What is the possible scale factor for the rectangle?

The lengths of pregnancies are normally distributed with a mean of 266 days and a standard deviation of 15 days. a. Find the pro

MrMuchimi

Answer:

a. 0.313% (0.003134842261), b. 237.79 days (237.788095878)

Step-by-step explanation:

In this case, the length of pregnancies is a normally distributed variable, with a mean of 266 days, and a standard deviation of 15 days.

A graph showing the distribution, with regions of interest for the answer, is presented below.

<h3>First Part: Find the probability of a pregnancy lasting 307 days or longer.</h3>

To answer the question regarding <em>the probability of a pregnancy lasting 307 days or longer</em>, it is necessary to calculate what the cumulative probability distribution value is at 307 days. By the way, according to the graph below, 307 days are quite far from the population mean (266 days).

Using the function <em>normaldist(266,15).cdf(307)</em>, from free Desmos software on Internet, we find that, at this length (307 days), the sum of all probabilities for all cases at this value is 99.69% (0.996865157739).

Considering that the total area of the curve is 1, then <em>the probability of pregnancy lasting 307 days or longer</em> is 1 - 0.996865157739 or 0.003134842261 (or 0.00313), approximately 0.313%, a very low probabilty.

This probability is showed as the "light blue" region at the right extreme of the graph.

<h3>Second Part: Find the length that separates premature babies from those who are not premature.</h3>

To find the length that separates premature babies from those who are not premature, it is a question about <em>find the days related with the probability of 3% (or 0.03)</em> to find such premature babies. So, it is a question of finding a percentile (or 100-quantiles): given the cumulative normal distribution curve, what is the value (length of pregnancies) that represents this 3%.

Using the function <em>quantile(normaldist(266,15), 0.03)</em>, from free Desmos software on Internet, we obtained a value of 237.79 days (237.788095878) for the length of pregnacies of premature babies. In other words, those babies whose mothers have a length of pregnancy <em>lower</em> than 237.79 days are considered premature, or this is "the length that separates premature babies from those who are not premature".

The area below 237.79 days is the blue shaded region in the graph below, at the left extreme of it.

4 0

3 years ago

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Of the 9-letter passwords formed by rearranging the letters AAAABBCCC (4 A's, 2 B's, and 3 C's), I select one at random. Determi

Tanya [424]

Answer:

a) 3

b) (8!/9!)-(7!/9!)

c) (1-(8!/9!))*(7!/9!)

Step-by-step explanation:

a)With 4 As ; 2Bs and 3Cs it is possible to get a palindrome if you fixed the letters C according to: (2) in the extremes of the word and the other one at the center therefore you only have palindrome in the following cases

To fill in the gaps we have 4 letters A and 2 letters B, wich we have two divide in two palindrome gaps,

AAB and BAA the palindrome is C AAB C BAA C

BAA and AAB " " is C BAA C AAB C

ABA and ABA " " is C ABA C ABA C

b) 4 A ; 2B ; 3C

We have the total number of elements 9, so the total number of possible outcomes is : 9!

Total events: 9!

if we fixed 3 C we have (the group of 3 Cs becoming one element) so the total amount of events with 3 adjacent Cs is: 7!

Therefore the probability of having 3 adjacent Cs is: 7!/9!

If we fixed only 2 Cs we have:

4 A ; 2 B ; 2C : 1C

Total number of words (events) in this case is 8! (2C becomes 1 element)

so the total numbers of events is 8! the probability in this case is 8!/9!(this value includes cases of adjacent 3 Cs previous calculated ) so this value minus the case of 3 adjacent Cs ) give us 2 adjacent C and the other no next to them

Probability (of words with 2 adjacent Cs and the other no next to them is); 8!/9! - 7!/9!

c) Probability of B apart from each other is the whole set of events minus those where 2 B are adjacent or (become 1 element)

4 A ; 2B ; 3C

Total of events 9! and events with adjacent B is 8!/9!

Therefore the probability of words with 3 adjacent Cs and 2 B separeted is

the probability of 3 adjacent Cs (7!/9!) times probability of words with no adjacent Bs wich is (1-(8!/9!))*(7!/9!)

5 0

3 years ago

A ratio of 2 to 10 means

shtirl [24]

Answer:

1:5

Step-by-step explanation:

If you simplify ( I think that's what you mean by "means")

10 is divisible by 2

2=1

10=5

1:5

3 0

3 years ago

Triangle LPQ has side lengths of 9, 20, and 15. Classify the triangle by its sides

Aneli [31]

Since all of the side lengths are different, it is a scalene triangle

5 0

3 years ago

a quarter weighs about 0.2 ounce. Naoki has 2 pounds (32 ounces) of quarters. About how many quarters does he have?

Sidana [21]

Answer:

160

Step-by-step explanation:

8 0

3 years ago

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