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Lilit [14]
3 years ago
6

Also can anyone help with these problems too?

Mathematics
1 answer:
d1i1m1o1n [39]3 years ago
7 0

Answer:

6. a

------

7. b

------

8. d

------

9. d

------

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Answer:

4

Step-by-step explanation:

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Sean buys a packet of perfumed candles that contains 12 lavender, 8 vanilla, and 20 rose-perfumed candles. He randomly takes
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05.

Step-by-step explanation:

trust me dude i just took this question :D

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3 years ago
Find the number of elements in A 1 ∪ A 2 ∪ A 3 if there are 200 elements in A 1 , 1000 in A 2 , and 5, 000 in A 3 if (a) A 1 ⊆ A
lina2011 [118]

Answer:

a. 4600

b. 6200

c. 6193

Step-by-step explanation:

Let n(A) the number of elements in A.

Remember, the number of elements in A_1 \cup A_2 \cup A_3 satisfies

n(A_1 \cup A_2 \cup A_3)=n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)

Then,

a) If A_1\subseteq A_2, n(A_1 \cap A_2)=n(A_1)=200, and if A_2\subseteq A_3, n(A_2\cap A_3)=n(A_2)=1000

Since A_1\subseteq A_2\; and \; A_2\subseteq A_3, \; then \; A_1\cap A_2 \cap A_3= A_1

So

n(A_1 \cup A_2 \cup A_3)=\\=n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)=\\=200+1000+5000-200-200-1000-200=4600

b) Since the sets are pairwise disjoint

n(A_1 \cup A_2 \cup A_3)=\\n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)=\\200+1000+5000-0-0-0-0=6200

c) Since there are two elements in common to each pair of sets and one element in all three sets, then

n(A_1 \cup A_2 \cup A_3)=\\=n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)=\\=200+1000+5000-2-2-2-1=6193

8 0
2 years ago
As a store manger you would like to find out the average time it takes to unload the truck which delivers the merchandise for yo
Georgia [21]

Answer:

The upper boundary of the 95% confidence interval for the average unload time is 264.97 minutes

Step-by-step explanation:

We have the standard deviation for the sample, but not for the population, so we use the students t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 35 - 1 = 35

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 34 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.9}{2} = 0.975([tex]t_{975}). So we have T = 2.0322

The margin of error is:

M = T*s = 2.0322*30 = 60.97

The upper end of the interval is the sample mean added to M. So it is 204 + 60.97 = 264.97

The upper boundary of the 95% confidence interval for the average unload time is 264.97 minutes

4 0
3 years ago
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He had 6 golf balls to begin with
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