ABCD 1234 AEIOU = meow meow i’m not a cow who who
Answer:
The answer is 3
<u>F</u><u>o</u><u>r</u><u> </u><u>t</u><u>h</u><u>e</u><u> </u><u>s</u><u>t</u><u>e</u><u>p</u><u>s</u><u>,</u><u> </u><u>I</u><u>m</u><u> </u><u>l</u><u>e</u><u>a</u><u>v</u><u>i</u><u>n</u><u>g</u><u> </u><u>i</u><u>t</u><u> </u><u>b</u><u>l</u><u>a</u><u>n</u><u>k</u><u> </u><u>f</u><u>o</u><u>r</u><u> </u><u>y</u><u>o</u><u>u</u><u> </u><u>t</u><u>o</u><u> </u><u>f</u><u>i</u><u>g</u><u>u</u><u>r</u><u>e</u><u> </u><u>i</u><u>t</u><u> </u><u>o</u><u>u</u><u>t</u><u> </u><u>f</u><u>i</u><u>r</u><u>s</u><u>t</u><u>.</u>
<u>I</u><u>f</u><u> </u><u>y</u><u>o</u><u>u</u><u> </u><u>r</u><u>e</u><u>a</u><u>l</u><u>l</u><u>y</u><u> </u><u>n</u><u>e</u><u>e</u><u>d</u><u> </u><u>h</u><u>e</u><u>l</u><u>p</u><u>,</u><u> </u><u>a</u><u>s</u><u>k</u><u> </u><u>a</u><u>n</u><u>y</u><u>t</u><u>i</u><u>m</u><u>e</u><u> </u><u>a</u><u>n</u><u>d</u><u> </u><u>I</u><u>'</u><u>l</u><u>l</u><u> </u><u>r</u><u>e</u><u>p</u><u>l</u><u>y</u><u> </u><u>a</u><u>s</u><u> </u><u>s</u><u>o</u><u>o</u><u>n</u><u> </u><u>a</u><u>s</u><u> </u><u>i</u><u> </u><u>c</u><u>a</u><u>n</u><u> </u><u>:</u><u>)</u>
Using the normal distribution, it is found that there is a 0.0436 = 4.36% probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters.
<h3>Normal Probability Distribution</h3>
The z-score of a measure X of a normally distributed variable with mean
and standard deviation
is given by:

- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
In this problem, the mean and the standard deviation are given, respectively, by:
.
The probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters is <u>one subtracted by the p-value of Z when X = 4</u>, hence:


Z = 1.71
Z = 1.71 has a p-value of 0.9564.
1 - 0.9564 = 0.0436.
0.0436 = 4.36% probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters.
More can be learned about the normal distribution at brainly.com/question/24663213
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Answer:
Addition of 323+412 = 1240 at base 5
Step-by-step explanation:
We have to add 323+412 on base 5
As we have to use base 5 for addition so we can use only numbers 0,1,2,3.4 and 5 as we use 0,1,2,3,4,5,6,7,8,9,10 in decimal when base is 10
Means in addition of sum is greater than 5 then we have to take carry as in base 10 we take carry if the sum is greater than 10
So sum 323+412 = 1240
I believe the answer is A, the standard deviation is preferable to the range as a measure of variation because the standard deviation takes into account all of the observations, whereas the range considers only the largest and the smallest. Range gives an overall spread of data from the lowest to the largest and thus can be influenced by anomalies, standard deviation on the other hand, takes into account the variable data/spread about the mean and allows for statistical use so inferences can be made.<span />