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Ber [7]
3 years ago
13

Find a6: 1, 1/2, 1/6, 1/24, 1/120, a6, 1/5,040, ...

Mathematics
2 answers:
stepan [7]3 years ago
8 0

Answer:

Is the #6 in the row. Is this a riddle

sertanlavr [38]3 years ago
7 0

Answer:

its 1/720 i just took it

Step-by-step explanation:

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You and your friend are sharing 3/8 of a bag of candy how much of the bag will each of you get share it evenly. Is it going to b
olga_2 [115]

Answer:

You and your friend both get 3/16 of the bag.

Step-by-step explanation:

3/8 divided by 2 friends

3/8 / 2 = 3/16

6 0
3 years ago
F(x) = 3x - 5 and g(x) = x + 3, find (1-9)(x).
snow_lady [41]

f(x)= 3x - 5 is x = 5/3

g(x) = x +3 is x = -3

(1-9)(x) is -8x

3 0
3 years ago
Solve for x. Round to the nearest tenth, if necessary. 1.8 & x
Valentin [98]

Answer:

x ≈ 1.4

Step-by-step explanation:

Using the sine ratio in the right triangle

sin50° = \frac{opposite}{hypotenuse} = \frac{RS}{RT} = \frac{x}{1.8} ( multiply both sides by 1.8 )

1.8 × sin50° = x , then

x ≈ 1.4 ( to the nearest tenth )

7 0
3 years ago
PLEASE HELP MUST BE DONE TODAY PLEEEEAAASEEEE!!!!
Nikolay [14]

B because the table shows of stories and height

7 0
3 years ago
A study found that a 95% confidence interval for the mean μ of a particular population was computed from a random sample of 1200
kotykmax [81]

Answer:

95% confidence interval for the mean μ is (6,14)

The Population mean μ lies between ( 6, 14 )

Step-by-step explanation:

<u><em>Explanation</em></u>:-

Given random sample 'n' = 1200

95% confidence interval for the mean μ is determined by

        (x^{-} -Z_{\alpha } \frac{S.D}{\sqrt{n} } , x^{-} +Z_{\alpha } \frac{S.D}{\sqrt{n} } )

Level of significance = 95% 0r 0.05

Z₀.₀₅ = 1.96

   (x^{-} -Z_{\alpha } \frac{S.D}{\sqrt{n} } , x^{-} +Z_{\alpha } \frac{S.D}{\sqrt{n} } ) = 10 ± 4

Mean of the small sample = 10

95% of confidence intervals are

                         ( 10 ±4 )

                  ( 10 -4 , 10+4)

                  ( 6 , 14 )

95% confidence interval for the mean μ lies between ( 6, 14 )

8 0
3 years ago
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