Answer:
ai) n(E⋂C) = ∅ = null
n(E⋂G) = 4
aii) see attachment
bi) n(C⋂G) = x = 1
bii) n(G) only = 3
Step-by-step explanation:
Let chemistry = C
Economic = E
Government = G
n(E) = 12
n(G) = 8
n(C) = 7
ai) number of pupils for economics and chemistry = 0
number of pupils for economics and government = 4
The set notation for both:
n(E⋂C) = ∅ = null
n(E⋂G) = 4
aii) find attached the Venn diagram
bi) n(C⋂G) = ?
Let number of n(C⋂G) = x
From the Venn diagram
n(C) only = 12-4 = 8
n(G) only = 8-(4+x) = 4-x
n(E) only = 7-x
n(E⋂C⋂G) = 0
n(E⋂C) = 0
n(E⋂G) = 4
Total: 8+ 4-x + 7-x + x + 0+0+4 = 22
23 -x = 22
23-22 = x
x = 1
n(C⋂G) = x = 1
Number of pupils that take both chemistry and government = 1
(bii) government only = n(G) only = 4-x
n(G) only = 4-1 = 3
Number of students that take government only = 3
Answer:
<em>A) (-5,7)</em>
Step-by-step explanation:
<u>Functions and Relations</u>
A set of values A can have a relation with another set B as long as at least one element of A has at least one image in B. Functions are special relations where each element of A (the domain of the function) has one and only one image on B (the range of the function).
By looking at the options, we can see that x=9, x=-8, and x=-1 already have defined values in Y, so if we define another value for any of them the relation will stop being a function. The only possible choice to preserve the function is the option
To find the specification limit such that only 0.5% of the bulbs will not exceed this limit we proceed as follows;
From the z-table, a z-score of -2.57 cuts off 0.005 in the left tail; given the formula for z-score
(x-μ)/σ
we shall have:
(x-5000)/50=-2.57
solving for x we get:
x-5000=-128.5
x=-128.5+5000
x=4871.50
B) 65.84 add 67.5 and 6.443 and divide sum by 1.123
