5z>15-----subtract 2 on both sides
z>3----divide by 5
therefore, z must equal -10, or -4
Answer:
18a(2bc + 3d)
Step-by-step explanation:
36abc + 54ad
Step 1: Find the Highest Common Factor of each
36abc = 2×2×3×3×a×b×c = 18a × 2bc
54ad = 2×3×3×3×a×d = 18a × 3d
HCF = 2×3×3×a = 18a
Step 2: Factor out with HCF
18a(2bc + 3d)
Answer:
the gas 4$
Step-by-step explanation:
5x=20
divide 5/5 to cancel the 5 then 20/5 and there you have it 4
Answer:
7. 20 + 5 + 3/10 + 6/100
8. 7 + 4/10 + 2/100 + 5/1000
Step-by-step explanation:
We are writing two numbers in expanded form, to help, we can use the guide above both numbers. Expanded form is all about breaking apart a number by place value and putting them in an additional sequence.
Let's look at 7, 25.36.
There are the tens, ones, tenths, and hundredths place being used, therefore :
20 + 5 + 3/10 + 6/100
We can do the same for 8, 7.425.
There are ones, tenths, hundredths, and thousandths being used, therefore :
7 + 4/10 + 2/100 + 5/1000
Answer:
a) S = {1, 2, 3}
b) P(odd number) = 
c) No
d) Yes
Step-by-step explanation:
a) The sample space is the set of all possible outcomes. By definition, the elements of a set should not be repeated. Hence, the sample space S = {1, 2, 3}
However, the sample is not equiprobable because each element has different probabilities.
b) P(odd number) = 
Note that the odd numbers are 1 (on three faces) and 3 (on one face).
c) The fact the die has been biased does not change the possible outcomes. It only changes the probability of getting any given number.
d) Because the 3-face has been loaded, this probability changes. In fact, it is calculated thus:
Let's assume the probability for 1 or 2 is 
. Then that of 3 is 
(because it is twice the others). The sum of probabilities must be 1.
P(odd number) = 
Prob(1) + Prob(3)
= 
= 
