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lbvjy [14]
3 years ago
6

Suppose the weekly amount of money spent on maintenance and repairs by a company is NOT approximately normally distributed. Howe

ver, its mean and standard deviation are known to be $400 and $20, respectively. If $450 is budgeted for next week, what can you say about the probability that the actual costs will exceed the budgeted amount
Mathematics
1 answer:
UNO [17]3 years ago
8 0

Answer:

Probability = 0.00621

Step-by-step explanation:

We are given;

Population mean; μ = 400

Population standard deviation; σ = 20

Raw score; x¯ = 450

the probability that the actual costs will exceed the budgeted amount will be the p-value.

Thus, let's first find the z-score after which we will find the p-value.

z = (x¯ - μ)/σ

z = (450 - 400)/20

z = 50/20

z = 2.5

From online p-value from z-score calculator attached, using z = 2.5; one tailed hypothesis; significance level of 0.05, we have;

P-value = 0.00621

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(a) If G is a finite group of even order, show that there must be an element a = e, such that a−1 = a (b) Give an example to sho
Dahasolnce [82]

Answer:

See proof below

Step-by-step explanation:

First, notice that if a≠e and a^-1=a, then a²=e (this is an equivalent way of formulating the problem).

a) Since G has even order, |G|=2n for some positive number n. Let e be the identity element of G. Then A=G\{e} is a set with 2n-1 elements.

Now reason inductively with A by "pairing elements with its inverses":

List A as A={a1,a2,a3,...,a_(2n-1)}. If a1²=e, then we have proved the theorem.

If not, then a1^(-1)≠a1, hence a1^(-1)=aj for some j>1 (it is impossible that a^(-1)=e, since e is the only element in G such that e^(-1)=e). Reorder the elements of A in such a way that a2=a^(-1), therefore a2^(-1)=a1.

Now consider the set A\{a1,a2}={a3,a4,...,a_(2n-1)}. If a3²=e, then we have proved the theorem.

If not, then a3^(-1)≠a1, hence we can reorder this set to get a3^(-1)=a4 (it is impossible that a^(-1)∈{e,a1,a2} because inverses are unique and e^(-1)=e, a1^(-1)=a2, a2^(-1)=a1 and a3∉{e,a1,a2}.

Again, consider A\{a1,a2,a3,a4}={a5,a6,...,a_(2n-1)} and repeat this reasoning. In the k-th step, either we proved the theorem, or obtained that a_(2k-1)^(-1)=a_(2k)

After n-1 steps, if the theorem has not been proven, we end up with the set A\{a1,a2,a3,a4,...,a_(2n-3), a_(2n-2)}={a_(2n-1)}. By process of elimination, we must have that a_(2n-1)^(-1)=a_(2n-1), since this last element was not chosen from any of the previous inverses. Additionally, a_(2n1)≠e by construction. Hence, in any case, the statement holds true.

b) Consider the group (Z3,+), the integers modulo 3 with addition modulo 3. (Z3={0,1,2}). Z3 has odd order, namely |Z3|=3.

Here, e=0. Note that 1²=1+1=2≠e, and 2²=2+2=4mod3=1≠e. Therefore the conclusion of part a) does not hold

7 0
3 years ago
Subtract 23.4 – 8.78 =_____
anastassius [24]

Answer:

14.62

Step-by-step explanation:

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3 years ago
ACCORDING to the rate of electricity charge, the minimum charge up to 20 unit is Rs 80 and the cost per unit from 21 units to 25
azamat

Answer:

180

Step-by-step explanation:

let x = total units of electricity after 250 units

the total units of electricity consumed can be determined using the formula below

80 + 7.3x = 1248

7.3x = 1248 - 80

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divide both sides of the equation by 7.3

x = 1168 / 7.3

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total electricity consumed = 160 + 20 = 180

160

6 0
3 years ago
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