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Goshia [24]
3 years ago
9

Here is a list of questions about the students and teachers at a school. Select the questions that are statistical questions.

Mathematics
1 answer:
Gwar [14]3 years ago
4 0

Answer: 1. 2. 4. 5. 6.

Step-by-step explanation:

Im pretty sure these are because you have to collect data.

You might be interested in
Solve the formula V = pi r2h for r
Andrei [34K]

Answer:

V=\pi \times r^{2}\times h\\\\\Longrightarrow V=r^{2}\times \left( \pi \times h\right)  \\\\\Longrightarrow r^{2}=\frac{V}{\pi \times h} \\\\\Longrightarrow r=\sqrt{\frac{V}{\pi \times h} }

6 0
2 years ago
Read 2 more answers
What is the slope intercept form of the equation of the line 5x-6y=36
irina1246 [14]
It's an easy problems. what equations are you trying to find?
3 0
3 years ago
Can anybody help me with this ?
BabaBlast [244]

Answer:

1st scenario: 163.3°

2nd scenario: 125.2°

3rd scenario: 142.3°

Step-by-step explanation:

1st scenario: angles 1 and 2 form a linear pair and add up to 180

2nd scenario: angles 1 and 4 are same-side interior and add up to 180

3rd scenario: angles 1 and 8 are corresponding and equal; if angle 7 is 37.7 then angle 8 is 142.3 (they form a linear pair). Since angles 1 and 8 are equal then angle 1 is also 142.3

6 0
3 years ago
Alex wants to fence in an area for a dog park. He has plotted three sides of the fenced area at the points E (1, 5), F (3, 5), a
hram777 [196]

Answer:

(1,1)

Step-by-step explanation:

Given: E, F, G, H denote the three coordinates of the area fenced

To find: coordinates of point H

Solution:

According to distance formula,

length of side joining points (x_1,y_1)\,,\,(x_2,y_2) is equal to \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

So,

EF=\sqrt{(3-1)^2+(5-5)^2}=2\,\,units\\FG=\sqrt{(6-3)^2+(1-5)^2}=\sqrt{9+16}=5\,\,units\\GH=\sqrt{(x-6)^2+(y-1)^2}\\EH=\sqrt{(x-1)^2+(y-5)^2}

Perimeter of a figure is the length of its outline.

EF+FG+GH+EH=16\\2+5+\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16-2-5\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=9

Put (x,y)=(1,1)

\sqrt{(1-6)^2+(1-1)^2}+\sqrt{(1-1)^2+(1-5)^2}=9\\\sqrt{25}+\sqrt{16}=9\\5+4=9\\9=9

This is true.

So, the point (1,1) satisfies the equation \sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=9

So, point H is (1,1).

7 0
3 years ago
Six cakes cost $2.40 so how much do 10 cakes cost?
Gemiola [76]
6c...........2.40\$ \\ 10c...........x \\\\ x=\frac{2.40*10}{6}= \frac{24.0}{6} \\\\ \boxed{x=4\$}
5 0
3 years ago
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