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babymother [125]
3 years ago
14

Why is it necessary to check that n modifyingabove p with caret greater than or equals 5np≥5 and n modifyingabove q with caret g

reater than or equals 5nq≥5​?
Mathematics
1 answer:
Nookie1986 [14]3 years ago
4 0

Both of these conditions must be true in order for the assumption that the binomial distribution is approximately normal. In other words, if np \ge 5 and nq \ge 5 then we can use a normal distribution to get a good estimate of the binomial distribution. If either np or nq is smaller than 5, then a normal distribution wouldn't be a good model to use.

side note: q = 1-p is the complement of probability p

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so 1/5 * 20 = 4

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Which expressions are equal to the expression _2.7 _ 16.4?
Genrish500 [490]
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8 0
3 years ago
Read 2 more answers
Please find the exact length of the midsegment of trapezoid JKLM with vertices J(6, 10), K(10, 6), L(8, 2), and M(2, 2). Thank y
I am Lyosha [343]

Answer:

the exact length of the midsegment of trapezoid JKLM  = \mathbf{ = 3 \sqrt{5} } i.e 6.708 units on the graph

Step-by-step explanation:

From the diagram attached below; we can see a graphical representation showing the mid-segment of the trapezoid JKLM. The mid-segment is located at the line parallel to the sides of the trapezoid. However; these mid-segments are X and Y found on the line JK and LM respectively from the graph.

Using the expression for midpoints between two points to determine the exact length of the mid-segment ; we have:

\mathbf{ YX = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} }

\mathbf{ YX = \sqrt{(8-5)^2+(8-2)^2} }

\mathbf{ YX = \sqrt{(3)^2+(6)^2} }

\mathbf{ YX = \sqrt{9+36} }

\mathbf{ YX = \sqrt{45} }

\mathbf{ YX = \sqrt{9*5} }

\mathbf{ YX = 3 \sqrt{5} }

Thus; the exact length of the midsegment of trapezoid JKLM  = \mathbf{ = 3 \sqrt{5} } i.e 6.708 units on the graph

8 0
3 years ago
Help please very much
Norma-Jean [14]

Answer:

D

Step-by-step explanation:

3 0
3 years ago
Please can someone help me with my maths? I’m so confused
Blababa [14]
The y intercept is 0.5 because thats where the line starts.
the slope/gradient of the line is 1.5.
this would be written in the form: y=1.5x+0.5
5 0
3 years ago
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