On a map, Julia's home is 8.5 in. from the library. If the map scale is 1 in.: 0.25 mi, how many miles from the library does Julia live?

Answers

Answer 1

Answer: Just multiply.

$\sf8.5*0.25=\boxed{\sf2.125}$

Answer 2

Answer: If the map scale is 1 inch: 0.25 miles, then we can multiply 0.25 x 8.5 to get our answer. 0.25 x 8.5 = 2.125, so Julia's home is 2.125 miles away from the library. :)

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An inequality is shown.12+11/6x≤ 5+3x
Select the statement(s) and number line(s) that can represent the inequality. Click all that apply.
a. The solution set is {6, ∞} for x ∈ R.
b. The solution set is {6, 7, 8, …} for x ∈ N.
c. 6 ≤ x
d. The value of a number substituted for x is greater than 6.
(more options below.)

Answers

The right answer for the question that is being asked and shown above is that: "d. The value of a number substituted for x is greater than 6.; b. The solution set is {6, 7, 8, …} for x ∈ N." These are the statements and number lines that can represent the inequality.

Answer:

Step-by-step explanation:

56y

What is the mean for the data set? {29, 40, 46, 6, 29, 6}

Answers

To get the Mean, you have to add all the ages of the men together:
$29+40+46+6+29+6=156$
Next you divide $156$ by the amount of people:
$156/6=26$
Final Answer: 26

What are the dimensions of the rectangle?

Answers

the dimensions would be 18 by 5.5. because you know the area is 99, and you know the area formula is length multiplied by width, you set the dimensions to 99, and continue to solve for x. then you plug the x value into the length’s dimension to get 18.

How do you input this into the calculator

Answers

$\sin\theta=(b)/(c)\ therefore\ b=c\cdot\sin\theta\n\n\cos\theta=(a)/(c)\ therefore\ a=c\cdot\cos\theta$

$b=32\cdot\sin55^o=\sin55^o\cdot32\n\na=32\cdot\cos55^o=\cos55^o\cdot32$

The data set below has an outlier.2, 10, 10, 11, 11, 12, 12, 12, 13, 14, 14

Which value is changed the most by removing the outlier?

A: the range
B: the median
C: the lower quartile
D: the interquartile range

Answers

Data set: 2, 10, 10, 11, 11, 12, 12, 12, 13, 14, 14

range: 14 - 2 = 12
median: 12
lower quartile = 10
interquartile range = 4

Data set w/o outlier 2: 10, 10, 11, 11, 12, 12, 12, 13, 14, 14

range: 14 - 10 = 4
median: 12
lower quartile: 10.5
interquartile range = 3.5

The value that change the most by removing the outlier is A.) THE RANGE.

To answer the problem above, we must define what an outlier is. An outlier is a number in a set of data that is smaller or larger than the rest of the data. In this problem, the outlier is 2. To know which value changed the most we solve each with and without the outlier.
A. range: 12(with), 4(without)
B. median: 12 (with), 12 (without)
C. lower quartile: 10.5(with), 11(without)
D. interquartile range: 2(with), 2(without)

The answer is A. range.

Kaleb has 5 pink, 4 yellow, 3 orange, and 3 red Starburst. What is the probability Kaleb will randomly pick a yellow Starburst, eat it, and then randomly pick another yellow Starburst?

Answers

The awnser would be 4/15 for the yellow starburst

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