Answer:
Probability that at least 490 do not result in birth defects = 0.1076
Step-by-step explanation:
Given - The proportion of U.S. births that result in a birth defect is approximately 1/33 according to the Centers for Disease Control and Prevention (CDC). A local hospital randomly selects five births and lets the random variable X count the number not resulting in a defect. Assume the births are independent.
To find - If 500 births were observed rather than only 5, what is the approximate probability that at least 490 do not result in birth defects
Proof -
Given that,
P(birth that result in a birth defect) = 1/33
P(birth that not result in a birth defect) = 1 - 1/33 = 32/33
Now,
Given that, n = 500
X = Number of birth that does not result in birth defects
Now,
P(X ≥ 490) =
=
+ .......+
= 0.04541 + ......+0.0000002079
= 0.1076
⇒Probability that at least 490 do not result in birth defects = 0.1076
Answer: y= −12
Step-by-step explanation:
450
10*12= 120 the bottom of the box
10*7.5= 75 one side
75*2= 150 two sides
12*7.5= 90 another side
90*2= 180 two other sides
120+150+180 = 450
Hope this helped :)
2 + 2 + 2 + 2 + 2 = 10
Five days
2 + 2 + 2 + 2 + 2 + 2 + 2 = 14
Full week
10 x 2 = 20
14 x 2 = 28
The bacteria will double every day for a total of 20/28 bacteria on the last day.
<em>Answer:</em>
<em> a2 - 5ab + 7b2</em>
<em>Step-by-step explanation:</em>
<em></em>
<em>Changes made to your input should not affect the solution:
</em>
<em> </em><em>(1):</em><em> "b2" </em><em>was replaced by</em><em> "b^2". 3 </em><em>more similar replacements.</em>
<em></em>
<em>= (((2•(a2))-3ab)+(2•(b2)))-(((a2)+2ab)-5b2) </em>
<em>= (((2•(a2))-3ab)+2b2)-(a2+2ab-5b2)</em>
<em>= ((2a2 - 3ab) + 2b2) - (a2 + 2ab - 5b2)</em>
<em><u> factor Into multi variable polynomial :</u></em>
<em> = 4.1 Factoring a2 - 5ab + 7b2 </em>
<em />
<em>Your answer should be : </em>
<em><u>= a2 - 5ab + 7b2</u></em>