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Arada [10]
3 years ago
10

Answer the ones you want to. At least answer one…

Mathematics
1 answer:
Artyom0805 [142]3 years ago
7 0

Answer:

9. First Five terms

If n=0. Then,

\frac{4(0) - 5}{2}

\frac{0 - 5}{2}  =  \frac{ - 5}{2}

If n=1. Then,

\frac{4(1) - 5}{2}

\frac{4 - 5}{2}  =  \frac{ - 1}{2}

If n=2. Then,

\frac{4(2) - 5}{2}

\frac{8 - 5}{2}  =  \frac{3}{2}

If n=3. Then,

\frac{4(3) - 5}{2}

\frac{12 - 5}{2}  =  \frac{7}{2}

If n=4. Then,

\frac{4(4) - 5}{2}

\frac{16 - 5}{2}  =  \frac{11}{2}

If n=5. Then,

\frac{4(5) - 5}{2}

\frac{20 - 5}{2}  =  \frac{15}{2}

If n=6. Then,

\frac{4(6) - 5}{2}

\frac{24 - 5}{2}  =  \frac{19}{2}

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Si la probabilidad de que un recién nacido sea niño es de 1 de dos opciones, representado en fracción 1/2, ¿Cuál es la probabili
Leno4ka [110]

Answer:

1/2 o 50%

Step-by-step explanation:

debemos partir de la presuncion que un bebe puede ser niño o niña uinicamente, o sea la P(niño) + P(niña) = 1

no existen los bebes que son niños y niñas al mismo tiempo, eso quiere decir que es una opcion o la otra

P(niña) = 1 - P(niño) = 1 - 1/2 = 1/2 = 0.5 = 50%

8 0
2 years ago
Find the distance.<br> -3,5 and 7,-1
Brilliant_brown [7]

Answer:

≈ 11.66 units

Step-by-step explanation:

<u>Given points:</u>

  • (-3,5) and (7,-1)

<u>To find:</u>

  • The distance between the given points

<u>The distance between two points is calculated by formula:</u>

  • d= √((x2-x1)² + (y2-y1)²)
  • d= √(((7-(-3))² + (-1-5)²) = √(10²+(-6)²)= √136 ≈ 11.66 units

<u>Answer is</u> ≈ 11.66 units

8 0
3 years ago
A production process fills containers by weight. Weights of containers are approximately normally distributed. Historically, the
forsale [732]

Answer:

option (c) n = 201

Step-by-step explanation:

Data provided in the question:

Standard deviation, s = 5.5 ounce

Confidence level = 99%

Length of confidence interval = 2 ounces

Therefore,

margin of error, E = (Length of confidence interval ) ÷ 2

= 2 ÷ 2

= 1 ounce

Now,

E = \frac{zs}{\sqrt n}

here,

z = 2.58 for 99% confidence interval

n = sample size

thus,

1 = \frac{2.58\times5.5}{\sqrt n}

or

n = (2.58 × 5.5)²

or

n = 201.3561 ≈ 201

Hence,

option (c) n = 201

5 0
3 years ago
Can someone answer this please
choli [55]
The awenser of the problem is A
5 0
3 years ago
Read 2 more answers
According to a Yale program on climate change communication survey, 71% of Americans think global warming is happening.† (a) For
SpyIntel [72]

Answer:

a) 0.2741 = 27.41% probability that at least 13 believe global warming is occurring

b) 0.7611 = 76.11% probability that at least 110 believe global warming is occurring

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

p = 0.71

(a) For a sample of 16 Americans, what is the probability that at least 13 believe global warming is occurring?

Here n = 16, we want P(X \geq 13). So

P(X \geq 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 13) = C_{16,13}.(0.71)^{13}.(0.29)^{3} = 0.1591

P(X = 14) = C_{16,14}.(0.71)^{14}.(0.29)^{2} = 0.0835

P(X = 15) = C_{16,15}.(0.71)^{15}.(0.29)^{1} = 0.0273

P(X = 16) = C_{16,16}.(0.71)^{16}.(0.29)^{0} = 0.0042

P(X \geq 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16) = 0.1591 + 0.0835 + 0.0273 + 0.0042 = 0.2741

0.2741 = 27.41% probability that at least 13 believe global warming is occurring

(b) For a sample of 160 Americans, what is the probability that at least 110 believe global warming is occurring?

Now n = 160. So

\mu = E(X) = np = 160*0.71 = 113.6

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{160*0.71*0.29} = 5.74

Using continuity correction, this is P(X \geq 110 - 0.5) = P(X \geq 109.5), which is 1 subtracted by the pvalue of Z when X = 109.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{109.5 - 113.6}{5.74}

Z = -0.71

Z = -0.71 has a pvalue of 0.2389

1 - 0.2389 = 0.7611

0.7611 = 76.11% probability that at least 110 believe global warming is occurring

3 0
3 years ago
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