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Ivanshal [37]
2 years ago
8

Compare the values of the 2s and 5s in 55,220

Mathematics
1 answer:
bagirrra123 [75]2 years ago
6 0
The five digit number, 55,220 contains to 5's and two 2's.  The 5's are in the ten thousand and one thousand columns. This means that one five represents 50,000 and the second represents 5, 000.  The two's are in the hundreds and tens columns meaning one 2 represents 200 and the other 2 represents 20. In a direct comparison of these numbers the 5's equal 55,000 and the 2's equal 220.
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The equation of a line is y = 1.5x − 2. What are its slope and y-intercept?
irakobra [83]
Slope: 1.5
y-intercept: -2

Explanation:
The slope is the number attached to 'x' in the equation. In this case, 1.5 would be the slope because it's attached to x.

The y-intercept is the constant the number not attached to any variables. In this case, -2 is all by itself.

Hope that helps!

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Look it up on photomath
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A direct variation function contains the points (2,14) and (4,38) which equation represents the function
Dmitry [639]
You are given two points on a line, and told that the function is "direct variation."  You can save a lot of time by recognizing that the y-intercept here is zero (0).  Thus, the line passes through the origin.

A line through (2,14) and (4,38) is NOT direct variation, because the line connecting those two points does NOT pass through the origin, and thus has a vertical intercept.
3 0
2 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
2 years ago
Which decimal number is equal to 3/18 ? 0.167 0.16 0.16 0.167
Rus_ich [418]

3/18 = 1/6 =.16666666.....

.16(repeating)

your choice A and D are the same

and choice B and C are the same


5 0
2 years ago
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