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taurus [48]
2 years ago
9

The price, p, for different size orders of printed programs for a musical production, n, is given in the table.

Mathematics
2 answers:
Leya [2.2K]2 years ago
8 0
The answer is B, hope this helps!
soldier1979 [14.2K]2 years ago
8 0

Answer:

A linear equation can not be used for this problem.

Step-by-step explanation:

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What is 4 ten thousands 4 thousands times 10
cluponka [151]
4,000*10= 40,000 which is forty thousands which is also 4 ten thousands 
6 0
3 years ago
Jacob's favorite movie is 1 5/6 hours long. He says he has watch the movie 5 and a half times. If that is true. How many hours h
34kurt
1\frac{5}{6}=\frac{11}{6}, and when you multiply it by 5 it turns into \frac{55}{6}, which is 9\frac{1}{6}

So in conclusion he watched 9 & 10 minutes watching the movie
3 0
3 years ago
Plz help solve this please and thx u<br><br>​
kow [346]
2 (an - 1) = 40
2an - 2 =40
2an = 40 - 2
2an = 38
2an/2 = 38/2
an = 19
3 0
2 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.

6 0
3 years ago
HELP ME PLEASE ASAP SAT
Nimfa-mama [501]

The answer is 4

4/x=6/x+2

⇒x=4

Not that hard :VV

4 0
2 years ago
Read 2 more answers
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