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Dahasolnce [82]
3 years ago
12

Find the percent of change:

Mathematics
2 answers:
valentinak56 [21]3 years ago
8 0
The answer might be B or C. If I’m not mistaken.
dedylja [7]3 years ago
8 0

Answer:

B. (Reason rounding to the nearest whole is it 67 and it's a negative/decreasing)

Step-by-step explanation:

From 120 bananas to 40 bananas is -66.67%

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What is the area of the circle?
hodyreva [135]

Diameter = radius x 2

So, the radius is half of the denominator.

10 / 2 = 5

A = pi x r^2

A = pi x 5^2

A =  3 x 25

A = 75 ft^2

Hope this helps!

4 0
3 years ago
A radar is designed to report the track of an aircraft every 2 seconds. If this radar reports 15 reports 15 tracks in one minute
meriva

Answer:

Step-by-step explanation:

Since radar reports 30 tracks in one minute in which 60 seconds makes 1 minute

Let calculate using this formula

Using this formula

Percentage of time=Tracks reported in minutes/60 seconds*100

Let plug in the formula

Percentage of time=30 tracks/60 seconds*100

Percentage of time=0.5*100

Percentage of time=50%

Inconclusion The percentage of the time that  radar track the aircraft is 50%

5 0
3 years ago
Read 2 more answers
The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 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
HELP When does expanding and simplifying a(b + c) result in a positive value for ac?
Reil [10]

Answer:

When c and a are positive or both are negative

Step-by-step explanation:

negative times negative equals positive so if a was negative and c was negative -a*-c= ac

Of course positive times positive equals negative so if a and c were positive we could do

a*c= ac

Hope I could help !

7 0
2 years ago
What is the decimal for 2/3
lakkis [162]
0.6 repeating (AKA 0.6666666-----)
7 0
3 years ago
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