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3 years ago

Solve the Inequality. Please help answer needed fast! :) m - 3 > -2

Mathematics

2 answers:

shtirl [24]3 years ago

8 0

Answer:

Step-by-step explanation:

Ksju [112]3 years ago

6 0

Answer:

$m - 3 > - 2 \\ m > - 2 + 3 \\ m > 1$

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Considering the following statements, what is X → Z? X → Y: If the sum of the interior angles of a shape is 180°, then it's a tr

tresset_1 [31]

Answer:

Step-by-step explanation:z=180

Interior angle =180

8 0

3 years ago

A sample of 5 buttons is randomly selected and the following diameters are measured in inches. Give a point estimate for the pop

Helen [10]

Answer:

$s^2 = \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}$

But we need to calculate the mean with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_I}{n}$

And replacing we got:

$\bar X = \frac{ 1.04+1.00+1.13+1.08+1.11}{5}= 1.072$

And for the sample variance we have:

$s^2 = \frac{(1.04-1.072)^2 +(1.00-1.072)^2 +(1.13-1.072)^2 +(1.08-1.072)^2 +(1.11-1.072)^2}{5-1}= 0.00277\ approx 0.003$

And thi is the best estimator for the population variance since is an unbiased estimator od the population variance $\sigma^2$

$E(s^2) = \sigma^2$

Step-by-step explanation:

For this case we have the following data:

1.04,1.00,1.13,1.08,1.11

And in order to estimate the population variance we can use the sample variance formula:

$s^2 = \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}$

But we need to calculate the mean with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_I}{n}$

And replacing we got:

$\bar X = \frac{ 1.04+1.00+1.13+1.08+1.11}{5}= 1.072$

And for the sample variance we have:

$s^2 = \frac{(1.04-1.072)^2 +(1.00-1.072)^2 +(1.13-1.072)^2 +(1.08-1.072)^2 +(1.11-1.072)^2}{5-1}= 0.00277\ approx 0.003$

And thi is the best estimator for the population variance since is an unbiased estimator od the population variance $\sigma^2$

$E(s^2) = \sigma^2$

3 0

4 years ago

How are the GCF and the LCM alike and different?

damaskus [11]

Alike: They both find what's in common.
Difference: LCM finds the least multiple while GCF finds the greatest factor.

4 0

3 years ago

Read 2 more answers

An individual repeatedly attempts to pass a driving test. Suppose that the probability of passing the test with each attempt is

vladimir1956 [14]

Answer:

a) Our random variable X="number of tests taken until the individual passes" follows a geomteric distribution with probability of success p=0.25

For this case the probability mass function would be given by:

$P(X= k) = (1-p)^{k-1} p , k = 1,2,3,...$

b) $P(X \leq 3) = P(X=1) +P(X=2) +P(X=3)$

$P(X= 1) = (1-0.25)^{1-1} *0.25 = 0.25$

$P(X= 2) = (1-0.25)^{2-1} *0.25 = 0.1875$

$P(X= 3) = (1-0.25)^{3-1} *0.25 = 0.1406$

And adding the values we got:

$P(X \leq 3) =0.25+0.1875+0.1406=0.578$

c) $P(X \geq 5) = 1-P(X$

And we can find the individual probabilities:

$P(X= 1) = (1-0.25)^{1-1} *0.25 = 0.25$

$P(X= 2) = (1-0.25)^{2-1} *0.25 = 0.1875$

$P(X= 3) = (1-0.25)^{3-1} *0.25 = 0.1406$

$P(X= 4) = (1-0.25)^{4-1} *0.25 = 0.1055$

$P(X \geq 5) = 1-[0.25+0.1875+0.1406+0.1055]= 0.316$

Step-by-step explanation:

Previous concepts

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

$P(X=x)=(1-p)^{x-1} p$

Let X the random variable that measures the number of trials until the first success, we know that X follows this distribution:

$X\sim Geo (1-p)$

Part a

Our random variable X="number of tests taken until the individual passes" follows a geomteric distribution with probability of success p=0.25

For this case the probability mass function would be given by:

$P(X= k) = (1-p)^{k-1} p , k = 1,2,3,...$

Part b

We want this probability:

$P(X \leq 3) = P(X=1) +P(X=2) +P(X=3)$

We find the individual probabilities like this:

$P(X= 1) = (1-0.25)^{1-1} *0.25 = 0.25$

$P(X= 2) = (1-0.25)^{2-1} *0.25 = 0.1875$

$P(X= 3) = (1-0.25)^{3-1} *0.25 = 0.1406$

And adding the values we got:

$P(X \leq 3) =0.25+0.1875+0.1406=0.578$

Part c

For this case we want this probability:

$P(X \geq 5)$

And we can use the complement rule like this:

$P(X \geq 5) = 1-P(X$

And we can find the individual probabilities:

$P(X= 1) = (1-0.25)^{1-1} *0.25 = 0.25$

$P(X= 2) = (1-0.25)^{2-1} *0.25 = 0.1875$

$P(X= 3) = (1-0.25)^{3-1} *0.25 = 0.1406$

$P(X= 4) = (1-0.25)^{4-1} *0.25 = 0.1055$

$P(X \geq 5) = 1-[0.25+0.1875+0.1406+0.1055]= 0.316$

3 0

3 years ago

Your Turn!

Pachacha [2.7K]

B don’t quote me if I’m wrong

5 0

3 years ago