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oksian1 [2.3K]
3 years ago
14

27/45=_/5? 2/6=_/3?jjjjjj

Mathematics
2 answers:
Ahat [919]3 years ago
5 0

Answer:

27/45=3/5

2/6=1/3

Step-by-step explanation:

VikaD [51]3 years ago
4 0

Answer:

27/45 = 3/5

2/6 = 1/3

hope dis helps

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The annual precipitation amounts in a certain mountain range are normally distributed with a mean of 91 inches, and a standard d
Vesna [10]
The sample std. dev. will be (14 inches) / sqrt(49), or (14 inches) / 7, or 2 inches.

Find the z score for 93.8 inches:
       93.8 inches - 91.0 inches          2.8 inches
z = ------------------------------------- = ----------------- = 1.4
                    2 inches                         2 inches

Now find the area under the standard normal curve to the left of z = +1.4.

My calculator returns the following:

normalcdf(-100,1.4) = 0.919.  This is the probability that the mean annual precipitation during those 49 years will be less than 93.8 inches.
 
4 0
3 years ago
Help me out please!
Marina86 [1]

Answer: Depends on what you come up with . /:

Step-by-step explanation:

3x-11+x+9= if u no the form for this... -x with 3x and -9 -11 should to a division problem so u divide from both sides leaving x = your sum.

4 0
2 years ago
can some one also please help my one that i should do after i finds the area of the square aka how to find the area of the trian
Andre45 [30]

Answer:

6 square units

Step-by-step explanation:

Given shape is of a trapezoid:

area \: of \: trapezoid  \\ =  \frac{1}{2}  \times (4 + 2) \times 2 \\  \\  = 6 \:  {units}^{2}

Area of shape = area of square + area of triangle

= 2^2 + \frac{1}{2} \times 2\times 2\\= 4 + 2\\= 6\: units^2 \\

4 0
3 years ago
Given that the series kcoskt kº +2 k=1 converges, suppose that the 3rd partial sum of the series is used to estimate the sum of
3241004551 [841]

Answer:

c

Step-by-step explanation:

Given that:

\sum \limits ^{\infty}_{k=1} \dfrac{kcos (k\pi)}{k^3+2}

since cos (kπ) = -1^k

Then, the  series can be expressed as:

\sum \limits ^{\infty}_{k=1} \dfrac{(-1)^kk)}{k^3+2}

In the sum of an alternating series, the best bound on the remainder for the approximation is related to its (n+1)^{th term.

∴

\sum \limits ^{\infty}_{k=1} \dfrac{(-1)^{(3+1)}(3+1))}{(3+1)^3+2}

\sum \limits ^{\infty}_{k=1} \dfrac{(-1)^{(4)}(4))}{(4)^3+2}

= \dfrac{4}{64+2}

=\dfrac{2}{33}

5 0
2 years ago
Write as an expression:<br> the total of n and 16
Mekhanik [1.2K]
Total is just add
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4 0
3 years ago
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