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3 years ago

Which figure can serve as a counterexample to the conjecture below?

1 answer:

8 0

When we say conjecture, in math, this is defined as a statement that appears to be true but is not yet proven. Now, the question asked for the counterexample, which means, the opposite of the given statement above. Rectangle, rhombus and square are considered parallelograms. Therefore, the figure that can serve as a counterexample to the conjecture given is the TRAPEZOID. The answer would be option D.

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A merchant mixed 8 lb of a cinnamon tea with 3 lb of spice tea. the 11-pound mixture cost $29. a second mixture included 14 lb o

zimovet [89]

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4 years ago

Anyone know this?????

uranmaximum [27]

Answer: x=13
Explanation:

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step 2 - simplify both sides
step 3 - subtract 1.4 from both sides
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3 years ago

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3 years ago

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borishaifa [10]

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4 years ago

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The data shown represent the number of runs made each year during Bill Mazeroski’s career. Check for normality.

const2013 [10]

Answer:

The given data is not normal.

Step-by-step explanation:

We are given the following data:

30, 59, 69, 50, 58, 71, 55, 43, 3, 66, 52, 56, 62, 36, 13, 29, 17, 31

Condition for normality:

Mean = Mode = Median

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{800}{18} = 44.44$

Mode is the most frequent observation of the data.

Since all the value appeared once, there is no mode.

$Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}$

Sorted data: 3, 13, 17, 29, 30, 31, 36, 43, 50, 52, 55, 56, 58, 59, 62, 66, 69, 71

Median =

$=\dfrac{9^{th}+10^{th}}{2} = \dfrac{50+52}{2}=51$

Since the mean, mode and median of data are not equal, the data is not normal.

7 0

4 years ago