The box plots below show the distribution of grades in one class on two tests. 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
Test 1 Grades
72 73 74 75 76 77 78 79 80 81 82
3 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
Test 2 Grades
Which measures of center and variability would be best to use when making comparisons of the two data sets?
mean and MAD
mean and IQR
O median and MAD
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i use my brain for prombles like this maybe you could try it

step-by-step explanation:

yes

step-by-step explanation:

so of course you want to start by plugging the 5 in which goes to 7(5)+3(5)=50 because 7*5=35 and 3*5=15 and then add them together, then plug in the 5 to the other equation as well 9(5)+5=50 because 9*5= 45 and then add 5 and so yes they are equivalent.

answer:   the paragraph proof is mentioned below.

step-by-step explanation:

given: in a δxwy where

such that, ∠wzx ≅ ∠yzx and zw≅zy

we have to prove that: segment zx is a perpendicular bisector of segment wy.

prove: since, zw≅zy ⇒ ∠wzy is a supplementary angle.

then, ∠wzy= 180°

⇒∠wzx + ∠yzx = 180° ( because xz divides ∠wzy into two parts)

but, ∠wzx ≅ ∠yzx ( given)

⇒∠wzy+∠wzy=180° ( by substitution)

⇒2∠wzy=180°

⇒∠wzy=90°

⇒∠wzy=∠yzx =90°

therefore, xz⊥wy where wz=zy

⇒zx is a perpendicular bisector of wy.

since we know that the two functions are equal, f (x) = g

(x), therefore when we equate the two equations the result should be similar to

3x = x – 4:

a. f(x) = 3x, g(x) = x – 4

equating the two functions:

3x = x – 4             (correct)

so we now got our answer (a)