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lora16 [44]
3 years ago
13

What type of number is -17i

Mathematics
1 answer:
emmainna [20.7K]3 years ago
3 0

Answer:

See below

Step-by-step explanation:

-17i        ' i '  stands for 'imaginary'  

<u>   this is a negative imaginary number </u>

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A school paid $31.25 for each calculator. If the school bought x calculators, how much did they pay?
Katyanochek1 [597]

Answer:

31.25×

Step-by-step explanation:

suppose that the value of x was 3 . that would mean that they bought 3 calculators .

if one calculator is $31.25 then 3 would be $31.25× 3

the same rule would apply for x

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Help??? What is 3x+5x=67y?
ruslelena [56]
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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

7 0
3 years ago
Someone please help or i’ll be failing geometry this year
Vlad [161]

Step-by-step explanation:

I don't know what constructions you were taught.

a "similar" triangle is a triangle with exactly the same angles as the other triangle, but the lengths of all sides are stretched or shortened by the same scaling factor f.

by saying 1:2 she means the second triangle should have sides with twice the lengths of the first triangle (f=2).

and the extra challenge - same basic thing. she allows you to pick one of the two triangles as reference. and then you need to draw a third triangle (again with the same angles) with the side lengths extended by the scaling factor f of 4/3.

I would draw the triangles right on top of each other with the same starting corner (let's call it A) for all 3.

we would get the triangles ABC, AMN and AXY.

the points B and C would be then halfway on AM and AN.

and M and N would then a bit before X and Y on AX and AY.

the beauty is, you only need to construct 2 sides of every new triangle down to the new endpoints. the third side is automatically scaled correctly, and you only need to connect these new endpoints.

let's assume you draw a triangle (just very simple) ABC with all side lengths being 3. so, AB=3, AC=3, BC=3.

now you draw AMN by extending AB and AC to a side length of 6 (f=2) creating M and N, and you connect M and N.

and then you can create the third triangle AXY by extending AM and AN by a factor of 4/3 to side lengths of 8 (4/3 × 6 = 8) creating new end points X and Y. and you connect X and Y.

and that is it. all 3 triangles are similar (the same angles), and all sides of a triangle have the same length ratio to the sides of the other triangle(s).

4 0
3 years ago
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