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IgorLugansk [536]

2 years ago

6

Two galaxies are shown. Galaxy U: Dust, gas, and stars in the shape of a pinwheel, relatively small in size. Galaxy NGC 1427A: D

ust, gas, and stars in a nonuniform arrangement spread out over a large area. How are NGC 1427A and U different? How are they the same?

2 answers:

Pani-rosa [81]2 years ago

8 0

Answer:

NGC 1427A has no general shape, so it is an irregular galaxy. U has a bulge in the center and arms, so it is a spiral galaxy. They are similar in that both contain plenty of dust and gas. Both also have active star-forming sites.

Step-by-step explanation: Got it right on the test

anzhelika [568]2 years ago

6 0

Answer:

NGC 1427A has no general shape, so it is an irregular galaxy. U has a bulge in the center and arms, so it is a spiral galaxy. They are similar in that both contain plenty of dust and gas. Both also have active star-forming sites.

Step-by-step explanation:

i got it right.

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To evaluate the effect of a treatment, a sample of n=8 is obtained from a population with a mean of μ=40 , and the treatment is

Westkost [7]

Answer:

Step-by-step explanation:

a)

Test statistic:

$t=\frac{35-40}{\sqrt{\frac{32}{8} } }$

$t=-2.5$

$df=8-1=7$

$critical, t ,values=+/-2.365$

here test statistic lie in rejection region,that why null hypothesis fails

so Yes, its significant.

b)

Test statistic:

$t=\frac{35-40}{\sqrt{\frac{72}{8} } }$

$t=-1.67$

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c)

sample variability increases, therefore likelihood of rejecting the null hypothesis decreases.

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1 year ago

To go on the class trip, chang needs at least $125. he has saved $45, and he makes $7 an hour at his job. what is the minimum nu

Murljashka [212]

The total amount that Chang will received from his wages is the product of his hourly wage times the number of hours he has worked (n). His total savings should not be less than $125. The equation that best describes the situation is,
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1 year ago

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(5y + 2) + 14x = 5y + (2 + 14x)

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1 year ago

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The square of a positive number is 24 more than 5 times a number. What is the value of the number?

satela [25.4K]

First, let's construct an algebraic equation to formally find the answer:

Let x represent (the value of) the unknown number.

x²= 5x + 24 --- This equation shows that the (positive) number's square equals 24 more than 5 times the number.

Now we can solve for x (the number):

x² = 5x + 24
x² - 5x - 24 = 0
(x + 3)(x - 8) = 0
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Therefore, this means that x (the number ) can be either -3 OR 8. Both will satisfy the description.
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1 year ago

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A bank gets an average of 12 customers per hour. Assume the variable follows a Poisson distribution. Find the probability that t

Alina [70]

Answer:

75.76% probability that there will be 10 or more customers at this bank in one hour.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

A bank gets an average of 12 customers per hour.

This means that $\mu = 12$

Find the probability that there will be 10 or more customers at this bank in one hour.

Either there are less than 10 customers, or there are 10 or more. The sum of the probabilities of these events is 1. Then

$P(X < 10) + P(X \geq 10) = 1$

We want $P(X \geq 10)$

Then

$P(X \geq 10) = 1 - P(X < 10)$

In which

$P(X < 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9)$

So

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-12}*12^{0}}{(0)!} \approx 0$

$P(X = 1) = \frac{e^{-12}*12^{1}}{(1)!} = 0.0001$

$P(X = 2) = \frac{e^{-12}*12^{2}}{(2)!} = 0.0004$

$P(X = 3) = \frac{e^{-12}*12^{3}}{(3)!} = 0.0018$

$P(X = 4) = \frac{e^{-12}*12^{4}}{(4)!} = 0.0053$

$P(X = 5) = \frac{e^{-12}*12^{5}}{(5)!} = 0.0127$

$P(X = 6) = \frac{e^{-12}*12^{6}}{(6)!} = 0.0255$

$P(X = 7) = \frac{e^{-12}*12^{7}}{(7)!} = 0.0437$

$P(X = 8) = \frac{e^{-12}*12^{8}}{(8)!} = 0.0655$

$P(X = 9) = \frac{e^{-12}*12^{9}}{(9)!} = 0.0874$

$P(X < 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) = 0 + 0.0001 + 0.0004 + 0.0018 + 0.0053 + 0.0127 + 0.0255 + 0.0437 + 0.0655 + 0.0874 = 0.2424$

Then

$P(X \geq 10) = 1 - P(X < 10) = 1 - 0.2424 = 0.7576$

75.76% probability that there will be 10 or more customers at this bank in one hour.

3 0

2 years ago