Answer:
Table 2
Step-by-step explanation:
We have the tables:
<u>Table 1:</u>
x: 1 2 3 4
y: 2 4 6 8
<u>Table 2:</u>
x: 1 2 3 4
y: 2 4 8 16
<u>Table 3:</u>
x: 1 2 3 4
y: 2 4 7 11
<u>Table 4:</u>
x: 1 2 3 4
y: 2 4 6 10
An exponential growth data set will show a common ratio between y values. Let's look at each of the ratios from each table.
<u>Table 1:</u>
8/6 = 4/3
6/4 = 3/2
Already, we can see that 4/3 ≠ 3/2, which means that this doesn't have a common ratio. So Table 1 is wrong.
<u>Table 2:</u>
16/8 = 2
8/4 = 2
4/2 = 2
The common ratio here is 2, so we know this is correct.
<u>Table 3:</u>
11/7 = 1.57
7/4 = 1.75
Again, we can see that 1/57 ≠ 1.75, so this is wrong.
<u>Table 4:</u>
10/6 = 1.67
6/4 = 1.5
Again, there is no common ratio here, so this is wrong.
The answer is thus Table 2.
Is that all the question says ??
4.36 I think I hope this helps
Answer:
a) probability that is cracked=1/30 (3.33%)
b) probability that is discoloured = 29/600 (4.83%)
c) probability that is cracked and discoloured = 11/600 (1.83%)
Step-by-step explanation:
assuming that each stone is equally likely to be chosen then defining the events C= the stone is cracked , D= the stone is discoloured , N= the stone is neither cracked or discoloured, then
P(C)= number of favourable outcomes/total number of outcomes = 20 stones/600 stones = 1/30 (3.33%)
P(D)= number of favourable outcomes/total number of outcomes = 29 stones/600 stones = 29/600 (4.83%)
the probability that is discoloured and cracked is P(C∩D) , where
P(C∪D)=P(C) + P(D)-P(C∩D)
and
P(C∪D)= 1- P(N)
thus
1- P(N)=P(C) + P(D)-P(C∩D)
P(C∩D)= P(N)+P(C)+P(D) -1
replacing values
P(C∩D)= P(N)+P(C)+P(D)=562/600 + 20/600 + 29/600 -1= 611/600 -1 = 11/600
thus
P(C∩D)= 11/600 (1.83%)
