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3 years ago

PLEASE HELP ASAP!!!! IT WILL BE DEEPLY APRECIATED AND NO LINKS PLEASE

1 answer:

6 0

Answer: Parallel lines are lines in a plane which do not meet; that is, two straight lines in a plane that do not intersect at any point are said to be parallel. Colloquially, curves that do not touch each other or intersect and keep a fixed minimum distance are said to be parallel.

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The mean of 100 observation is found to be 40. If at the time of compution two items were wrongly taken as 30 and 27 instead of

Colt1911 [192]

Answer:

40.18

Step-by-step explanation:

According to this question, the mean of 100 observation is found to be 40. This means that the number of values in the set of data is 100. However, at the time of computation two items were wrongly taken as 30 and 27 instead of 3 and 72.

The mean is derived by dividing the sum of values in the data (Σx) by the number of observations (n) i.e.

Mean = Σx/n

For the computation error:

Mean = 40

Let x represent the sun of other 98 values except 30 and 27

n = 100

40 = x + 30 + 27/100

40 = 57 + x/100

Cross multiply

4000 = 57 + x

x = 4000 - 57

x = 3943.

Sum of other 98 values = 3943.

Now, let's find the correct mean with correct values of 3 and 72;

Mean = Σx/n

Mean = 3943 + 3 + 72/100

Mean = 4018/100

Mean = 40.18

7 0

2 years ago

Here are yesterday's high temperatures (in Fahrenheit) in 12 U.S. cities. 48, 50, 54, 56, 63, 63, 64, 68, 74, 74, 79, 80 Notice

irinina [24]

For the given data set

Minimum = 48

Lower quartile = 55

Median = 63.5

Upper quartile = 74

Maximum = 80

Interquartile range = 19

<h3>Measures of a Data </h3>

From the question, we are to determine the minimum, lower quartile, median, upper quartile, maximum, and interquartile range of the given data set

The given data set is

48, 50, 54, 56, 63, 63, 64, 68, 74, 74, 79, 80

Minimum = 48

Lower quartile = (54+56)/2

Lower quartile = 110/2

Lower quartile = 55

Median = (63+64)/2

Median = 127/2

Median = 63.5

Upper quartile = (74+74)/2

Upper quartile = 148/2

Upper quartile = 74

Maximum = 80

Interquartile range = Upper quartile - Lower quartile

Interquartile range = 74 - 55

Interquartile range = 19

Hence, for the given data set

Minimum = 48

Lower quartile = 55

Median = 63.5

Upper quartile = 74

Maximum = 80

Interquartile range = 19

Learn more on Measures of a Data here: brainly.com/question/15097997

#SPJ1

5 0

2 years ago

Graph the line with the equation y= 4/5 x+2.

lawyer [7]

Answer:

x are -5/2 and 0 . Y are 0 and 2.

5 0

3 years ago

What is the volume of a shipping cube with dimensions of 2 and 1/2 feet

Fantom [35]

For this case we have that by definition, the volume of a cube is given by:

$V = l ^ 3$

Where:

l: It's the side of the cube

According to the statement data:

$l = 2 \frac {1} {2} = \frac {2 * 2 + 1} {2} = \frac {5} {2}$

Substituting in the formula we have:

$V = (\frac {5} {2}) ^ 3\\V = \frac {5 ^ 3} {2 ^ 3}\\V = \frac {125} {8}\\V = 15.625$

Thus, the shipping cube volume is $15.625 \ ft ^ 3$

Answer:

$15.625 \ ft ^ 3$

6 0

3 years ago

PLEASE HELP MEEEEE!!!!!!!

OverLord2011 [107]

Answer:

Cheese: $4\frac{3}{8}$
Milk: $1\frac{2}{3}$
Pasta: $6\frac{1}{4}$

Step-by-step explanation:

Cheese:

$1\frac{3}{4} = \frac{7}{4}$
Set up a proportion: $\frac{\frac{7}{4} }{4} = \frac{x}{10}$
Cross multiply, then divide: 10 × 7/4 = 17.5, 17.5 ÷ 4 = 4.375
$4.375 = 4\frac{3}{8}$

Milk:

Set up a proportion: $\frac{\frac{2}{3} }{4} = \frac{x}{10}$
Cross multiply, then divide: 10 × 2/3 = $6\frac{2}{3}$ , $6\frac{2}{3}$ ÷ 4 = $1\frac{2}{3}$

Pasta:

$2\frac{1}{2} = \frac{5}{2}$
Set up a proportion: $\frac{\frac{5}{2} }{4} = \frac{x}{10}$
Cross multiply, then divide: 10 × 5/2 = 25, 25 ÷ 4 = 6.25
$6.25 = 6\frac{1}{4}$

I hope this helps!

4 0

3 years ago