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docker41 [41]
2 years ago
8

27. A 35 foot long tube is cut into two pieces with ratio 4:5 Find the length of the shorter piece.

Mathematics
2 answers:
puteri [66]2 years ago
8 0
The ratio 4:5> 4x+ 5x=35> 9x=35> x=3.8 and then you do (3.8)3.8=14.7. Therefore there is not a correct answer.
polet [3.4K]2 years ago
7 0

Answer:

x= 16 ft.

Step-by-step explanation:

Let 'x' and 'y' are the lengths of the two pieces then:

1.) x + y = 36

2.) x : y = 4 : 5 cross multiply

3.) 5x = 4y

by solving the system of equations

x + y = 36

5x = 4y

we find

x = 16 ft

y = 20 ft

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.<br> A. 139<br> B. 147°<br> C. 155<br> D. 157<br> PLEASE HELP ASAAAPP!
enyata [817]

Answer:

\huge\boxed{157 \ \ \text{degrees}}

Step-by-step explanation:

In order to find the value of ∠EFB here, we have to note our angle relationships.

We know that ∠CFE is already 90°. We also know that ∠CFA is 90°. Angle ∠AFB is inside ∠CFA. Since we know the measure of ∠AFB, we can find the measure of ∠BFC.

23 + x = 90\\\\x = 90-23\\\\x = 67

Now that we know ∠CFE and ∠BFC, which together make ∠BFE, we can add these angles up.

90+67=157

Hope this helped!

5 0
3 years ago
It describes itself &lt;3
irina [24]

Answer:

432 ft²

Step-by-step explanation:

The formula of finding the area of a trapezoid: (base 1 + base 2/ 2) x h

(base 1 + base 2/ 2) x h

= (20 + 28/2) x 18

= 24 x 18

= 432 ft²

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
Pls help me asapp quick easy 10 points
tekilochka [14]
It’s goes by 7.5
Just by looking at the chart you can see x-axis to y- axis for example, point for 2 ,you can divide to find out the answer. So it’s 7.5
6 0
3 years ago
The math team is selling snacks after school to raise money. If chips are $0.50 a bag and sodas are $0.75 a can, write an inequa
amid [387]
0.50x+0.75b is greater than or equal to 200
(idk how to put the greater than or equal to sign but make sure you put the sign)


X is the bag of chips


B is the sodas
7 0
3 years ago
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