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Rzqust [24]
3 years ago
11

Find the 11th term of this sequence -2, 10, -50

Mathematics
1 answer:
mrs_skeptik [129]3 years ago
8 0

Answer:

See below

Step-by-step explanation:

<em><u>Formula</u></em>

The general term we will use is

t = a r^(n - 1)

<em><u>Givens</u></em>

a = - 2

r = - 5

n = 11

t_n = a*r^(n - 1)

t_n = -2*(-5)^(11 - 1)

t_11 = -2*(-5)^10

t_11 = -2 (9,765,625)

t_11 = 19,531,250

<em><u>Note</u></em>

Before accepting this answer, you should check to see that the formula works with a value you know the answer to.

t_3 = -2(-5)^(3-1)

t_3 = -2(-5)^2

t_3 = -2 * 25

t_3 = - 50

Which is what your series shows.

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A random sample is to be selected from a population that has a proportion of successes p=.60. When n=400, what is the probabilit
Nata [24]

Answer:

0.45134

Step-by-step explanation:

Given that :

p = 0.6

n = 400

Probability that sample. Proportion falls between 0.59 and 0.62

Using Normal approximation :

Mean (m) = n * p = 400 * 0.6 = 240

Standard deviation (s) = sqrt(pq/n)

q = 1 - p = 1 - 0.6 = 0.4

s = sqrt((0.6 * 0.4) / 400) = 0.0244948

P(0.59 < p < 0.62) :

(x - m) / s

P((0.59 - 0.6) / 0.0244948) < p < P((0.62 - 0.6) / 0.0244948)

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Using the Z probability calculator :

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3 0
3 years ago
The equation of the hyperbola with center at (0, 0) opening horizontally, with a = 2, b = 5 is:
tresset_1 [31]

Answer:

\frac{x^2}{4}-\frac{y^2}{25} =1

Step-by-step explanation:

we are given

center =(0,0)

we can compare with

center=(h,k)=(0,0)

so, h=0 and k=0

a=2 and b=5

now, we can use equation of hyperbola formula

\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2} =1

now, we can plug values

\frac{(x-0)^2}{2^2}-\frac{(y-0)^2}{5^2} =1

we can simplify it

and we get equation of hyperbola as

\frac{x^2}{4}-\frac{y^2}{25} =1


5 0
3 years ago
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