There are 11 girls and 7 boys what is the ratio of boys to all students.

A farmer leans a 12-ft ladder against a barn. The base of the ladder is 3 ft from the barn. To the nearest tenth, how high on th

DENIUS [597]

The barn, the ground, and the ladder form a right triangle. If we represent the height on the barn by y, then the Pythagorean theorem tells us

... hypotenuse² = (side1)² + (side2)²

... (12 ft)² = (3 ft)² + y² . . . . . . . . . . . . fill in given information

... 144 ft² - 9 ft² = y² = 135 ft² . . . . . .subtract (3 ft)²

... y = 3√15 ft ≈ 11.6 ft . . . . . . . . . . . take the square root

The ladder reaches 11.6 feet high on the barn.

4 0

3 years ago

Read 2 more answers

The box-and-whisker plot below shows the numbers of text messages received in one day by students in the seventh and eighth grad

Leya [2.2K]

<h3><u>Answer with explanation:</u></h3>

From the box and whisker plot of seventh grade we have:

The minimum value =6

First quartile or lower quartile i.e. $Q_1$ = 14

Median or second quartile i.e. $Q_2$ = 18

Third quartile or upper quartile i.e. $Q_3$ =22

and maximum value = 26

From the box and whisker plot of eighth grade we have:

The minimum value =22

First quartile or lower quartile i.e. $Q_1$ = 26

Median or second quartile i.e. $Q_2$ = 30

Third quartile or upper quartile i.e. $Q_3$ = 34

and maximum value = 38

a)

The overlap of the two sets of data is as follows.

The upper quartile or third quartile of seventh grade is same as the minimum value of the data of eighth grade.
And the maximum value of seventh grade is same as the lower quartile of eighth grade.

b)

IQR is calculated as the difference of the Upper quartile and the lower quartile

i.e. $Q_3-Q_1$

so, IQR of seventh grade is:

22-14=8

IQR of seventh grade=8

IQR of eighth grade is:

34-26=8

Hence, IQR of eighth grade=8

c)

The difference of the median of the two data sets is:

30-18=12

Hence, the difference of median is: 12

d)

As the IQR of both the sets is same i.e. 8.

Hence, the number that must be multiplied by IQR so that it is equal to the difference between the medians of the two sets is:

$8\times n=12\\\\n=\dfrac{12}{8}\\\\n=\dfrac{3}{2}\\\\n=1.5$

Hence, the number is : 1.5

6 0

3 years ago

What's the answer to this question

frutty [35]

Step-by-step explanation:

there are several things to be considered to solve this.

the strategy is to caucuses the volume of the full "box", and then to deduct the 3-dimensional triangle that is "cut off" from the oil.

then we know the volume of the pool.

next step is to calculate the volume of a "box" with the same layout but is only 20 cm deep.

then we know how much volume can be pumped out in 30 minutes.

and then we divide the whole volume by that small volume to see how many 30-minues units are needed to empty the whole pool.

and then we transform the 30- minutes units into actual minutes or hours.

so, let's begin :

the whole box's volume is

2×1×1.4 = 2.8 m³

the volume of the 3-D triangle we need to subtract is the area of the triangle × width of the pool (this pool width is the height of the 3-D triangle).

2×(1.4 - 0.6) × 1 / 2 = 2×0.8 × 1 / 2 = 1.6/2 = 0.8 m³

so, the volume of the pool is

2.8 - 0.8 = 2 m³

now, how much volume can the pump eliminate in 30 minutes ?

the content of a box

1×2×0.2 = 0.4 m³

as 20 cm = 0.2 m

so, the pump pumps out 0.4 m³ in 30 minutes.

and we need to empty 2 m³.

that means we need

2 / 0.4 = 5

30-minues units = 5×30 = 150 minutes = 2.5 hours.

Colin has to wait 2.5 hours (150 minutes) to empty the complete pool.

6 0

3 years ago

The ratio of students who wear glasses to the total class is 2: 5. If there

Dahasolnce [82]

25 students
hope this helps

4 0

2 years ago

36 x 69 in area model

inysia [295]

ANSWER: 2484

(Alternative forms): 2.484 × 10³

8 0

2 years ago