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Alecsey [184]
3 years ago
15

Which set of measurements could be the side lengths of a triangle? please help!

Mathematics
1 answer:
Rom4ik [11]3 years ago
3 0

All of them could be as A would be an equilateral B,C,D would be irregular triangles and 3 would be an Isosceles triangle but if you had to go with one it would be A

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A gallon of Moo Milk costs $5.12. What is the price, in dollars, of an 8 ounce glass of Moo Milk?
Lesechka [4]

Answer:

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8 0
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Solve:C=10d + 5n for n
katovenus [111]

n =  - 2d +  \frac{c}{5}
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5 0
3 years ago
8 divide by 1 1/4 KCF
Diano4ka-milaya [45]

Answer:

32/5

Step-by-step explanation:

K=Keep the first Fraction (You can rewrite 8 as 8/1 for when you multiply across.)

C=Change the Division Sign to a Multiplication Sign

F=Flip the Second Fraction

First, rewrite 1 1/4 as 5/4 (A mixed number can be changed to a fraction by multiplying the outside whole number by the denominator or bottom number of the fraction, then add the new rewritten whole number and the original fractional piece. In this case you would multiply the outside 1 by the bottom 4 and add it to the original 1/4. 4/4+1/4=5/4)

Second, Flip 5/4 to 4/5 and change your equation so it now reads: 8/1 x 4/5

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6 0
3 years ago
30 POINTS!!!!! WILL MARK BRAINLST!
bazaltina [42]

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Reduction

Step-by-step explanation:

3 0
3 years ago
Radical Expressions and Data Analysis Unit Portfolio
Gnoma [55]
TASK 1
a) Data and frequency table are attached in picture #1.

b) The histogram is attached in picture #2

c) As the intervals increase the number of pieces of mail received, the frequency decreases, therefore we can say that are received few pieces of mail <span>are received more often.

TASK 2
a) The list of breeds and weights are reported in picture #3. 
</span>
In order to find the mean, sum up all the weights and divide by 9:
m = (<span>8 + 10 + 15 + 29 + 32 + 75 + 77 + 80 + <span>85) / 9 = 45.67 lbs

In order to find the median, you need to select the central value of your distribution: since you have 9 values, the central one will be in the fifth position:
M = 32 lbs

In order to find the mode, you need to select the value which shows more frequently: in this case, every value shows only once, therefore the mode cannot be determined.

The data are best described by the mean because they are almost symmetrically distributed between the least and the biggest values, while the median is more towards the small-weight side of the distribution.
 
b) In order to have an average of 250 lbs, the tenth dog should weight 2089 lbs.

Indeed, if the average is 250 lbs for 10 dogs, it means that the total sum of their weights is 250 × 10 = 2500 lbs. We know that the sum of the first 9 dogs is 411 lbs, therefore, the tenth dog should weight 2500 - 411 = 2089 lbs.

TASK 3
The city picked is Baltimore, the year 2017.
The table is attached in picture #4.
</span></span>
a) The box-and-whisker plot is attached in picture #5.

b) The median is the central value of the distribution, which is: <span> 
42.4, 45.3, 46.8 | 54.7, 57.6, 66.0 | 69.1,  76.5,  80.6 | 85.8, 87.8, 90.0 </span>
where we marked with a tally the quartiles.

Since we have 12 values, the median will be the average between the two central values:
M = (66.0 + 69.1) / 2 = 67.1 °F

c) 7<span>5% of the temperatures are below 83.2 °F.
Indeed, this value represents the third quartile. The position of the third quartile can be found by the formula:
3/4 </span>· (n + 1) = 3/4 · (12 + 1) = 3/4 <span>· 13 = 9.75
Therefore, since we did not get in integer position, the third quartile will be the average between the numbers in position 9 and 10:
q</span>₀.₇₅ = (80.6 + 85.8) / 2 = 83.2 °F

d) 7<span>5% of the temperatures are above 50.8 °F.
Indeed, this value represents the first quartile. Similarly to point c), the position can be found by the formula:
1</span>/4 · (n + 1) = 1/4 <span>· 13 = 3.25
And therefore:
</span>q₀.₂₅ = (<span><span><span>46.8 + </span><span>54.7) / 2 = 50.8 °F.</span></span></span> 

e) Baltimore in 2017 had a median high temperature of 67.1 °F.
25% of the year the high temperatures were warmer than 83.2 °F, with no outlier above 90.0 °F which was the hottest temperature.
25% of the high temperatures were colder than 50.8 °F, with no outlier below 42.4 °F, which was the coldest high temperature.

6 0
3 years ago
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