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Tems11 [23]
3 years ago
5

In performing a chi-square goodness-of-fit test for a normal distribution, a researcher wants to make sure that all of the expec

ted cell frequencies are at least five. The sample is divided into 7 intervals. The second through the sixth intervals all have expected cell frequencies of at least five. The first and the last intervals have expected cell frequencies of 1.5 each. After adjusting the number of intervals, the degrees of freedom for the chi-square statistic is____________.a. 2b. 3c. 5d. 7
Mathematics
1 answer:
Sholpan [36]3 years ago
3 0

Answer:

The degrees of freedom for the chi square statistic is 7-4=3

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20 - 5 is her score after losing points:

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Answer:

\sf \:  \fbox{Option C) \: The  salary of B is 6000₹}

Step-by-step explanation:

Let the salary or A is x and salary of B is y.

now,

Condition one,The sum of 3/4th of A’s salary and 5/3rd of B’s salary is ₹16,000.

writing above statement in equation form,

\sf \frac{3x}{4}  +  \frac{5y}{3}  =  16000

Multiplying above equation with twelve,

\sf \frac{3x}{4}  +  \frac{5y}{3}  =  16000 \\ \sf \frac{12 \times 3x}{4}  +  \frac{12 \times 5y}{3}  =  16000 \times 12  \\  \sf  \fbox{9x + 20y = 192000} \rightarrow eq. 1

and condition two,

The difference of their salaries is ₹2000

\sf \: x - y = 2000

Multiplying above equation with 20

\sf \fbox{\: 20x -20y = 40000} \rightarrow eq.2

On adding equation 1 and equation 2,

\sf9x +  \cancel{20y} = 192000 \\ \sf 20x - \cancel{20y} = 40000 \\     \hline  \sf 29x  + 0y = 232000 \\  \sf x =  \frac{232000}{29}  \\ \sf \:  \fbox{x = 8000₹}

Substituting the value of x in,

\sf \: x - y = 2000 \\ \sf \: 8000 - y = 2000 \\ \sf \:  y = 8000 - 2000 \\ \sf  \fbox{y = 6000₹}

\sf \:  \fbox{The  salary of B is 6000₹}

<em><u>Thanks for joining brainly community!</u></em>

5 0
2 years ago
-3.08333333333 as a mixed number
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Answer:

-3\frac{1}{12}

Step-by-step explanation:

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6 0
3 years ago
Which point can be found using the vertical line x = –9 and the horizontal line y = 8?
Fudgin [204]
ANSWER

The point is
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EXPLANATION
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This is the unique solution to the two equations.

See graph.

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