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3 years ago

A diver at -60 feet stops every 5 feet before reaching the surface. How many stops will the

Mathematics

2 answers:

Art [367]3 years ago

5 0

The diver will make 12 stops before reaching the surface.

zepelin [54]3 years ago

4 0

You have to imagine -60 as a positive number, or in absolute value which is 60. After that, you divide 60 by 5 to find out how many times the diver will stop before finally making it up to the surface. 60/5= 12. The diver will stop 12 times.

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297.3 = 2hundred, 9tens, 7 ones, and 3 tenths. Regroup 297.3 using 2 hundred, 9 tens, 5 ones, and ? tenths

Mashcka [7]

23 would be the answer hope it helps

5 0

3 years ago

Determine the equation of the linear graph that passes through (-2,3) and (4,6). State your answer in the

Lena [83]

The linear equation is y = 4 + 0.5x

<h3>How to determine the equation?</h3>

The points are given as: (-2,3) and (4,6)

The equation is then calculated using:

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

Substitute the known values

$y = \frac{6-3}{4+2} *(x+2) + 3$

Evaluate

y = 0.5 *(x+2) + 3

Expand

y = 0.5x + 1 + 3

Evaluate the sum

y = 0.5x + 4

Rewrite as:

y = 4 + 0.5x

Hence, the equation is y = 4 + 0.5x

Read more about linear equations at:

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3 0

2 years ago

Work out the area of the shaded region.

Makovka662 [10]

Answer: Area of the shaded region is 6.867 square cm

Step-by-step explanation:

The shaded area corresponds to the difference of the area of the rectangle and the area of the semicircle.

Area of the rectangle = 4 times 8 = 32 cm^2

Area of the semicircle = 0.5 * pi * r^2 (where r is the radius, i.e., half of the diameters, i.e., 4 cm). So Area = 0.5*pi*4^2 = 25.133 cm^2

The difference: 32 - 25.133 = 6.867 cm^2 (area of the shaded region)

5 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

99% of all confidence intervals with a 99% confidence level should contain the population parameter of interest. true or false

tester [92]

The statement that 99% of all confidence intervals with a 99% confidence level should contain the population parameter of interest is false.

A confidence interval (CI) is essentially a range of estimates for an unknown parameter in frequentist statistics. The most frequent confidence level is 95%, but other levels, such 90% or 99%, are infrequently used for generating confidence intervals.

The confidence level is a measurement of the proportion of long-term associated CIs that include the parameter's true value. This is closely related to the moment-based estimate approach.

In a straightforward illustration, when the population mean is the quantity that needs to be estimated, the sample mean is a straightforward estimate. The population variance can also be calculated using the sample variance. Using the sample mean and the true mean's probability.

Hence we can generally infer that the given statement is false.

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6 0

1 year ago