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3 years ago

Help with geometry need fast

Mathematics

1 answer:

Temka [501]3 years ago

7 0

Answer:

Step-By-Step Explanation:

m<2 and m<3 are same side interior angles.

That means that the sum of them has to be 180 for the lines to be parallel.

They equal 190 in this problem.

So no, lines f and g are not parallel.

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write an equation that would allow you to solve the following problem. mart'a jump rope is 10 inches shorter than 3 times the le

mars1129 [50]

Answer:

Step-by-step explanation:

m=marta's rope

d=dilbert's rope

3d-10=m

m+d=120

Then you can substitute equation 1 into the second one

(3d-10)+d=120

3d-10+d=120

4d-10=120

4d=130

d=32.5

m+d=120

m+32.5=120

m=120-32.5

8 0

3 years ago

Suppose a simple random sample of size 50 is selected from a population with σ=10σ=10. Find the value of the standard error of t

bogdanovich [222]

Answer:

a) $\sigma_{\bar x} = 1.414$

b) $\sigma_{\bar x} = 1.414$

c) $\sigma_{\bar x} = 1.414$

d) $\sigma _{\bar x} = 1.343$

Step-by-step explanation:

Given that:

The random sample is of size 50 i.e the population standard deviation =10

Size of the sample n = 50

a) The population size is infinite;

The standard error is determined as:

$\sigma_{\bar x} = \dfrac{\sigma}{\sqrt{n}}$

$\sigma_{\bar x} = \dfrac{10}{\sqrt{50}}$

$\sigma_{\bar x} = 1.414$

b) When the population size N= 50000

n/N = 50/50000 = 0.001 < 0.05

Thus ; the finite population of the standard error is not applicable in this scenario;

Therefore;

The standard error is determined as:

$\sigma_{\bar x} = \dfrac{\sigma}{\sqrt{n}}$

$\sigma_{\bar x} = \dfrac{10}{\sqrt{50}}$

$\sigma_{\bar x} = 1.414$

c) When the population size N= 5000

n/N = 50/5000 = 0.01 < 0.05

Thus ; the finite population of the standard error is not applicable in this scenario;

Therefore;

The standard error is determined as:

$\sigma_{\bar x} = \dfrac{\sigma}{\sqrt{n}}$

$\sigma_{\bar x} = \dfrac{10}{\sqrt{50}}$

$\sigma_{\bar x} = 1.414$

d) When the population size N= 500

n/N = 50/500 = 0.1 > 0.05

So; the finite population of the standard error is applicable in this scenario;

Therefore;

The standard error is determined as:

$\sigma _{\bar x} = \sqrt{\dfrac{N-n}{N-1} }\dfrac{\sigma}{\sqrt{n} } }$

$\sigma _{\bar x} = \sqrt{\dfrac{500-50}{500-1} }\dfrac{10}{\sqrt{50} } }$

$\sigma _{\bar x} = 1.343$

7 0

3 years ago

the probability density for a particle in a box is an oscillatory function even for very large energies. Explain how the classic

77julia77 [94]

Answer:

This is achieved for the specific case when high quantum number with low resolution is present.

Step-by-step explanation:

In Quantum Mechanics, the probability density defines the region in which the likelihood of finding the particle is most.

Now for the particle in the box, the probability density is also dependent on resolution as well so for large quantum number with small resolution, the oscillations will be densely packed and thus indicating in the formation of a constant probability density throughout similar to that of classical approach.

8 0

3 years ago

Plss help if can!! Thanks!

Novosadov [1.4K]

The first one is 2 the second one is 3

5 0

3 years ago

M is the MIDPOINT OF FG. FG = 26 AND GM = (3X - 2)

vova2212 [387]

It is given in the question that

M is the MIDPOINT OF FG.

$FG = 26 \ AND \ GM = (3X - 2)$