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FromTheMoon [43]
3 years ago
11

HELP MEHHHHHHHHH. thanks.

Mathematics
2 answers:
jok3333 [9.3K]3 years ago
8 0
The Answer is A .....
maxonik [38]3 years ago
3 0

Answer:A

The mean is the average of all numbers, in this example it's to national averages which will have many numbers.

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I really need help with this chart !! 20 points !!
vichka [17]
Lean is 5 and the quasretiv
5 0
4 years ago
X3+2x=110
olchik [2.2K]
x^3+2x=110 \\  \\ test:4.5 \\ (4.5)^3+2(4.5)=110 \\ 91.125+9=110 \\ 100.125=110 \\ false \\  \\ test:4.75 \\ (4.75)^3+2(4.75)=110 \\ 107.17+9.5=110 \\ 116.67=110 \\ false \\  \\ test:4.6 \\ (4.6)^3+2(4.6)=110 \\ 97.33+9.2=110 \\ 106.53=110 \\ false \\  \\ test:4.65 \\ (4.65)^3+2(4.65)=110 \\ 100.7+9.3=110 \\ 110=110 \\ true \\  \\ solution:4.65
7 0
3 years ago
Simplify the expression
Vaselesa [24]
<h2>Answer; (-1024).(-16384)</h2><h2> = 16777216</h2>
3 0
3 years ago
Read 2 more answers
Find the probability that event A will appear at least three times in four independent tests if the probability of occurrence of
polet [3.4K]

This kind of problems are solved using Bernoulli's distribution. Everytime you have a win/lose scenario, and you know the probability p of winning, and you want k successes over n trials, you have the following probability:

\displaystyle P(k\text{ successes over }n\text{ trials}) = \binom{n}{k}p^k(1-p)^{n-k}

You want the probability of having at least three successes, i.e. you are interested in the cases k=3 and k=4. The corresponding probabilities are

\displaystyle P(3 \text{ successes}) = \binom{4}{3}0.6^3 \cdot 0.4 = 4\cdot 0.216 \cdot 0.4 = 0.3456

\displaystyle P(4 \text{ successes}) = \binom{4}{4}0.6^4 = 0.1296

So, the total probability is 0.3456+0.1296 = 0.4752

5 0
4 years ago
Given the following functions, find the indicated values.
schepotkina [342]

Answer:f(5)

Step-by-step explanation:

8 0
3 years ago
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