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Ivan
3 years ago
8

Sally works for mail workers stuffing envelopes. She is paid $0.20 for each envelope she stuffs. How much does she earn for a we

ek in which she stuffs 523 envelopes?
10.46
104.60
126.50
1,265.00
Mathematics
1 answer:
Ghella [55]3 years ago
7 0

104.60, hope this helps you

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The random variable X, representing the number of accidents in a certain intersection in a week, has the following probability d
ollegr [7]

Answer: 0.70

Step-by-step explanation:

Given : The random variable X, representing the number of accidents in a certain intersection in a week, has the following probability distribution:

       x       0       1         2        3       4       5

P(X = x) 0.20   0.30   0.20   0.15   0.10  0.05

Using the above probability distribution , the  the probability that in a given week there will be at most 3 accidents is given by :_

P(\leq3)=P(0)+P(1)+P(2)+P(3)\\\\=0.20+0.30+0.20=0.70

Hence, the required probability = 0.70

7 0
3 years ago
A ladder is placed 30 inches from a wall. It touches the wall at a height of 50 inches from the ground.
kati45 [8]
I don't know if you need the answer to how long the ladder is or what degree the ladder and ground form so I am going to give you both.

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7 0
2 years ago
Read 2 more answers
At a university, 60% of the 7,400 students are female. The student newspaper reports the results of a survey of a random sample
omeli [17]

Given Information:

Population mean = p  = 60% = 0.60

Population size = N = 7400

Sample size = n = 50

Required Information:

Sample mean = μ = ?

standard deviation = σ = ?

Answer:

Sample mean = μ = 0.60

standard deviation = σ = 0.069

Step-by-step explanation:

We know from the central limit theorem, the sampling distribution is approximately normal as long as the expected number of successes and failures are equal or greater than 10

np ≥ 10

50*0.60 ≥ 10

30 ≥ 10 (satisfied)

n(1 - p) ≥ 10

50(1 - 0.60) ≥ 10

50(0.40) ≥ 10

20 ≥ 10  (satisfied)

The mean of the sampling distribution will be same as population mean that is

Sample mean = p = μ = 0.60

The standard deviation for this sampling distribution is given by

\sigma = \sqrt{\frac{p(1-p)}{n} }

Where p is the population mean that is proportion of female students and n is the sample size.

\sigma = \sqrt{\frac{0.60(1-0.60)}{50} }\\\\\sigma = \sqrt{\frac{0.60(0.40)}{50} }\\\\\sigma = \sqrt{\frac{0.24}{50} }\\\\\sigma = \sqrt{0.0048} }\\\\\sigma =  0.069

Therefore, the standard deviation of the sampling distribution is 0.069.

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2 years ago
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