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vekshin1
3 years ago
5

Find the greatest common factor of 10, 30,10,30,10, comma, 30, comma and 454545.

Mathematics
1 answer:
borishaifa [10]3 years ago
5 0

Answer:

GCF = 5

Step-by-step explanation:

Given:

The numbers are given as:

10, 30 and 45

A factor of a number is a number by which the given number is evenly divisible.

Example: 2 is a factor of 4 as 4\div 2 =2. So, 4 is evenly divisible by 2 and hence 2 is a factor of 4.

Now, factors of 10 = 1, 2, <u>5</u>, 10.

Factors of 30 = 1, 2, 3, <u>5</u>, 6, 10, 15, 30.

Factors of 45 = 1, 3, <u>5</u>, 9, 15, 45.

So, the greatest common factor in all the three numbers is 5.

Hence, the greatest common factor of 10, 30 and 45 is 5.

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2 years ago
A manufacturing company has two retail outlets. It is known that 30% of all potential customers buy
aleksandrvk [35]

Answer:

(a) A y P(A) = 0.4 (b) \bar{B} y P(\bar{B})=0.5 (c) \bar{A}∪\bar{B} y P(\bar{A}∪\bar{B}) = 0.9 (d) \bar{A}∩\bar{B} y P(\bar{A}∩\bar{B})=0.2

Step-by-step explanation:

A was defined as the event that a potential customer, randomly chosen, buys from outlet 1 in the original problem statement. We know that B denotes the event that a randomly chosen customer buys from outlet 2. So

P(A\cap \bar{B}) = 0.3, P(B\cap \bar{A}) = 0.4 and P(A\cap B) = 0.1

(a) P(A) = P(A\cap (B\cup\bar{B})) = P(A\cap B) + P(A\cap \bar{B}) = 0.1 + 0.3 = 0.4

(b)  P(B) = P(B\cap (A\cup\bar{A})) = P(B\cap A) + P(B\cap \bar{A}) = 0.1 + 0.4 = 0.5

P( \bar{B}) = 1-P(B) = 1-0.5 = 0.5

(c) The customer does not buy from outlet 1 is the complement of A, i.e.,  \bar{A}, and the customer does not buy from outlet 2 is the complement of B, i.e.,  \bar{B}, so, the customer does not buy from outlet 1 or does not buy from outlet 2 is  \bar{A}∪ \bar{B} and P(\bar{A}∪ \bar{B}) = P((A\cap B)^{c}) by De Morgan's laws

P((A\cap B)^{c})  = 1-P(A∩B)=1-0.1=0.9

(d) The customer does not buy from outlet 1 is the complement of A, and the customer does not buy from outlet 2 is the complement of B, so we have that the statement in (d) is equivalent to \bar{A}∩\bar{B} and P( \bar{A}∩\bar{B}) = P((AUB)^{c}) by De Morgan's laws, and

P((AUB)^{c}) = 1-P(A∪B)=1-[P(A)+P(B)-P(A∩B)]=1-[0.4+0.5-0.1]=1-0.8=0.2

8 0
3 years ago
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Aleks04 [339]

Answer:

99% Confidence interval: (0.185,0.375)                                                  

Step-by-step explanation:

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Sample size, n = 150

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\hat{p} = \dfrac{x}{n} = \dfrac{42}{150} = 0.28

99% Confidence interval:

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0.28\pm 2.58(\sqrt{\frac{0.28(1-0.28)}{150}}) = 0.28\pm 0.095\\\\=(0.185,0.375)

The​ 99% confidence interval for the true proportion of new cars with faulty catalytic converters is​ (0.185,0.375)

8 0
3 years ago
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