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Tomtit [17]
3 years ago
15

The range for the Kentucky temperature is ____

Mathematics
2 answers:
natka813 [3]3 years ago
6 0

Answer:

1) 6.1

2)5.1

3)2.8

4)2.4

5)I don't know what the last one is asking for, are there answer choices

Step-by-step explanation:

1)Max-min=58.3-52.2=6.1

2)Max-min=55-48.9=5.1

3)Upper quartile- lower quartile= 57.3-54.5=2.8

4)Upper quartile- lower quartile= 53.3-50.9=2.4

5) I don't know what this one is asking

solmaris [256]3 years ago
5 0

Answer:

1) 6.1

2)5.1

3)2.8

4)2.4

5)More Variation for Kentuky

Step-by-step explanation:

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A boy flying a kite lets out 300 feet of string, which makes an angle of 38o with the ground. Assuming that the string is straig
Greeley [361]

Answer:

The answer is 184.7 ft

Step-by-step explanation:

You need to find the distance from kite to ground

sin38°/1 = x/300

x = 300 · sin 38

Therefore your answer is 184.7

This is the best way I could explain it....Hope this helps:)

4 0
2 years ago
Find the median mode range and any outliers. Then describe the data using them 77777,88,555,6,99,10101010,12
umka2103 [35]

Answer:

The median = 5

The mode  = 7

Range = 5

Outliers= None

Step-by-step explanation:

The median is the  value of the middle term.

The median is that value which divides the data into two equal halves.

7,7,7,7,7,8,8,5,5,5,6,9,9,10,10,10,10,12

There are 18 terms . So the middle term would be 9th term value= 5

The mode is the value which is repeated the most. = 7

Range is the difference between the highest and lowest value.= 12- 7= 5

The interquartile range, IQR= q3 - q1 where q3= 10 and q1= 7

IQR= 10 −7=3

The inner fences can be found out by

q1−(1.5*IQR) and q3+(1.5*IQR)

=7−(1.5*3) and 10+(1.5*3)

Inner fences: 2.5 and 14.5

The outer fences are q1−(3*IQR) and q3+(3*IQR)

=7−(3*3) and 10+(3*3)

Outer fences: -2 and 19

So if there are any outliers in this data set, they will either be less than -2 or greater than 19. But there are no outliers.

6 0
3 years ago
At Stephanie's new job, she spent $6.50, $8, $7.25, $13.50, and $9.75 on lunch the first week. In the second week, Stephanie spe
MaRussiya [10]

Answer:

The mean for the second week is $2 less than the first and in percentage it is 22% less.

Step-by-step explanation:

The mean is given by the sum of all individual values divided by the number of values. For the first week the sum is:

sum1 = 6.5 + 8 + 7.25 + 13.5 + 9.75

sum1 = 45

Since she spent 10 less in the second week the sum is:

sum2 = sum1 - 10 = 45 - 10 = 35

The mean for each week is:

mean1 = sum1/5 = 45/5 = 9

mean2 = sum2/5 = 35/5 = 7

difference = mean1 - mean2 = 9-7 = 2

difference(%) = [2/9]*100 = 0.22*100 = 22%

The mean for the second week is $2 less than the first and in percentage it is 22% less.

6 0
3 years ago
Can someone teach me how to solve this please
Alexeev081 [22]
Multiply it all together
7 0
3 years ago
A plane flew 360km in 3 hrs when flying with the wind.With no change in the wind,the return trip took 4 hrs.Find the speed of th
morpeh [17]
Speed of the plane: 250 mph
Speed of the wind: 50 mph
Explanation:
Let p = the speed of the plane
and w = the speed of the wind
It takes the plane 3 hours to go 600 miles when against the headwind and 2 hours to go 600 miles with the headwind. So we set up a system of equations.
600
m
i
3
h
r
=
p
−
w
600
m
i
2
h
r
=
p
+
w
Solving for the left sides we get:
200mph = p - w
300mph = p + w
Now solve for one variable in either equation. I'll solve for x in the first equation:
200mph = p - w
Add w to both sides:
p = 200mph + w
Now we can substitute the x that we found in the first equation into the second equation so we can solve for w:
300mph = (200mph + w) + w
Combine like terms:
300mph = 200mph + 2w
Subtract 200mph on both sides:
100mph = 2w
Divide by 2:
50mph = w
So the speed of the wind is 50mph.
Now plug the value we just found back in to either equation to find the speed of the plane, I'll plug it into the first equation:
200mph = p - 50mph
Add 50mph on both sides:
250mph = p
So the speed of the plane in still air is 250mph.
6 0
2 years ago
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