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weeeeeb [17]
2 years ago
12

Suppose you have developed a scale that indicates the brightness of sunlight . Each category in the tablet is 5 time brighter th

an the category above it. For example, a day that is dazzling is 5 times brighter than a day that is radiant . How many times brighter is a dazzling day than a dim day
Dim is 2
Mathematics
2 answers:
Ostrovityanka [42]2 years ago
8 0

If you could put in the tablet.

But the equation would be: a_n = a_1 (r)^(x-1)

with a_1 being the first category

r being 5 how much brighter than the day before is

and x is what nr. in the sequence it is

But I'm not sure as I cannot see the table

umka21 [38]2 years ago
4 0

Answer:

10 Times

Step-by-step explanation:

From the statement "Each category in the tablet is 5 time brighter than the category above it." It can be inferred that the brightest portion of the day must be 5 times the less bright day. This means that:

Dim = 2

Brightest = 5 times dim

               = 5 * 2

               = 10

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I WILL GIVE BRAINLIEST!!!!
-Dominant- [34]
The equation of the line shown is y = 1/3x - 1

Because the slope is rise/run and in this case you rise 1 and run 3. And the y-intercept is -1 because that is where the line crosses the y-axis.

hope this helps :)
8 0
3 years ago
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if a line contains a the point (0,-1) and has a slope of 2 which of the following points also lies on the line
Fudgin [204]

Answer:

I think 1 1

Step-by-step explanation:

7 0
3 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
Writetwo ways you could use to find 5+8
Igoryamba

5+8=13

or

13-8=5

Hope this helps!

5 0
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swat32
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8 0
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