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Alenkasestr [34]
3 years ago
6

here are some numbers 8 11 13 18 25 37 , 8 13 18 is an arithmetic progression and the rule is to add 5 use 3 of the numbers to m

ake a different arithmetic progression.
Mathematics
1 answer:
chubhunter [2.5K]3 years ago
3 0

Answer:

11, 18, 25

Step-by-step explanation:

add 7

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Irene can snap 48 times in 8 minutes at a constant rate. How many
vitfil [10]
6 snaps a minute if you multiply 9 x 6 is 54
8 0
3 years ago
Read 2 more answers
Given various values of the linear functions f(x) and g(x) in the table, determine the y-intercept of (f- g)(x).
lisabon 2012 [21]

The y-intercept of linear function (f- g)(x) is (0,9)

<h3>How to determine the y-intercept?</h3>

The table of values is given as:

x   -6 -4 -1 3 4

f(x) 15 11 5 -3 -5

g(x) -36 -26 -11 9 14

The equations of the functions is calculated using:

y = \frac{y_2 -y_1}{x_2-x_1} * (x -x_1) + y_1

So, we have:

f(x) = \frac{11 -15}{-4 + 6} * (x + 6) + 15

Evaluate

f(x) = -2x + 3

Also, we have:

g(x) = \frac{-26 + 36}{-4 + 6} * (x + 6) - 36

Evaluate

g(x) = 5x - 6

Next, we calculate (f - g)(x) using:

(f - g)(x) = f(x) - g(x)

This gives

(f - g)(x) = -2x + 3 - 5x + 6

Substitute 0 for x

(f - g)(0) = -2(0) + 3 - 5(0) + 6

Evaluate

(f - g)(0) = 9

Hence, the y-intercept of (f- g)(x) is (0,9)

Read more about linear functions at:

brainly.com/question/24896196

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7 0
2 years ago
Solve for n. 1/2n+3/4n=12
sveta [45]
First you add 1/2 and 3/4:
5/4n

5/4n = 12 so you divide both sides by 5/4
That gives you the answer of n= 9.6.

Hope this helped :)
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
Consider a t distribution with 7 degrees of freedom. Compute P(-1.29 &lt; t &lt; 1.29). Round your answer to at least three deci
marusya05 [52]

Answer:

Step-by-step explanation:

Given that there is a t distribution with 7 degrees of freedom.

P(-1.29 < t < 1.29)=1-0.2380

=0.7620

b) Now there is a different t distribution with 18 df.

When P(t\leq c)=0.05

c=-1.733

7 0
3 years ago
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