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3 years ago

What is the 20th term in the following arithmatic sequence 1/2, 1, 3/2, 2...

1 answer:

4 0

The 20th term in the following arithmetic sequence1/2, 1, 3/2, 2... is 10 because;

the number are divided by 2
and the 20th number would be 20/2 which is equal to 10

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The graph shows the equation x2 + y2 = 5. Use the slider for a to move the vertical line on the graph. Based on the vertical

lidiya [134]

Answer: This equation is not a function.

Step-by-step explanation:

Ok, a function is something like a "machine".

It takes an input value, x, and transforms it into an output value, y.

Such that for every input x, the function can transform it into only one value of y.

for example, if we have a "function" f(x)

such that f(2) = 3 and f(2) = 4

(for the same input, x = 2, we have two different outputs)

Then this is not a function.

Now, the given equation is:

x^2 + y^2 = 5

This is actually the equation for a circle, centered at the point (0, 0) and with a radius equal to √5.

Then, if we want to transform it into y(x), we can see a problem

If x = 0, we have:

0^2 + y^2 = 5

y^2 = 5

then we have two possible values for y:

y = +√5

y = -√5

Then for one value of x, we have two values of y.

This means that this is not a function.

4 0

4 years ago

Sixty percent of U.S. adults trust national newspapers to present the news fairly and accurately. You randomly select nine U.S.

Pavel [41]

Answer:

a) 0.251

b) 0.483

c) 0.0994

Step-by-step explanation:

We are given the following information:

We treat adult adults trusting national newspapers to present the news fairly and accurately as a success.

P(Adult trust) = 60% = 0.6

Then the number of adults follows a binomial distribution, where

$P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}$

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 9

a) exactly five

$P(x = 15) = \binom{9}{5}(0.6)^{5}(1-0.6)^{9-5} = 0.2508 \approx 0.251$

b) at least six

We have to evaluate

$P(x \geq 6) = P(x = 6) + P(x = 7) + P(x=8) + P(x=9) \\= \binom{9}{6}(0.6)^6(1-0.6)^3 + \binom{9}{7}(0.6)^7(1-0.6)^2 + \binom{9}{8}(0.6)^8(1-0.6)^1 +\binom{9}{9}(0.6)^9(1-0.6)^0\\= 0.482609 \approx 0.483$

c) less than four

We have to evaluate

$P(x < 4) = P(x = 0) + P(x = 1) + P(x=2) + P(x=3) \\= \binom{9}{0}(0.6)^0(1-0.6)^9 + \binom{9}{1}(0.6)^1(1-0.6)^8 + \binom{9}{2}(0.6)^2(1-0.6)^7 +\binom{9}{3}(0.6)^3(1-0.6)^6\\= 0.099352 \approx 0.0994$