Please help now I need to find the missing perimeter please show the work
Answers
Answer:
30 cm
Step-by-step explanation:
The first one it says 4cm. That means all sides equal to 4 cm.
The second one it says 5 cm. That means all sides equal to 5 cm.
Lets do the second shape.
Since you see 6 sides with 5cm.
You do 6 times 5. Which equals to 30.
You add the label, so 30cm.
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When 12 is subtracted from 3 times a number, the result is 24. Find the number.A) 8B) 12C) 4D) 6
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Answers
Answer:
Option D
Step-by-step explanation:
Given f(x) =
g(x) = 2x + 3
Since,
This function is defined for the denominator is not equal to zero.
(2x + 3) ≠ 0
x ≠
Therefore, Option D will be the correct option.
-16
12
-8
-20 -16 -12
-
-4
48
12
16 2024
-8
-12
16
If the graph of the second equation in the system passes through (-12, 20) and (4,12), which statement is true?
Answers
Answers
Answer:
0.0108
Step-by-step explanation:
Let X denote the number of uranium fission tracks occurring on the average 5 per square centimetre.We need to find the probability that a 2cm² sample of this zircon will reveal at most three tracks. X follows Poisson distribution, λ = 5 and s = 2.
k = λs = 5×2 = 10
Since we need to reveal at most three tracks the required probability is:
P (X≤3) = P (X =0) + P (X =1) + P (X =2) + P (X =3)
P (X≤3) = (((e^-10) × (10)⁰)/0!) + (((e^-10) × (10)¹)/1! + (((e^-10) × (10)²)/2! + (((e^-10) × (10)3)/3!
P (X≤3) = 0.0004 + 0.0005 +0.0023 +0.0076
P (X≤3) = 0.0108
Therefore, the probability that a 2cm² sample of this zircon will reveal at most three tracks is 0.0108
Answer:
p(x = 3, λ = 5) = 0.14044
Step-by-step explanation:
Given
λ = 5 (the average number of tracks per square centimeter)
ε = 2.718 (constant value)
x = 3 (the variable that denotes the number of successes that we want to occur)
p(x,λ) = probability of x successes, when the average number of occurrences of them is λ
We can use the equation
p(x,λ) = λˣ*ε∧(-λ)/x!
⇒ p(x = 3, λ = 5) = (5)³*(2.718)⁻⁵/3!
⇒ p(x = 3, λ = 5) = 0.14044
Answers
Answers
120.069 kilos times 0.36 pounds= 43.22 pounds raised
Answer:
£43.23
Step-by-step explanation:
8645 cans divided by 72 per kilo is 120.069 kilos of cans
120.069 kilos times 0.36 pounds
Rounded up to 43.23
Sample Space: Tutorial
Activity
In this exercise, you'll use the formula for the probability of the complement of an event.
Another game you've set up at casino night involves rolling a fair six-sided die followed by tossing a fair coin. In this game, players earn points
depending on the number they get on the die and which side of the coin turns up. For example, the player earns 5 points for getting (2, tails).
Question 1
Find the total number of possible outcomes in each trial of this game.
Answers
Answer:
The number of possible outcomes in each trial of this game is 12
Step-by-step explanation:
Given
Rolling of a 6 sided die followed by tossing of a fair coin
Required
Number of possible outcomes
The first step is to list out the possible outcomes of rolling a die and tossing a coin
Rolling a fair die = {1,2,3,4,5,6}
Tossing a coin = {Head, Tail}
Let Head be represented by H and Tail be represented by T;
So,
Rolling a fair die = {1,2,3,4,5,6}
Tossing a coin = {H, T}
The question states that a roll of a 6 sided die is followed by a toss of a fair coin
This means that each trial is {A roll of die and A toss of coin}
So, the sample space is as follows
Sample Space = {1H, 2H, 3H, 4H, 5H, 6H, 1T, 2T, 3T, 4T, 5T, 6T}
Number of outcomes in the sample space is 12.
Hence, the number of possible outcomes in each trial of this game is 12
Answer:
the total number of possible outcomes in each trial of this game is 12
Step-by-step explanation:
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