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3 years ago

Please Help Me ASAP I Need This For My grade

Mathematics

1 answer:

adell [148]3 years ago

3 0

Answer:

Option B: $x = \frac{sr-t}{2}$ is the correct answer.

Step-by-step explanation:

Given expression is:

$s=\frac{2x+t}{r}$

We have to find the value of x.

Multiplying both sides by r

$s*r =\frac{2x+t}{r}*r\\sr = 2x+t$

Now, we will subtract t from both sides of the equation

sr - t = 2x+t-t

sr-t = 2x

2x = sr-t

Dividing both sides of equation by 2

$\frac{2x}{2}=\frac{sr-t}{2}\\x = \frac{sr-t}{2}$

Hence,

Option B: $x = \frac{sr-t}{2}$ is the correct answer.

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4 years ago

If a new data point at 12 is added to the graph , which will be true ?

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Answer:

3. The mean will increase more than median, but both will increase.

Step-by-step explanation:

We have been given a dot plot. We are asked to find the true statement, if a new data point at 12 is added to our given dot plot.

Since we know that an large valued outlier affects mean more than median, so mean will increase more than median.

Let us check this using our given information. Our data set before adding 12 is:

1, 2, 2, 2, 3, 4, 4, 4, 5.

We can see that our data set has odd number (9) of data points, so median will be the value of 5th term, that is 3.

$\text{Mean}=\frac{1+2+2+2+3+4+4+4+5}{9}$

$\text{Mean}=\frac{27}{9}=3$

Therefore, mean of our data set is 3.

Upon adding a data point at 12, our new data set will be:

1, 2, 2, 2, 3, 4, 4, 4, 5, 12.

Now our data set has 10 terms, so median will be the average of 5th and 6th term.

$\text{Median}=\frac{3+4}{2}$

$\text{Median}=\frac{7}{2}=3.5$

Therefore, the median of new data set will be 3.5.

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$\text{Mean}=\frac{39}{10}$

$\text{Mean}=3.9$

Therefore, mean of new data set will be 3.9.

Upon looking at our given statements we can see that 3rd statement is true as mean and median both increased after adding 12 to our data set, but mean increased more that median. Mean and median before adding 12 was 3; but after adding 12 median became 3.5, while mean became 3.9 (which is greater than mean).

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