Step 1
Given; If Body temperatures of healthy adults have a bell-shaped distribution with a mean of 98.20°F and a standard deviation of 0.93°F.
Using the empirical rule, what is the approximate percentage of healthy adults with body temperatures between 96.34°F and 100.06°F?
Step 2
![P(96.34The empirical rule states that 95% of the data fall within two standard deviations of the mean. Thus, 95% of healthy individuals have body temperatures between 96.34F and 100.06F.Answer;[tex]95\text{\%}](https://tex.z-dn.net/?f=P%2896.34The%20empirical%20rule%20states%20that%2095%25%20of%20the%20data%20fall%20within%20two%20standard%20deviations%20of%20the%20mean.%20Thus%2C%2095%25%20of%20healthy%20individuals%20have%20body%20temperatures%20between%2096.34F%20and%20100.06F.%3Cp%3E%3C%2Fp%3E%3Cp%3E%3Cstrong%3EAnswer%3B%3C%2Fstrong%3E%3C%2Fp%3E%5Btex%5D95%5Ctext%7B%5C%25%7D)
Answer:
As a fraction, the answer is exactly 86/969
In decimal form, the answer is approximately 0.08875
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Work Shown:
The assumption is that no replacements are made for each selection.
5 red, 6 blue, 8 green
5+6+8 = 19 total
A = P(3 red) = (5/19)*(4/18)*(3/17) = 10/969
B = P(3 blue) = (6/19)*(5/18)*(4/17) = 20/969
C = P(3 green) = (8/19)*(7/18)*(6/17) = 56/969
D = P(3 all same color)
D = A+B+C
D = 10/969 + 20/969 + 56/969
D = (10+20+56)/969
D = 86/969
D = 0.08875
The question is asking us to determine the number of rows in which Sam can arrange his 12 black and white and 18 color photos. Sam wants to put them in the equal rows so that each row has either black and white photos only or color photos only. We have to find the GCF ( the Greatest Common Factor ) of 18 and 12. GCF ( 18, 12 ) = 6. So there should be 6 photos in each row if they have to be equal ( the equal number of photos in every row ). 18 = 6 * 3 and 12 = 6 * 2. Therefore the number of the rows is : 3 + 2 = 5. Answer: Sam can arrange his photos in 5 rows<span>. </span>
Answer: g(h(x))=6x +1 and h(g(x)) = 6x+7
Step-by-step explanation: Basically , this is a substitution exercise.
g(h(x)) = g(2x-1) . Now substitute x = 2x-1 in for x in g(x)
hence g(2x-1) = 3(2x-1)+4 = 6x - 3 + 4 = 6x + 1
Similarly : h(g(x)) = h(3x+4) substitute x = 3x+4 in h(x)
hence h(3x+4) = 2(3x + 4 ) - 1 = 6x + 8 - 1 = 6x + 7
