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Arada [10]
3 years ago
13

Csqrt%7B6%7D%20%7D%20%7D%20%7D%20" id="TexFormula1" title=" 3 \sqrt{2 \times 2 \sqrt{8 \times \sqrt{3 \times \sqrt{6} } } } " alt=" 3 \sqrt{2 \times 2 \sqrt{8 \times \sqrt{3 \times \sqrt{6} } } } " align="absmiddle" class="latex-formula">
Help? ​

Mathematics
1 answer:
vazorg [7]3 years ago
4 0
The answer to your question is 12.95

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Simplify 4x^2 / 3 X 6/ 8x
Leokris [45]

Answer:

x

Step-by-step explanation:

The answer to

\frac{4 {x}^{2} }{3}  \times  \frac{6}{8x}

= x

3 0
2 years ago
Read 2 more answers
Solve the equation:<br> <img src="https://tex.z-dn.net/?f=%5Csqrt%7B%28y%2B2%29%5E%7B2%7D%20%7D%20%3D%20c" id="TexFormula1" titl
r-ruslan [8.4K]

Answer:

Not quite sure what you're meant to solve for, but my solution is below.

Step-by-step explanation:

The square root of a value squared is the absolute value of that value, because squaring removes all negatives. Therefore:

\sqrt{(y+2)^2}=c

\mid y+2\mid=c

\mid y\mid =c-2

Hope this helps!

7 0
3 years ago
Samir is an expert marksman. When he takes aim at a particular target on the shooting range, there is a 0.950.950, point, 95 pro
Vinvika [58]

Answer:

40.1% probability that he will miss at least one of them

Step-by-step explanation:

For each target, there are only two possible outcomes. Either he hits it, or he does not. The probability of hitting a target is independent of other targets. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

0.95 probaiblity of hitting a target

This means that p = 0.95

10 targets

This means that n = 10

What is the probability that he will miss at least one of them?

Either he hits all the targets, or he misses at least one of them. The sum of the probabilities of these events is decimal 1. So

P(X = 10) + P(X < 10) = 1

We want P(X < 10). So

P(X < 10) = 1 - P(X = 10)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 10) = C_{10,10}.(0.95)^{10}.(0.05)^{0} = 0.5987

P(X < 10) = 1 - P(X = 10) = 1 - 0.5987 = 0.401

40.1% probability that he will miss at least one of them

7 0
2 years ago
Consider the functions f(x)=x−7 and g(x)=4x^3.
maxonik [38]

Answer:

I think A is correct

7 0
3 years ago
In a classroom of 3333 students, the ratio of boys to girls is 3:83:8. How many boys are in the class?
Paladinen [302]
3x+8x=3333
11x=3333
X= 303
Boys 3x,so 3*303=909
6 0
3 years ago
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