Write the measure 29° 6' 6" as decimal to the nearest thousandth.

1 answer:

6 0

This angle is "29 degrees 6 minutes 6 seconds".

-- There are 60 minutes in 1 degree.
   So 1 minute = 1/60 degree.

-- There are 60 seconds in 1 minute.
      So 1 second = 1/60 minute = 1/3600 degree.

29 6' 6" = (29)° + (6/60)° + (6/3600)°

   = (29)° + (0.1)° + (0.001666...)°

   = 29.101666...°

Rounded to the nearest thousandth of a degree: 29.102°

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In a beach town, 13% of the residents own boats. A random sample of 100 residents was selected. What is the probability that les

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Answer:

Approximately $0.038$ (or equivalently $3.8\%$ ,) assuming that whether each resident owns boats is independent from one another.

Step-by-step explanation:

Assume that whether each resident of this town owns boats is independent from one another. It would be possible to model whether each of the $n = 100$ selected residents owns boats as a Bernoulli random variable: for $k = \lbrace 1,\, \dots,\, 100\rbrace$ , $X_k \sim \text{Bernoulli}(\underbrace{0.13}_{p})$ .

$X_k = 0$ means that the $k$ th resident in this sample does not own boats. On the other hand, $X_k = 1$ means that this resident owns boats. Therefore, the sum $(X_1 + \cdots + X_{100})$ would represent the number of residents in this sample that own boats.

Each of these $100$ random variables are all independent from one another. The mean of each $X_k$ would be $\mu = 0.13$ , whereas the variance of each $X_k\!$ would be $\sigma = p\, (1 - p) = 0.13 \times (1 - 0.13) = 0.1131$ .

The sample size of $100$ is a rather large number. Besides, all these samples share the same probability distribution. Apply the Central Limit Theorem. By this theorem, the sum $(X_1 + \cdots + X_{100})$ would approximately follow a normal distribution with:

mean $n\, \mu = 100 \times 0.13 = 13$ , and
variance $\sigma\, \sqrt{n} = p\, (1 - p)\, \sqrt{n} = 0.1131 \times 10 = 1.131$ .

$11\%$ of that sample of $100$ residents would correspond to $11\% \times 100 = 11$ residents. Calculate the $z$ -score corresponding to a sum of $11$ :

$\begin{aligned}z &= \frac{11 - 13}{1.131} \approx -1.77 \end{aligned}$ .

The question is (equivalently) asking for $P( (X_1 + \cdots + X_{100}) < 11)$ . That is equal to $P(Z < -1.77)$ . However, some $z$ -tables list only probabilities like $P(Z > z)$ . Hence, convert $P(Z < -1.77)\!$ to that form: