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Vlad [161]
3 years ago
9

What is the amount of change of 200-125

Mathematics
1 answer:
kati45 [8]3 years ago
7 0
The answer to that would be 0.125 because you do 25 divided by 200= which equals 0.125.
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Consider the following recursive definition of a set S: 2,3, and 4 are members of S, so is 6 added to any member of S. Consider
neonofarm [45]

Answer:

Detailed solution is given below:

8 0
3 years ago
You wash dishes for a chemistry laboratory to make extra money for laundry. You earn 12 dollars/hour, and each shift lasts 75 mi
Leni [432]

Answer:

2 shifts

Step-by-step explanation:

Amount earned washes dishes = 12 dollars / hour

Shift duration = 75 minutes

Cost of laundry = 12 quarters per load

Cost of 10 loads of laundry = (10 * 12 quarters) = 120 quarters

1 quarter = 25 cents

120 quarters = (25 * 120) = 3000 cents

100 cents = $1

3000 cents = 3000/ 100 = $30

Hence, to make $30 from washing dishes :

$12 = 1 hour

$30 = x

x = 30 / 12

x = 2.5 hours

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4 0
3 years ago
What is 7 times 2/8th
IgorLugansk [536]
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Hope this helps! (:
3 0
3 years ago
Suppose that a sample of size 100 is to be drawn from a population with standard deviation 10.
larisa86 [58]

Answer:

a) 68% probability that the sample mean will be within 1 of the value of μ.

b)

1)

Approximately 95% of the time, x will be within 2 of μ.

2)

Approximately 0.3% of the time, x will be farther than 3 from μ.

The last:

n = 40

n = 65

n = 130

n = 520

Step-by-step explanation:

To solve this problem, it is important to know two concepts: The Empirical Rule and the Central Limit Theorem.

Empirical Rule

The Empericial Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measuers are within 2 standard deviations of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

Central Limit Theorem

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

Suppose that a sample of size 100 is to be drawn from a population with standard deviation 10.

So \sigma = 10, n = 100, s = \frac{10}{\sqrt{100}} = 1

(a) What is the probability that the sample mean will be within 1 of the value of μ?

Within 1 is within one standard deviation of the mean \mu.

So there is a 68% probability that the sample mean will be within 1 of the value of μ.

(b) For this example (n = 100, σ = 10), complete each of the following statements by computing the appropriate value. (Round the answers to the nearest whole number.)

(1) Approximately 95% of the time, x will be within___ of μ.

By the empirical rule, 95% of the measures are within 2 standard deviations of the mean. In our sample, the standard deviation is 1.

So

Approximately 95% of the time, x will be within 2 of μ.

(2) Approximately 0.3% of the time, x will be farther than___ from μ.

By the empirical rule, 99.7% of the measures are within 3 standard deviations of the mean. In the other 0.3% of the time, the measures are farther than 3 standard deviations of the mean. In our sample, the standard deviation is 1.

So:

Approximately 0.3% of the time, x will be farther than 3 from μ.

A random sample is selected from a population with mean μ = 100 and standard deviation σ = 10. For which of the sample sizes would it be reasonable to think that the xsampling distribution is approximately normal in shape? (Select all that apply.)

As we saw above, in the central Limit theorem, we should use a sample size of at least 30. So

n = 40

n = 65

n = 130

n = 520

7 0
3 years ago
Find the distance between points p(2,2) q(7,4) to the nearest tenth
Greeley [361]
If you know the horizontal distances between the two points is a positive 5, and the vertical distance is a positive 2. By using the distance formula, you can come up with sqrt(5^2+2^2) = <span>5.38516480713, which to the nearest tenth would be 5.4</span>
7 0
3 years ago
Read 2 more answers
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