Using the information given, it is found that the class width for this frequency distribution table is of 1.
In this problem, these following classes are given:
0 – 1 14
2 – 3 1
4 – 5 8
6 – 7 12
8 – 9 12
The classes not given, which are 1 - 2, 3 - 4 and 5 - 6, have values of 0.
The <u>difference between the bounds of the classes is of 1</u>, thus, the class width is of 1.
A similar problem is given at brainly.com/question/24701109
Answer:
I’m pretty sure 60 or 56
Step-by-step explanation:
12+12+12+14+16 should be each square/shape I wrong so I’m so sorry I really hope I helped I think my equation is wrong but I hope my answer isn’t
The areas of the circle will represent the population, so:
Area = πr²
5² : 19,465,000
C² : 37,691,000
California circle radius = 6.96 cm
For Texas:
5² : 19,465,000
T² : 25,675,000
Texas circle radius = 5.74 cm
Answer:
Step-by-step explanation:
ST = w + 6,
PR = w
From the diagram given, we can deduce that PR is the midsegment of ∆QST. Therefore, according to the midsegment theorem:
PR = ½ of ST
Plug in the values into the equation and solve for w.
(distributive property of equality)
(subtraction property of equality)
(multiplication property of equality)
(subtraction property of equality)
Divide both sides by -1
Rhombus, <span>opposite angles are congruent
One angle is 35 so the opposite angle will be 35 also.
Sum of interior angles in a rhombus equal 360
360 - 35 - 35 = 290
290/2 = 145 (The other two angles will be congruent)
Answer:
The remaining three angles will be 35</span>°<span>, 145</span>°<span>, 145</span>°<span>
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