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3 years ago

Multiply (4.7 x 10^4) by (6.4 x 10^15). Express your answer in scientific notation.

Mathematics

2 answers:

umka2103 [35]3 years ago

8 0

Steps to solve:

(4.7 x 10^4)(6.4 x 10^15)

~Combine like terms

(4.7 x 6.4)(10^4 x 10^15)

~Simplify

30.08 x 10^4+15

~Simplify

30.08 x 10^19

~Divide by 10 to get a number less then 10

30.08/10 x 10^19

~Simplify

3.008 x 10^20

Therefore, the answer is A) 3.008 x 10^20

Best of Luck!

Katen [24]3 years ago

6 0

4.7 x 10^4= 47,000
6.4 x 10^15= 6,400,000,000,000,000
(yay) 47, 000 x 6,400,000,000,000,000 = 256,000,000,000,000,000,000 + 44,800,000,000,000,000,000 = 3,00,800,000,000,000,000,000 = 3.008 x 10^20
Hope this helped!

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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

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Oksanka [162]

Hi!

Positive discriminants will give you two solutions.

Discriminants equal to zero will give you one solution.

Negative discriminants will give you no solutions.

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In the first graph, we can see it intersects the graph at two points: (2, 0) and (6, 0). Since there are two solutions it is a positive discriminant.

In the second graph, we can see it intersects at one point: the origin, or (0, 0). Since there is one solution it is a discriminant equal to zero.

In the third graph, we can see it doesn't intersect; it is above the x-axis. Since there are no solutions it is a negative discriminant.

For similar problems, see:

brainly.com/question/4592351

brainly.com/question/19936101

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