraction i guess
4 0
3 years ago
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