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gayaneshka [121]
2 years ago
12

Which is the graph of x Greater-than-or-equal-to 98.6? A number line going from 95 to 103. A closed circle is at 98.6. Everythin

g to the right of the circle is shaded. A number line going from 95 to 103. An open circle is at 98.6. Everything to the right of the circle is shaded. A number line going from 95 to 103. An open circle is at 98.6. Everything to the left of the circle is shaded. A number line going from 95 to 103. A closed circle is at 98.6. Everything to the left of the circle is shaded.
Mathematics
2 answers:
DENIUS [597]2 years ago
3 0

Answer:

A number line going from 95 to 103. A closed circle is at 98.6. Everything to the right of the circle is shaded.

Step-by-step explanation:

We want to determine the graph of x\geq 98.6.

Since we have the sign, \geq, we use a closed circle to show that the number 98.6 is included in the domain of definition.

Then since the sign is greater than, we shade to the right of 98.6.

The correct option is A.

scZoUnD [109]2 years ago
3 0

I will say it fast the answer is A

Step-by-step explanation:

And hi

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There are often fees associated with investments that reduce the return you actually have. True False
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3 years ago
Cheryl collected data for her mathematics project. She noted that the data set was approximately normal.
nadya68 [22]

Answer:

X_(r) >> X_(n)

The mean for this case would increase since is defined as:

\bar X= \frac{\sum_{i=1}^n X_i}{n}

The interquartile range would not change since the definition for the IQR is IQR =Q_3 -Q_1 and the quartiles are the same.

The standard deviation would not remain the same since by definition is:

s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

And since we change the largest value the deviation would increase considerably.

And for the last option is not always true since if we select a value so much higher then the distribution would be skewed to the right.

So the best option for this case is:

Mean would increase.

Step-by-step explanation:

For this case we assume that we have a random sample given X_(1), X_(2) ,..., X_(n) and for each observation X_i \sim N(\mu, \sigma) since the problem states that the data is approximately normal.

Let's assume that the largest value on this sample is X_(n) and for this case we are going to replace this value by another one extremely higher so we satisfy this condition:

X_(r) >> X_(n)

The mean for this case would increase since is defined as:

\bar X= \frac{\sum_{i=1}^n X_i}{n}

The interquartile range would not change since the definition for the IQR is IQR =Q_3 -Q_1 and the quartiles are the same.

The standard deviation would not remain the same since by definition is:

s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

And since we change the largest value the deviation would increase considerably.

And for the last option is not always true since if we select a value so much higher then the distribution would be skewed to the right.

So the best option for this case is:

Mean would increase.

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3 years ago
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Anastasy [175]

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6 0
2 years ago
What output is twice the sum of the input 3 and 4
Andreas93 [3]

Answer:

Required output = 14

Step-by-step explanation:

Sum of the input 3 and 4 = 3 + 4

Twice the sum means two times.

So, to find twice the sum of the input 3 and 4, we need to multiply the sum by 2.

2(3 + 4)

Perform the addition inside the parentheses.

2(3 + 4) = 2(7)

Now, multiply.

2(3 + 4) = 14

Hence, required output = 14

5 0
3 years ago
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