Answer: The question is that of Line WX and YZ are parallel or perpendicular, not WZ and XY unlike the other given answer.
THE ANSWER IS NEITHER
Step-by-step explanation:
See moc up i made. The two lines don't even intersect, I.E the answer is neither!
Answer:
You have not mentioned the joining points so I have joined the points WZ and XY and plotted in the graph. It seems that the lines WZ and XY are perpendicular.
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16.
For the function h(x) = 6 - X the value of (hoh)(10) is 10.
A relation is a function if it has only One y-value for each x-value.
The given function is h(x) = 6 - X
We have to find the value of (hoh)(10)
The composite function (hoh)(10) can be written as h(h(10))
h(h(10))
Plug in the value 10 in place of x
=h(6-10)
=h(-4)
Now let us find h(-4)
=6-(-4)
=6+4
=10
Hence, for the function h(x) = 6 - X the value of (hoh)(10) is 10.
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Answer:
h(h(10))---> h(-4)---->10 ......hOh(10)=10
Step-by-step explanation:
Step-by-step explanation:
The greatest common factor (GCF) of 27 and 45 is 9.
27 + 45
9 (3 + 5)
9 (8)
72
The expression 27 + 45 is rewritten using the Greatest Common Factor (GCF) and the distributive property as 9*(3 + 5) = 72.
The given expression is 27 + 45. To rewrite this using the Greatest Common Factor (GCF) and the distributive property, we first find the GCF of the two numbers. The GCF of 27 and 45 is 9. We then divide each number by the GCF and rewrite the expression using the distributive property.
So, 27 + 45 = 9*(3 + 5) = 9*8 = 72.
The above step is the application of the distributive property, where a*(b + c) = ab + ac.
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Answer:
The expected number of tests, E(X) = 6.00
Step-by-step explanation:
Let us denote the number of tests required by X.
In the case of 5 individuals, the possible value of x are 1, if no one has the disease, and 6, if at least one person has the disease.
To find the probability that no one has the disease, we will consider the fact that the selection is independent. Thus, only one test is necessary.
Case 1: P(X=1) = [P (not infected)]⁵
= (0.15 - 0.1)⁵
P(X=1) = 3.125*10⁻⁷
Case 2: P(X=6) = 1- P(X=1)
= 1 - (1 - 0.1)⁵
P(X=6) = (1 - 3.125*10⁻⁷) = 0.999999
P(X=6) = 1.0
We can then use the previously determined values to compute the expected number of tests.
E(X) = ∑x.P(X=x)
= (1).(3.125*10⁻⁷) + 6.(1.0)
E(X) = E(X) = 6.00
Therefore, the expected number of tests, E(X) = 6.00
mL
mg
kL
kg
600 liters of soda is equal to 600,000 milliliters.
To convert liters to milliliters, we multiply by 1000 since there are 1000 milliliters in one liter. So, 600 liters of soda is equal to 600,000 milliliters.
kyle regalara 16 figuras y le quedara 3/4 de ellas