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3 years ago

Find the quotient of 5/31 divided by 15/23. Reduce your answer to the lowest fraction

Mathematics

2 answers:

cupoosta [38]3 years ago

8 0

The answer is 23/93 because I just took the test.

LekaFEV [45]3 years ago

3 0

5/31 / 15/23 5/31*23/15 23/93

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blsea [12.9K]

The circumference can be obtained fairly easily by simply substituting the d in c = πd for the colony's given diameter of 12 mm. Performing that calculation using the approximation of π ≈ 3.14, we obtain a circumference of 12 x 3.14 = 37.68 mm.

To find the radius, remember how the diameter and radius of a circle are defined. The radius is a length extending from the center of a circle to a point on its circumference, and a diameter is a line extending from one point on the circle's circumference to an opposite point, passing through the circle's center along the way. The diameter can, in this way, be defined as twice the length of the radius, which means we can find the radius of a circle by taking half of its diameter. In this case, our diameter is 12 mm, so our radius would be 6 mm.

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3 years ago

This table gives points on a line. What is the equation of the line?

vitfil [10]

Answer:

y = -2x + 5

Step-by-step explanation:

Assuming this is linear, the slope would be constant. For every 1 x, y decreases by 2. This means that the slope is -2.

y = -2x + b

Now all we need is the y-intercept.

If we plug in the first point (1,3), we get:

3 = -2*1+b

3 = -2+b

b = 5

This is the y-intercept.

y = -2x + 5

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3 years ago

Can someone help solve?

icang [17]

Answer:

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2 years ago

Read 2 more answers

Which equation represents this statement? One-third of a number is 21.

Musya8 [376]

This states that a number divided by 3 is 21. the equation is: n/3 = 21

5 0

3 years ago

Let X1 and X2 be independent random variables with mean μand variance σ².

My name is Ann [436]

Answer:

a) $E(\hat \theta_1) =\frac{1}{2} [E(X_1) +E(X_2)]= \frac{1}{2} [\mu + \mu] = \mu$

So then we conclude that $\hat \theta_1$ is an unbiased estimator of $\mu$

$E(\hat \theta_2) =\frac{1}{4} [E(X_1) +3E(X_2)]= \frac{1}{4} [\mu + 3\mu] = \mu$

So then we conclude that $\hat \theta_2$ is an unbiased estimator of $\mu$

b) $Var(\hat \theta_1) =\frac{1}{4} [\sigma^2 + \sigma^2 ] =\frac{\sigma^2}{2}$

$Var(\hat \theta_2) =\frac{1}{16} [\sigma^2 + 9\sigma^2 ] =\frac{5\sigma^2}{8}$

Step-by-step explanation:

For this case we know that we have two random variables:

$X_1 , X_2$ both with mean $\mu = \mu$ and variance $\sigma^2$

And we define the following estimators:

$\hat \theta_1 = \frac{X_1 + X_2}{2}$

$\hat \theta_2 = \frac{X_1 + 3X_2}{4}$

Part a

In order to see if both estimators are unbiased we need to proof if the expected value of the estimators are equal to the real value of the parameter:

$E(\hat \theta_i) = \mu , i = 1,2$