Es la segunda media es 1.65, mediana es 1.65, y modo es 1.61
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:
Step-by-step explanation:
The y intercept form for the equation of a line is
You should note that <em>c</em> represents the y-intercept of the line (where the line touches the y-axis)
Answer:
they both have 5 in common
if you were to reduce the fraction it would be 11/16 because
55 divided by 5 is 11 and 80 divided by 5 is 16
Step-by-step explanation:
Answer:
AC = 5 cm
BD = 12.5 cm (3 sf) [or 2 × root 39]
BE = 6.93 cm (3 sf) [or 4 × root 3]
Step-by-step explanation:
CE = 8cm [CE is radius of circle]
AC + 3 = 8
<u>A</u><u>C</u><u> </u><u>=</u><u> </u><u>5</u><u> </u><u>c</u><u>m</u>
BC = 8cm [BC is a radius of circle]
(AC)^2 + (AB)^2 = (BC)^2 [Pythagoras theorem]
25 + (AB)^2 = 64
AB = 6.2450 cm (5 sf) [or root 39]
BD = 2(BA)
= 2(6.2450)
<u>B</u><u>D</u><u> </u><u>=</u><u> </u><u>1</u><u>2</u><u>.</u><u>5</u><u> </u><u>c</u><u>m</u><u> </u><u>(</u><u>3</u><u> </u><u>s</u><u>f</u><u>)</u><u> </u><u>[</u><u>o</u><u>r</u><u> </u><u>2</u><u> </u><u>×</u><u> </u><u>r</u><u>o</u><u>o</u><u>t</u><u> </u><u>3</u><u>9</u><u>]</u>
(BA)^2 + (AE)^2 = (BE)^2 [Pythagoras theorem]
39 + 9 = (BE)^2
<u>B</u><u>E</u><u> </u><u>=</u><u> </u><u>6</u><u>.</u><u>9</u><u>3</u><u> </u><u>c</u><u>m</u><u> </u><u>(</u><u>3</u><u> </u><u>s</u><u>f</u><u>)</u><u> </u><u>[</u><u>o</u><u>r</u><u> </u><u>4</u><u> </u><u>×</u><u> </u><u>r</u><u>o</u><u>o</u><u>t</u><u> </u><u>3</u><u>]</u>