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koban [17]
3 years ago
10

2cos(3pi/4) – 4 sin(7pi/6)

Mathematics
1 answer:
dangina [55]3 years ago
7 0

Answer:

0.58578643762

Step-by-step explanation:

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The average number of customers arriving at an ATM machine is 27 per hour during lunch hours. Use equation to denote the number
Helga [31]

Answer:

The probability that exactly 2 customers arrive in a given 5 minute interval =  0.2667

Step-by-step explanation:

Given -

The average number of customers arriving at an ATM machine is 27 per hour during lunch hours. then the average number of customers arriving at an ATM machine n a 5 minute time interval = \frac{27}{60} \times 5 = 2.25

average number of customers arriving at an ATM machine n a 5 minute time interval (\lambda ) = 2.25

Let X denote the no of customer arrivals in a 5 minute time interval

The probability that exactly 2 customers arrive in a given 5 minute interval =

P( X = 2 )  = \frac{e^{-\lambda }\lambda ^{X}}{X!}            ( Using poision distribution )

               =  \frac{e^{-2.25} (2.25)^2}{2!}

               = \frac{.1054 \times 5.0625}{2}

               =   0.2667

4 0
3 years ago
I’ll give brainliest answer if correct
STatiana [176]

Answer:

x = 1.7272727272

Step-by-step explanation:

8 0
3 years ago
Are the expressions you developed in Part A and Part C the same? Why or why not?
pshichka [43]

Answer:

It is the same answer, Because b is the same as a but in fraction form, and c is the same as b but in improper fraction form, so that means they are the same answer

Step-by-step explanation:

3 0
2 years ago
The mean life of a television set is 119119 months with a standard deviation of 1414 months. If a sample of 7474 televisions is
Pepsi [2]

Answer:

0.5034 = 50.34% probability that the sample mean would differ from the true mean by less than 1.1 months

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean

In this problem, we have that:

\mu = 119, \sigma = 14, n = 74, s = \frac{14}{\sqrt{74}} = 1.6275

What is the probability that the sample mean would differ from the true mean by less than 1.11 months?

This is the pvalue of Z when X = 119 + 1.1 = 120.1 subtracted by the pvalue of Z when X = 119 - 1.1 = 117.9. So

X = 120.1

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

By the Central Limit Theorem

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

Z = \frac{120.1 - 119}{1.6275}

Z = 0.68

Z = 0.68 has a pvalue of 0.7517

X = 117.9

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

Z = \frac{117.9 - 119}{1.6275}

Z = -0.68

Z = -0.68 has a pvalue of 0.2483

0.7517 - 0.2483 = 0.5034

0.5034 = 50.34% probability that the sample mean would differ from the true mean by less than 1.1 months

3 0
3 years ago
Find the 15th term of the arithmetic sequence -5x+3−5x+3, -9x-2−9x−2, -13x-7,.......???
Anna35 [415]

Answer:

- 61x - 67

Step-by-step explanation:

The n th term of an arithmetic sequence is

a_{n} = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

d = - 9x - 2 - (- 5x + 3)

  = - 9x - 2 + 5x - 3

  = - 4x - 5 and a₁ = - 5x + 3, thus

a_{15} = - 5x + 3 + 14(- 4x - 5) = - 5x + 3 - 56x - 70 = - 61x - 67

5 0
3 years ago
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