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pychu [463]
3 years ago
11

1, 3/2 , 9,4 ... Find the 9th term HELP ASAP

Mathematics
1 answer:
yarga [219]3 years ago
4 0
This is a geometric sequence since there is a common ratio between each term. In this case, multiplying the previous term in the sequence by
2
3
2
3
gives the next term. In other words,
a
n
=
a
1
⋅
r
n
−
1
a
n
=
a
1
⋅
r
n
-
1
.
Geometric Sequence:
r
=
2
3
r
=
2
3
This is the form of a geometric sequence.
a
n
=
a
1
r
n
−
1
a
n
=
a
1
r
n
-
1
Substitute in the values of
a
1
=
1
2
a
1
=
1
2
and
r
=
2
3
r
=
2
3
.
a
n
=
(
1
2
)
⋅
(
2
3
)
n
−
1
a
n
=
(
1
2
)
⋅
(
2
3
)
n
-
1
Apply the product rule to
2
3
2
3
.
a
n
=
1
2
⋅
2
n
−
1
3
n
−
1
a
n
=
1
2
⋅
2
n
-
1
3
n
-
1
Multiply
1
2
1
2
and
2
n
−
1
3
n
−
1
2
n
-
1
3
n
-
1
.
a
n
=
2
n
−
1
2
⋅
3
n
−
1
a
n
=
2
n
-
1
2
⋅
3
n
-
1
Cancel the common factor of
2
n
−
1
2
n
-
1
and
2
2
.
Tap for more steps...
a
n
=
2
n
−
2
3
n
−
1
a
n
=
2
n
-
2
3
n
-
1
Substitute in the value of
n
n
to find the
n
n
th term.
a
5
=
2
(
5
)
−
2
3
(
5
)
−
1
a
5
=
2
(
5
)
-
2
3
(
5
)
-
1
Simplify the numerator.
Tap for more steps...
a
5
=
8
3
(
5
)
−
1
a
5
=
8
3
(
5
)
-
1
Simplify the denominator.
Tap for more steps...
a
5
=
8
81
a
5
=
8
81
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8 0
3 years ago
Measurements of the sodium content in samples of two brands of chocolate bar yield the following results (in grams):
Tpy6a [65]

Answer:

98% confidence interval for the difference μX−μY = [ 0.697 , 7.303 ] .

Step-by-step explanation:

We are give the data of Measurements of the sodium content in samples of two brands of chocolate bar (in grams) below;

Brand A : 34.36, 31.26, 37.36, 28.52, 33.14, 32.74, 34.34, 34.33, 29.95

Brand B : 41.08, 38.22, 39.59, 38.82, 36.24, 37.73, 35.03, 39.22, 34.13, 34.33, 34.98, 29.64, 40.60

Also, \mu_X represent the population mean for Brand B and let \mu_Y represent the population mean for Brand A.

Since, we know nothing about the population standard deviation so the pivotal quantity used here for finding confidence interval is;

        P.Q. = \frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}  } } ~ t_n__1+n_2-2

where, Xbar = Sample mean for Brand B data = 36.9

            Ybar = Sample mean for Brand A data = 32.9

              n_1  = Sample size for Brand B data = 13

              n_2 = Sample size for Brand A data = 9

              s_p = \sqrt{\frac{(n_1-1)s_X^{2}+(n_2-1)s_Y^{2}  }{n_1+n_2-2} } = \sqrt{\frac{(13-1)*10.4+(9-1)*7.1 }{13+9-2} } = 3.013

Here, s^{2}_X and s^{2} _Y are sample variance of Brand B and Brand A data respectively.

So, 98% confidence interval for the difference μX−μY is given by;

P(-2.528 < t_2_0 < 2.528) = 0.98

P(-2.528 < \frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}  } } < 2.528) = 0.98

P(-2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} < (Xbar -Ybar) -(\mu_X-\mu_Y) < 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} ) = 0.98

P( (Xbar - Ybar) - 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} < (\mu_X-\mu_Y) < (Xbar - Ybar) + 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} ) = 0.98

98% Confidence interval for μX−μY =

[ (Xbar - Ybar) - 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} , (Xbar - Ybar) + 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} ]

[ (36.9 - 32.9)-2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9} , (36.9 - 32.9)+2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9} ]

[ 0.697 , 7.303 ]

Therefore, 98% confidence interval for the difference μX−μY is [ 0.697 , 7.303 ] .

                     

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Daniel deposited $500 into a savings account and after 8 years, his investment is worth $807.07. The equation A = d(1.005)12t mo
shtirl [24]

Answer:

Step-by-step explanation:

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We can confirm by inputting the given values

t = 8 years

d = 509

Therefore,

A = 500(1.005)12 × 8

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Zolol [24]

Answer:

36

Step-by-step explanation:

Given

x² + 12x = 11

To complete the square

add ( half the coefficient of the x- term )² to both sides

x² + 2(6)x + 36 = 11 + 36

(x + 6)² = 47

7 0
3 years ago
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