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11111nata11111 [884]
3 years ago
10

What is the following product?

Mathematics
1 answer:
allochka39001 [22]3 years ago
4 0

The product is 10√14 +8√21 +30√3 +36√2

Step-by-step explanation:

Given the question as;

( 2√7 +3√6)(5√2+4√3)

Distribute

2√7(5√2+4√3) + 3√6 (5√2+4√3)

open brackets

2√7 × 5√2 + 2√7 × 4√3 + 3√6×5√2+3√6×4√3

2×5×√7×√2 + 2×4×√7×√3+3×5×√6×√2 + 3×4×√6×√3

10×√14 +8√21+15√12+12√18-----------------------------simplify √12 and √18 further

10√14+  8√21 + 15×√4×√3 + 12×√9×√2

10√14 +8√21 +15×2×√3 + 12×3×√2

10√14 +8√21 +30√3 +36√2

Learn More

Surds operation : brainly.com/question/13658479

Keyword : Product

#LearnwithBrainly

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PLS HELP WILL MARK BRAINLIEST!
Allisa [31]

Answer:

-41

Step-by-step explanation:

if you plug -2 everywhere you see an x and 1 everywhere you see a y, and then simplify you get -41

4 0
2 years ago
Read 2 more answers
Find the probability that the person is frequently or occasionally involved in charity work.
Schach [20]
Given the table below which shows the result of a survey that asked 2,881 people whether they are involved in any type of charity work.

\begin{tabular}
{|c|c|c|c|c|c|}
 &Frequently&Occassionally&Not at all&Total\\[1ex]
Male&227&454&798&1,479\\
Female &205&450&747&1,402\\
Total&432&904&1,545&2,881
\end{tabular}

Part A:

If a person is chosen at random, the probability that the person is frequently or occassinally involved in charity work is given by

P(being \ frequently \ involved \ or \ being \ occassionally \ involved)\\ \\= \frac{432}{2881} + \frac{904}{2881} = \frac{1336}{2881}=\bold{0.464}



Part B:

If a person is chosen at random, the probability that the person is female or not involved in charity work at all is given by

P(being
 \ female \ or \ not \ being \ involved)\\ \\= 
\frac{1402}{2881} + \frac{1545}{2881}-\frac{747}{2881} = 
\frac{2200}{2881}=\bold{0.764}



Part C:

If a person is chosen at random, the probability that the person is male or frequently involved in charity work is given by

P(being
 \ male \ or \ being \ frequently \ involved)\\ \\= 
\frac{1479}{2881} + \frac{432}{2881}-\frac{227}{2881} = 
\frac{1684}{2881}=\bold{0.585}



Part D:

If a person is chosen at random, the probability that the person is female or not frequently involved in charity work is given by

P(being
 \ female \ or \ not \ being \ frequently \ involved)\\ \\= 
\frac{1402}{2881} + \frac{904}{2881} + \frac{1545}{2881}-\frac{450}{2881}-\frac{747}{2881} = 
\frac{2654}{2881}=\bold{0.921}



Part E:

The events "being female" and "being frequently involved in charity work" are not mutually exclusive because being a female does not prevent a person from being frequently involved in charity work.

Indeed from the table, there are 205 females who are frequently involved in charity work.

Therefore, the answer to the question is "No, because 205 females are frequently involved charity work".
4 0
3 years ago
Click on all the points that are solutions to y = x^2 - 9.
Oliga [24]

9514 1404 393

Answer:

  B) (2, -5)

  D) (3, 0)

Step-by-step explanation:

I find it convenient to let a graphing calculator plot the points and the graph.

The only two points on the graph of the curve are ...

  (2, -5) and (3, 0)

3 0
3 years ago
How do you fine out if y is a linear function of x on a graph
mote1985 [20]

Answer: If a vertical line can be drawn to touch the graph of a function in more than one place, then is NOT a function of . If it is not possible to draw a vertical line to touch the graph of a function in more than one place, then y is a function of x.

6 0
2 years ago
A normally distributed random variable with mean 4.5 and standard deviation 7.6 is sampled to get two independent values, X1 and
mr Goodwill [35]

Answer:

Bias for the estimator = -0.56

Mean Square Error for the estimator = 6.6311

Step-by-step explanation:

Given - A normally distributed random variable with mean 4.5 and standard deviation 7.6 is sampled to get two independent values, X1 and X2. The mean is estimated using the formula (3X1 + 4X2)/8.

To find - Determine the bias and the mean squared error for this estimator of the mean.

Proof -

Let us denote

X be a random variable such that X ~ N(mean = 4.5, SD = 7.6)

Now,

An estimate of mean, μ is suggested as

\mu = \frac{3X_{1} + 4X_{2}  }{8}

Now

Bias for the estimator = E(μ bar) - μ

                                    = E( \frac{3X_{1} + 4X_{2}  }{8}) - 4.5

                                    = \frac{3E(X_{1}) + 4E(X_{2})}{8} - 4.5

                                    = \frac{3(4.5) + 4(4.5)}{8} - 4.5

                                    = \frac{13.5 + 18}{8} - 4.5

                                    = \frac{31.5}{8} - 4.5

                                    = 3.9375 - 4.5

                                    = - 0.5625 ≈ -0.56

∴ we get

Bias for the estimator = -0.56

Now,

Mean Square Error for the estimator = E[(μ bar - μ)²]

                                                             = Var(μ bar) + [Bias(μ bar, μ)]²

                                                             = Var( \frac{3X_{1} + 4X_{2}  }{8}) + 0.3136

                                                             = \frac{1}{64} Var( {3X_{1} + 4X_{2}  }) + 0.3136

                                                             = \frac{1}{64} ( [{3Var(X_{1}) + 4Var(X_{2})]  }) + 0.3136

                                                             = \frac{1}{64} [{3(57.76) + 4(57.76)}]  } + 0.3136

                                                             = \frac{1}{64} [7(57.76)}]  } + 0.3136

                                                             = \frac{1}{64} [404.32]  } + 0.3136

                                                             = 6.3175 + 0.3136

                                                              = 6.6311

∴ we get

Mean Square Error for the estimator = 6.6311

6 0
3 years ago
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