Answer:
Look at the place after the thousandths, it is less than 5.
0.8344
So don't add +1 to the thousandths place.
Rounded to the nearest thousandths would be:
<h2>0.834</h2>
The total area of the face of the watch to the nearest tenth of a square centimemter is 9.0 cm²
Since an electronics company is designing a watch with a face that is in the shape of a hexagon and two congruent trapezoids attached. The heights of the trapezoids and the apothem of the hexagon measure 2 centimeters each, and the length of the shorter base of each trapezoid is 1.5 centimeters, the radii of the hexagon, and the base of the trapezoid form a triangle of
- height, h = apothem of the hexagon = 2 cm and
- base, b = length of shorter base of trapezoid.
<h3>Area of the triangle</h3>
So, the area of this triangle is A = 1/2bh
= 1/2 × 1. 5 cm × 2 cm
= 1.5 cm × 1 cm
= 1.5 cm²
<h3>Area of the hexagon</h3>
Since there are 6 of such triangles in the hexagon, the area of the hexagon, A' = 6A
= 6 × 1.5 cm²
= 9.0 cm²
So, the total area of the face of the watch to the nearest tenth of a square centimemter is 9.0 cm²
Learn more about area of a hexagon here:
brainly.com/question/369332
Answer:
how many bags because you just said bags?
Answer:
D. There were no significant effects.
Step-by-step explanation:
The table below shows the representation of the significance level using the two-way between subjects ANOVA.
Source of Variation SS df MS F P-value
Factor A 10 1 10 0.21 0.660
Factor B 50 2 25 0.52 0.6235
A × B 40 2 20 0.42 0.6783
Error 240 5 48 - -
Total 340 10 - - -
From the table above , the SS(B) is determined as follows:
SS(B) = SS(Total)-SS(Error-(A×B)-A)
= 340-(240-40-10)
= 50
A researcher computes the following 2 x 3 between-subjects ANOVA;
k=2
n=3
N(total) = no of participants observed in each group =11
df for Factor A= (k-1)
=(2-1)
=1
df for Factor B = (n-1)
=(3-1)
=2
df for A × B
= 2 × 1
= 2
df factor for total
=(N-1)
=11-1
=10
MS = SS/df
Thus, from the table, the P-Value for all data is greater than 0.05, therefore we fail to reject H₀.
