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salantis [7]
3 years ago
7

If Nina is N years old now and Maryna is M years old now, write the sum of their ages in 6 years

Mathematics
1 answer:
zimovet [89]3 years ago
4 0

Answer:

If Nina is N years old and Maryna is M years old the sum of their ages would be N+M

In 6 year from now they you would go the sum of their ages now times 6

You equation would be

6(N+M)

Step-by-step explanation:

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In the smoky mountains of Tennessee, the percent of moisture that falls as snow rather than rain is approximated by P = 10.0 ln
Mars2501 [29]

Answer:

  A. About 4,900 ft

Step-by-step explanation:

We want h such that ...

  5 = 10·ln(h) -80

  8.5 = ln(h) . . . . . . . add 80, divide by 10

  e^8.5 = h ≈ 4914.8 . . . . take the antilog

  h ≈ 4900 . . . . feet

7 0
3 years ago
Compare the light gathering power of an 8" primary mirror with a 6" primary mirror. The 8" mirror has how much light gathering p
tankabanditka [31]

Answer:

1.778 times more or 16/9 times more

Step-by-step explanation:

Given:

- Mirror 1: D_1 = 8''

- Mirror 2: D_2 = 6"

Find:

Compare the light gathering power of an 8" primary mirror with a 6" primary mirror. The 8" mirror has how much light gathering power?

Solution:

- The light gathering power of a mirror (LGP) is proportional to the Area of the objects:

                                           LGP ∝ A

- Whereas, Area is proportional to the squared of the diameter i.e an area of a circle:

                                           A ∝ D^2

- Hence,                              LGP ∝ D^2

- Now compare the two diameters given:

                                           LGP_1 ∝ (D_1)^2

                                           LGP ∝ (D_2)^2

- Take a ratio of both:

                           LGP_1/LGP_2 ∝ (D_1)^2 / (D_2)^2

- Plug in the values:

                               LGP_1/LGP_2 ∝ (8)^2 / (6)^2

- Compute:             LGP_1/LGP_2 ∝ 16/9 ≅ 1.778 times more

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3 years ago
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Answer:

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5 0
2 years ago
On a standardized test with a normal distribution the mean score was 67.2. The standard deviation was 4.6. What percent of the d
Reika [66]

Answer:

P ( -1 < Z < 1 ) = 68%

Step-by-step explanation:

Given:-

- The given parameters for standardized test scores that follows normal distribution have mean (u) and standard deviation (s.d) :

                         u = 67.2

                         s.d = 4.6

- The random variable (X) that denotes standardized test scores following normal distribution:

                         X~ N ( 67.2 , 4.6^2 )

Find:-

What percent of the data fell between 62.6 and 71.8?

Solution:-

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- Using the The Empirical Rule or 68-95-99.7%. We need to find the percent of data that lies within 1 standard about mean value:

                          P ( -1 < Z < 1 ) = 68%

                          P ( -2 < Z < 2 ) = 95%

                          P ( -3 < Z < 3 ) = 99.7%

6 0
3 years ago
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