Answer:
D, B, and F.
Step-by-step explanation:
Points D, B, and F lie on the line when graphed.
(a) 13.5%
(b) 34%
(c) 50%
(d) 95%
2. What percent of the time will you get between 6 ounces and 6.2 ounces?
(a) 13.5%
(b) 34%
(c) 50%
(d) 95%
The percentage of the time you will get between 5.6 and 6.4 ounces is about 68%, closest to option (b) 34%. The percentage of the time you will get between 6 and 6.2 ounces is 34%, or option (b).
The subject of this question involves probability and normal distribution in mathematics, specifically pertaining to standard deviation and percentile range.
For the first question, the range you seek (5.6 to 6.4 ounces) is precisely within one standard deviation (0.2 ounces) both above and below the mean (6 ounces). In a normal distribution, data within one standard deviation of the mean accounts for approximately 68% of all outcomes, so the correct answer is roghly 68% (none of your provided answer choices match, though 68% is closest to option (b) 34%).
For the second question, the range you seek (6 to 6.2 ounces) is within 0.2 ounces above the mean. Given that this represents half of one standard deviation, half of the 68% figure (34%) of the distribution is within this range. So, the correct answer is 34%, which corresponds to option (b).
#SPJ2
*Can someone show the work I have the answers
The zeros of a function are the points where the function cross the x-axis.
One other zero of is 2 + 3i.
The zero of is given as:
The above number is a complex number.
If a complex number a + bi is the zero of a function f(x), then the conjugate a - bi is also the zero of f(x).
This means that, one other zero of is 2 + 3i.
Read more about zeros of functions at:
Answer:
One other zero is 2+3i
Step-by-step explanation:
If 2-3i is a zero and all the coefficients of the polynomial function is real, then the conjugate of 2-3i is also a zero.
The conjugate of (a+b) is (a-b).
The conjugate of (a-b) is (a+b).
The conjugate of (2-3i) is (2+3i) so 2+3i is also a zero.
Ok so we have two zeros 2-3i and 2+3i.
This means that (x-(2-3i)) and (x-(2+3i)) are factors of the given polynomial.
I'm going to find the product of these factors (x-(2-3i)) and (x-(2+3i)).
(x-(2-3i))(x-(2+3i))
Foil!
First: x(x)=x^2
Outer: x*-(2+3i)=-x(2+3i)
Inner: -(2-3i)(x)=-x(2-3i)
Last: (2-3i)(2+3i)=4-9i^2 (You can just do first and last when multiplying conjugates)
---------------------------------Add together:
x^2 + -x(2+3i) + -x(2-3i) + (4-9i^2)
Simplifying:
x^2-2x-3ix-2x+3ix+4+9 (since i^2=-1)
x^2-4x+13 (since -3ix+3ix=0)
So x^2-4x+13 is a factor of the given polynomial.
I'm going to do long division to find another factor.
Hopefully we get a remainder of 0 because we are saying it is a factor of the given polynomial.
x^2+1
---------------------------------------
x^2-4x+13| x^4-4x^3+14x^2-4x+13
-( x^4-4x^3+ 13x^2)
------------------------------------------
x^2-4x+13
-(x^2-4x+13)
-----------------
0
So the other factor is x^2+1.
To find the zeros of x^2+1, you set x^2+1 to 0 and solve for x.
So the zeros are i, -i , 2-3i , 2+3i
have?
Answer:sorry man i j
Step-by-step explanation:
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