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3 years ago

A student solved the following problem and made an error:

Mathematics

1 answer:

liubo4ka [24]3 years ago

8 0

All the steps were correct except the final statement. The mistake was in Line 6.

Line 6 triangle ABC is congruent to triangle EFD by SAS.

<span>This does not follow. The SAS postulate states that if two sides and the included angle of one triangle is congruent to two sides and the included angle of another triangle. The student only proved that one side of the triangle (AC) is congruent to the side of another triangle (EF) .</span>

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Step-by-step explanation:

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As of 2008 the longest car ever built was 30.5 meters long. A meter is a little longer than a yard. Estimate the length of this

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3 years ago

There are 1 × 1014 good bacteria in the human body. There are 2.6 × 1018 good bacteria among the spectators in a soccer stadium.

grandymaker [24]

Answer:

There are $2.6\times 10^{4}$ spectators in the soccer stadium.

Step-by-step explanation:

From this statement, we can calculate the number of spectators in a soccer stadium by dividing the amount of bacteria among the spectators in a soccer stadium by the amount of bacteria in the human body. That is:

$x = \frac{2.6\times 10^{18}\,bacteria}{1\times 10^{14}\,bacteria}$

$x = \frac{2.6}{1}\times \frac{10^{18}}{10^{14}}$

$x = 2.6\times 10^{18-14}$

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There are $2.6\times 10^{4}$ spectators in the soccer stadium.

5 0

3 years ago

A bag contains red marbles, white marbles, and blue marbles. Randomly choose two marbles, one at a time, and without replacement

dsp73

Answer:

$P(First\ White\ and\ Second\ Blue) = \frac{3}{28}$

$P(Same) = \frac{67}{210}$

Step-by-step explanation:

Given (Omitted from the question)

$Red = 7$

$White = 9$

$Blue = 5$

Solving (a): $P(First\ White\ and\ Second\ Blue)$

This is calculated using:

$P(First\ White\ and\ Second\ Blue) = P(White) * P(Blue)$

$P(First\ White\ and\ Second\ Blue) = \frac{n(White)}{Total} * \frac{n(Blue)}{Total - 1}$

<em>We used Total - 1 because it is a probability without replacement</em>

So, we have:

$P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{21 - 1}$

$P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{20}$

$P(First\ White\ and\ Second\ Blue) = \frac{9*5}{21*20}$

$P(First\ White\ and\ Second\ Blue) = \frac{45}{420}$

$P(First\ White\ and\ Second\ Blue) = \frac{3}{28}$

Solving (b) $P(Same)$

This is calculated as:

$P(Same) = P(First\ Blue\ and Second\ Blue)\$ $or\ P(First\ Red\ and Second\ Red)\ or\ P(First\ White\ and Second\ White)$

$P(Same) = (\frac{n(Blue)}{Total} * \frac{n(Blue)-1}{Total-1})+(\frac{n(Red)}{Total} * \frac{n(Red)-1}{Total-1})+(\frac{n(White)}{Total} * \frac{n(White)-1}{Total-1})$

$P(Same) = (\frac{5}{21} * \frac{4}{20})+(\frac{7}{21} * \frac{6}{20})+(\frac{9}{21} * \frac{8}{20})$

$P(Same) = \frac{20}{420}+\frac{42}{420} +\frac{72}{420}$

$P(Same) = \frac{20+42+72}{420}$

$P(Same) = \frac{134}{420}$

$P(Same) = \frac{67}{210}$

6 0

3 years ago