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Gre4nikov [31]
3 years ago
11

In monitoring lead in the air after the explosion at the battery factory, it is found that the amounts of lead over a 7 day peri

od had a standard error of 1.91. Find the margin of error that corresponds to a 95% confidence interval. (Round to 2 decimal places)
Mathematics
1 answer:
Andrew [12]3 years ago
7 0

Answer:

MOE=\pm \ 1.41

Step-by-step explanation:

-Given that \sigma=1.91, n=7

-We determine the z-value corresponding to the 95% confidence level:

z_{\alpha/2}=z_{0.025}=1.96

-The margin of error is the degree of error from the real value and is calculated as follows:

MOE=\pm z_{\alpha/2} (\frac{\sigma}{\sqrt{n}})\\\\=\pm z_{0.025}\times\frac{\sigma}{\sqrt{n}}\\\\=\pm 1.96\times \frac{1.91}{\sqrt{7}}\\\\=\pm 1.41

Hence, the margin of error is \pm 1.41

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Combine like terms.

2x + 6 + 6x - 1

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After multiplying, a calculator screen displays the answer shown below. How would you write this number in standard form? Enter
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Answer:

8.234 E14=8.234 \times 10^{14}

Step-by-step explanation:

The given number is 8.234 E14

We need to write this number in standard form.

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3 years ago
The graph of F(x) shown below resembles the graph of G(x) = x ^ 2 but it has been changed somewhat . Which of the following coul
sukhopar [10]

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f(x) = 3x^{2} + 2

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Given G(x) = x^{2}

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=> f(x) = ax^{2} + 2

As you can see, the graph of f(x) is stretched vertically by a dilation factor of 3

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3 years ago
The random variable x has the following probability distribution: x f(x) 0 .25 1 .20 2 .15 3 .30 4 .10 a. Is this probability di
Reptile [31]

Answer and Explanation:

Given : The random variable x has the following probability distribution.

To find :

a. Is this probability distribution valid? Explain and list the requirements for a valid probability distribution.

b. Calculate the expected value of x.

c. Calculate the variance of x.

d. Calculate the standard deviation of x.

Solution :

First we create the table as per requirements,

x      P(x)         xP(x)           x²            x²P(x)

0    0.25           0               0                0

1     0.20        0.20             1              0.20

2    0.15          0.3               4             0.6

3    0.30         0.9               9             2.7

4    0.10          0.4               16             1.6

   ∑P(x)=1     ∑xP(x)=1.8               ∑x²P(x)=5.1

a) To determine that table shows a probability distribution we add up all five probabilities if the sum is 1 then it is a valid distribution.

\sum P(X)=0.25+0.20+0.15+0.30+0.10

\sum P(X)=1

Yes it is a probability distribution.

b) The expected value of x is defined as

E(x)=\sum xP(x)=1.8

c) The variance of x is defined as

V=\sum x^2P(x)-(\sum xP(x))^2\\V=5.1-(1.8)^2\\V=5.1-3.24\\V=1.86

d) The standard deviation of x is  defined as

\sigma=\sqrt{V}

\sigma=\sqrt{1.86}

\sigma=1.136

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ch4aika [34]

Answer:

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6 0
2 years ago
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