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3 years ago

QUICKLY ANSWER :Order the following numbers from least to greatest.

Mathematics

2 answers:

KonstantinChe [14]3 years ago

8 0

Answer:

1.4, 1 1/3, 127/100

Step-by-step explanation:

1 1/3 = 1.3333...

127/100 = 1.27

1.4 = 1.4

1.4 is the largest, 1 1/3 is the second largest, and 127/100 is the smallest.

andreev551 [17]3 years ago

3 0

Answer:

________________________________________________

From least to greatest:

The 3 (three) numbers given are ordered as follows:

________________________________________________

→ " 127/100 < 1 $\frac{1}{3}$ < 1.4 " .

________________________________________________

Step-by-step explanation:

________________________________________________

→ 1 $\frac{1}{3}$ = 1.333333333333333333.... ;

→ 127/100 = 1.27 ;

→ 1.4 = 1.4 ;

_________________________________________________

Of the three numeric values, the least value is:

→ " 1.27 " ; {that is: "127/100" .}.

The second least vaiue is: " 1.33333333..." ;

{that is: → " 1 $\frac{1}{3}$ " .}.

The greatest value; or third least value of the 3 (three) numbers is: " 1.4 " .

_________________________________________________

{<u>Note</u>: The three (3) numbers given are all greater than one and less than 2 (two). They can be written in decimal form. In the case of each of these 3 (three) values, we look to the value of the right number following the decimal point of each number. In the case of the particular 3 (three) numbers presented, each number has a different value to the right of the decimal point; one has a "2" ; one has a "3" and one has a "4" ; and these represent the values of the numbers given from least to greatest, respectively.}.

__________________________________________________

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Answer:

$L(t)=7600e^{0.2273t}$

Step-by-step explanation:

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$P$ is the population after t days.
$P_o$ is the initial population given as 7600
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-The population increases by a factor of 5 every 22 days. therefore at any time instance, t will be divided by 22 to get the effective time for calculations.

Hence, the exponential growth function will be expressed as:

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4 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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kvasek [131]

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NNADVOKAT [17]

Answer:

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Step-by-step explanation:

We will use this formula to find the slope⤵⤵⤵

$\frac{y _{2} - y _{1} }{x _{2} - x _{1} }$

(Sorry if it's a bit small btw)

Now we plug in the values accordingly:

-3 - (-3) = -3 + 3 = 0

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So we have 0/-10, which means our slope is actually just 0.

(This also means that the line on the graph will be horizontal.)

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