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3 years ago

The students in Marly’s math class recorded the dimensions of their bedrooms in a frequency table.

Mathematics

2 answers:

Alex3 years ago

5 0

Answer:

The histogram is right-skewed, so the mean is greater than the median.

Step-by-step explanation:

34kurt3 years ago

4 0

Answer:

the answer is B on edge

Step-by-step explanation:

the histogram is right-skewed, so the mean is greater than the median

You might be interested in

A teacher was interested in knowing the amount of physical activity that his students were engaged in daily. He randomly sampled

klasskru [66]

Answer:

The standard error of the mean is 4.5.

Step-by-step explanation:

As we don't know the standard deviation of the population, we can estimate the standard error of the mean from the standard deviation of the sample as:

$\sigma_{\bar{x}}\approx\frac{s}{\sqrt{n}}$

The sample is [30mins, 40 mins, 60 mins, 80 mins, 20 mins, 85 mins]. The size of the sample is n=6.

The mean of the sample is:

$\bar{x}=\frac{1}n} \sum x_i =\frac{30+40+60+80+20+85}{6}=52.5$

The standard deviation of the sample is calculated as:

$s=\sqrt{\frac{1}{n-1}\sum (x_i-\bar x)^2} \\\\ s=\sqrt{\frac{1}{5}\cdot ((30-52.5)^2+(40-52.5)^2+(60-52.5)^2+(80-52.5)^2+(20-52.5)^2+(85-52.5)^2}\\\\s=\sqrt{\frac{1}{5} *3587.5}=\sqrt{717.5}=26.8$

Then, we can calculate the standard error of the mean as:

$\sigma_{\bar{x}}\approx\frac{s}{\sqrt{n}}=\frac{26.8}{6}= 4.5$

6 0

3 years ago

A certain square is to be drawn on a coordinate plane. One of the vertices must be on the origin, and the square is to have an a

Scrat [10]

Answer:

The answer is (C) 8

Step-by-step explanation:

First, let's calculate the length of the side of the square.

$A_{square}=a^2$ , where $a$ is the length of the side. Now, let's try to build the square. First we need to find a point which distance from (0, 0) is 10. For this, we can use the distance formula in the plane:

$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ which for $x_1=0$ and $y_1 = 0$ transforms as $d=\sqrt{(x_2)^2 + (y_2)^2}$ . The first point we are looking for is connected to the origin and therefore, its components will form a right triangle in which, the Pythagoras theorem holds, see the first attached figure. Then, $x_2$ , $y_2$ and 10 are a Pythagorean triple. From this, $x_2= 6$ or $x_2=8$ while $y_2= 6$ or $y_2=8$ . This leads us with the set of coordinates:

$(\pm 6, \pm 8)$ and $(\pm 8, \pm 6)$ . (A)

The next step is to find the coordinates of points that lie on lines which are perpendicular to the lines that joins the origin of the coordinate system with the set of points given in (A):

Let's do this for the point (6, 8).

The equation of the line that join the point (6, 8) with the origin (0, 0) has the equation $y = mx +n$ , however, we only need to find its slope in order to find a perpendicular line to it. Thus,

$m = \frac{y_2-y_1}{x_2-x_1} \\m = \frac{8-0}{6-0} \\m = 8/6$

Then, a perpendicular line has an slope $m_{\bot} = -\frac{1}{m} = -\frac{6}{8}$ (perpendicularity condition of two lines). With the equation of the slope of the perpendicular line and the given point (6, 8), together with the equation of the distance we can form a system of equations to find the coordinates of two points that lie on this perpendicular line.

$m_{\bot}=\frac{6}{8} = \frac{8-y}{6-x}\\ 6(6-x)+8(8-y)=0$ (1)

$d^2 = \sqrt{(y_o-y)^2+(x_o-x)^2} \\(10)^2=\sqrt{(8-y)^2+(6-x)^2}\\100 = \sqrt{(8-y)^2+(6-x)^2}$ (2)

This system has solutions in the coordinates (-2, 14) and (14, 2). Until here, we have three vertices of the square. Let's now find the fourth one in the same way we found the third one using the point (14,2). A line perpendicular to the line that joins the point (6, 8) and (14, 2) has an slope $m = 8/6$ based on the perpendicularity condition. Thus, we can form the system:

$\frac{8}{6} =\frac{2-y}{14-x} \\8(14-x) - 6(2-y) = 0$ (1)

$100 = \sqrt{(14-x)^2+(2-y)^2}$ (2)

with solution the coordinates (8, -6) and (20, 10). If you draw a line joining the coordinates (0, 0), (6, 8), (14, 2) and (8, -6) you will get one of the squares that fulfill the conditions of the problem. By repeating this process with the coordinates in (A), the following squares are found:

(0, 0), (6, 8), (14, 2), (8, -6)

(0, 0), (8, 6), (14, -2), (6, -8)
(0, 0), (-6, 8), (-14, 2), (-8, -6)
(0, 0), (-8, 6), (-14, -2), (-6, -8)

Now, notice that the equation of distance between the two points separated a distance of 10 has the trivial solution $(\pm10, 0)$ and $(0, \pm10)$ . By combining this points we get the following squares:

(0, 0), (10, 0), (10, 10), (0, 10)
(0, 0), (0, 10), (-10, 10), (-10, 0)
(0, 0), (-10, 0), (-10, -10), (0, -10)
(0, 0), (0, -10), (-10, -10), (10, 0)

See the attached second attached figure. Therefore, 8 squares can be drawn

8 0

3 years ago

Probability scenarios?

velikii [3]

Here are some probability scenarios.

- Flipping a coin and seeing if it is heads or tails.

- Thinking to yourself: "Is it going to rain or hail?"

- Spinning a spinner and seeing what color it lands on.

- Probability in winning the lottery.

- Having 5 pairs of black shoes and one pair of yellow shoes.

The probability of picking the yellow shoe is 1/5.

There are lots of more examples out there!

<em>Hope helps!-Aparri </em>

4 0

3 years ago

How many years did Vincent Laguardia Gambini tell Judge Chamberlain Haller he'd been practicing law, when asked? (Movie: My Cous

kotykmax [81]

The answer is 17 years

4 0

3 years ago

Solucion de ecuaciones lineales con dos variables

Reptile [31]

Si a, b y r son números reales (y si a y b no son iguales a 0), ax+by = r se denomina ecuación lineal en dos variables. (Las “dos variables” son la x y la y.) Los números a y b se denominan los coeficientes de la ecuación ax+by = r. El número r se denomina constante del eje de ecuación + por = r.

8 0

3 years ago