All categories

3 years ago

The box plot shows the total amount of time, in minutes, the students of a class spend studying each day:

Mathematics

1 answer:

Mademuasel [1]3 years ago

3 0

A. The lower quartile of data, which is represented by the first “whisker”

Not B or D because we don’t know the total, just where the data falls. Since we don’t know the total number of data points, we also can’t find the mean from C

You might be interested in

There are a total of 91 students in a drama club and a yearbook club. The drama club has 19 more students than the yearbook club

abruzzese [7]

Step-by-step explanation:

91-19=72

72/2=36

36+19=55=Drama Club students

36 =Yearbook Club Students

8 0

4 years ago

How do u solve 2/8=4

timofeeve [1]

Answer:

8/8 (then entire number) is 16

Step-by-step explanation:

4/2 = 2

2x8=16

so 8/8=16

8 0

4 years ago

Read 2 more answers

ILLL MAKE BRAINLIEST PLSS HELLPPP. Given the function f (2) = 3(x - 1)² + 2 , which of the following statements is true

nikdorinn [45]

Answer:

I'll choose the first one but don't know the right one

6 0

3 years ago

Gus, Verid and their three children went to see a show.,

seraphim [82]

Answer:

28.35 each adult

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Find the value of the variable y, where the sum of the fraction 2/y-3 and 6/y+3 is equal to the quotient.

NISA [10]

Answer:

Here we need to solve:

$\frac{2}{y - 3} + \frac{6}{y + 3 } = \frac{\frac{2}{y-3}}{\frac{6}{y + 3} }$

The sum of the fractions is equal to the quotient between the fractions.

Notice that the two values:

y = 3

y = -3

make the denominator equal to zero, so those values are restricted.

We can simplify the right side to get:

$\frac{2}{y - 3} + \frac{6}{y + 3 } = \frac{\frac{2}{y-3}}{\frac{6}{y + 3} } = \frac{2*(y + 3)}{6*(y - 3)} = 3*\frac{y + 3}{y - 3}$

Now we can multiply both sides by (y - 3)

$(y - 3)*(\frac{2}{y - 3} + \frac{6}{y + 3 }) = 3*(y + 3)\\2 + 6*\frac{y -3}{y + 3} = 3*(y + 3)$

Now we can multiply both sides by (y + 3)

$(2 + 6*\frac{y -3}{y + 3})*(y + 3) = 3*(y + 3)*(y + 3)$

$2*(y + 3) + 6*(y - 3) = 3*(y + 3)*(y + 3)\\\\2*y + 6 + 6*y - 18 = 3*(y^2 + 2*y*3 + 9)\\\\8*y - 12 = 3*y^2 + 6*y + 33\\\\0 = 3*y^2 + 6*y + 33 - 8*y + 12\\\\0 = 3*y^2 - 2*y + 45$

First, let's see the determinant of that quadratic equation:

$D = (-2)^2 - 4*3*45 = -536$

We can see that it is negative, thus, there are no real solutions of the equation.

Thus, there is no value of y such that the origina equation is true,

6 0

3 years ago