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timofeeve [1]
3 years ago
5

Use calculator to find the values of the following,correct to 3decimal places: 1. Cot 61,4

Mathematics
1 answer:
g100num [7]3 years ago
6 0

No such thing as cot (61,4). I think you meant to say cot (61.4°). See the difference?

So, cot (61.4°) is approximately -0.139844.

Rounded to 3 decimal places we get -0.140.

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The amount of syrup that people put on their pancakes is normally distributed with mean 63 mL and standard deviation 13 mL. Supp
andreyandreev [35.5K]

Answer:

(a) X ~ N(\mu=63, \sigma^{2} = 13^{2}).

    \bar X ~ N(\mu=63,s^{2} = (\frac{13}{\sqrt{43} } )^{2}).

(b) If a single randomly selected individual is observed, the probability that this person consumes is between 61.4 mL and 62.8 mL is 0.0398.

(c) For the group of 43 pancake eaters, the probability that the average amount of syrup is between 61.4 mL and 62.8 mL is 0.2512.

(d) Yes, for part (d), the assumption that the distribution is normally distributed necessary.

Step-by-step explanation:

We are given that the amount of syrup that people put on their pancakes is normally distributed with mean 63 mL and a standard deviation of 13 mL.

Suppose that 43 randomly selected people are observed pouring syrup on their pancakes.

(a) Let X = <u><em>amount of syrup that people put on their pancakes</em></u>

The z-score probability distribution for the normal distribution is given by;

                      Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = mean amount of syrup = 63 mL

            \sigma = standard deviation = 13 mL

So, the distribution of X ~ N(\mu=63, \sigma^{2} = 13^{2}).

Let \bar X = <u><em>sample mean amount of syrup that people put on their pancakes</em></u>

The z-score probability distribution for the sample mean is given by;

                      Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = mean amount of syrup = 63 mL

            \sigma = standard deviation = 13 mL

            n = sample of people = 43

So, the distribution of \bar X ~ N(\mu=63,s^{2} = (\frac{13}{\sqrt{43} } )^{2}).

(b) If a single randomly selected individual is observed, the probability that this person consumes is between 61.4 mL and 62.8 mL is given by = P(61.4 mL < X < 62.8 mL)

   P(61.4 mL < X < 62.8 mL) = P(X < 62.8 mL) - P(X \leq 61.4 mL)

  P(X < 62.8 mL) = P( \frac{X-\mu}{\sigma} < \frac{62.8-63}{13} ) = P(Z < -0.02) = 1 - P(Z \leq 0.02)

                                                           = 1 - 0.50798 = 0.49202

  P(X \leq 61.4 mL) = P( \frac{X-\mu}{\sigma} \leq \frac{61.4-63}{13} ) = P(Z \leq -0.12) = 1 - P(Z < 0.12)

                                                           = 1 - 0.54776 = 0.45224

Therefore, P(61.4 mL < X < 62.8 mL) = 0.49202 - 0.45224 = 0.0398.

(c) For the group of 43 pancake eaters, the probability that the average amount of syrup is between 61.4 mL and 62.8 mL is given by = P(61.4 mL < \bar X < 62.8 mL)

   P(61.4 mL < \bar X < 62.8 mL) = P(\bar X < 62.8 mL) - P(\bar X \leq 61.4 mL)

  P(\bar X < 62.8 mL) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{62.8-63}{\frac{13}{\sqrt{43} } } ) = P(Z < -0.10) = 1 - P(Z \leq 0.10)

                                                           = 1 - 0.53983 = 0.46017

  P(\bar X \leq 61.4 mL) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } \leq \frac{61.4-63}{\frac{13}{\sqrt{43} } } ) = P(Z \leq -0.81) = 1 - P(Z < 0.81)

                                                           = 1 - 0.79103 = 0.20897

Therefore, P(61.4 mL < X < 62.8 mL) = 0.46017 - 0.20897 = 0.2512.

(d) Yes, for part (d), the assumption that the distribution is normally distributed necessary.

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Answer:

14.8-16x

Step-by-step explanation:

2x+14-18x+0.8

14.8-16x

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John gave away some money. He has $12.00 left. If m represents the money he gave away, which expression represents the money he
makvit [3.9K]

Answer:

the answer would be d

Step-by-step explanation:

you have to divide $12.00 by the amount of money or m

3 0
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