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Luda [366]
2 years ago
7

How do I do this question, I'm very confused, I need an answer quickly

Mathematics
1 answer:
Sati [7]2 years ago
4 0
1) AB/AC = 1/2
(3x-4)/(2x+12) =1/2
2(3x-4) = 1(2x+12)
6x - 8 = 2x +12
4x = 20 And x = 5

2) Prove AC/BC 2/3 ===== (2x+12)/(7x-2) = 2/3;
Plug in x = 5:

(10+12)/(35-2) = 22/33. Simplify numerator & denominator by 11

(2)(11)/(3)(11) = 2/3
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7 Times 42 divided by 70,000×49×79 minus hundred
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The answer is -83.7418
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James sets off from home on his bike.The graph shows his journey.A)what is James average speed on his outward journey?B)How long
katen-ka-za [31]

Answer:

a) ~10.58 km/h

b) 45 min

c) 15 km/h

Step-by-step explanation:

a) 45km for 4 hours and 15min (4.25 hours) = 45 / 45.25 ~= 10.58 km/h

b) From 4:15 to 5 = 45min

c) 45km for 3 hours (5:00 to 8:00) = 45 / 3 = 15 km/h

NOTE!

The author actually wanted the answer on A) to be 10 km/h and the journey to be 4.5 hours (4 hours 30min) but instead they made it 4.25 (as shown on the graph, since 1 square is 30 minutes).

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3 years ago
Find the area of a triangle with a base of 10mm and a height of 8mm
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7 0
3 years ago
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23 20' 48" is the same as _____. Round to the nearest hundredth of a degree.
boyakko [2]
So this is read as 23 degrees, 20 minutes, and 48 seconds.  Each degree has 60 minutes and each minute has 60 seconds, somewhat like time.  You must start from right to left for this to work.  This may seem complicated but the way to find this is as follows:  
23+(20+(48/60))/60.  A simpler way to see this is by first taking 48/60 which is .8.  Now you take 20+.8 which is 20.8 and you divide it by 60 once more.  This comes out to be approximately .35.  Now you have converted the seconds and minutes to degrees so you add 23+.35 which is 23.35.  Therefore your answer is 23.35 degrees.  
8 0
3 years ago
Note: Enter your answer and show all the steps that you use to solve this problem in the space provided.
Vinil7 [7]

Answer:

a)

X[bar]_A= 71.8cm

X[bar]_B= 72cm

b)

M.A.D._A= 8.16cm

M.A.D._B= 5.4cm

c) The data set for Soil A is more variable.

Step-by-step explanation:

Hello!

The data in the stem-and-leaf plots show the heights in cm of Teddy Bear sunflowers grown in two different types of soil (A and B)

To read the data shown in the plots, remember that the first digit of the number is shown in the stem and the second digit is placed in the leaves.

The two data sets, in this case, are arranged in a "back to back" stem plot, which allows you to compare both distributions. In this type of graph, there is one single stem in the middle, shared by both samples, and the leaves are placed to its left and right of it corresponds to the observations of each one of them.

Since the stem is shared by both samples, there can be observations made only in one of the samples. For example in the first row, the stem value is 5, for the "Soil A" sample there is no leaf, this means that there was no plant of 50 ≤ X < 60 but for "Soil B" there was one observation of 59 cm.

X represents the variable of interest, as said before, the height of the Teddy Bear sunflowers.

a) To calculate the average or mean of a data set you have to add all observations of the sample and divide it by the number of observations:

X[bar]= ∑X/n

For soil A

Observations:

61, 61, 62, 65, 70, 71, 75, 81, 82, 90

The total of observations is n_A= 10

∑X_A= 61 + 61 + 62 + 65 + 70 + 71 + 75 + 81 + 82 + 90= 718

X[bar]_A= ∑X_A/n_A= 218/10= 71.8cm

For Soil B

Observations:

59, 63, 69, 70, 72, 73, 76, 77, 78, 83

The total of observations is n_B= 10

∑X_B= 59 + 63 + 69 + 70 + 72 + 73 + 76 + 77 + 78 + 83= 720

X[bar]_B= ∑X_B/n_B= 720/10= 72cm

b) The mean absolute deviation is the average of the absolute deviations of the sample. It is a summary of the sample's dispersion, meaning the greater its value, the greater the sample dispersion.

To calculate the mean absolute dispersion you have to:

1) Find the mean of the sample (done in the previous item)

2) Calculate the absolute difference of each observation and the sample mean |X-X[bar]|

3) Add all absolute differences

4) Divide the summation by the number of observations (sample size,n)

For Soil A

1) X[bar]_A= 71.8cm

2) Absolute differences |X_A-X[bar]_{A}|

|61-71.8|= 10.8

|61-71.8|= 10.8

|62-71.8|= 9.8

|65-71.8|= 6.8

|70-71.8|= 1.8

|71-71.8|= 0.8

|75-71.8|= 3.2

|81-71.8|= 9.2

|82-71.8|= 10.2

|90-71.8|= 18.2

3) Summation of all absolute differences

∑|X_A-X[bar]_A|= 10.8 + 10.8 + 9.8 + 6.8 + 1.8 + 0.8 + 3.2 + 9.2 + 10.2 + 18.2= 81.6

4) M.A.D._A=∑|X_A-X[bar]_A|/n_A= 81.6/10= 8.16cm

For Soil B

1) X[bar]_B= 72cm

2) Absolute differences |X_B-X[bar]_B|

|59-72|= 13

|63-72|= 9

|69-72|= 3

|70-72|= 2

|72-72|= 0

|73-72|= 1

|76-72|= 4

|77-72|= 5

|78-72|= 6

|83-72|= 11

3) Summation of all absolute differences

∑ |X_B-X[bar]_B|= 13 + 9 + 3 + 2 + 0 + 1 + 4 + 5 + 6 + 11= 54

4) M.A.D._B=∑ |X_B-X[bar]_B|/n_B= 54/10= 5.4cm

c)

If you compare both calculated mean absolute deviations, you can see M.A.D._A > M.A.D._B. As said before, the M.A.D. summary of the sample's dispersion. The greater value obtained for "Soil A" indicates this sample has greater variability.

I hope this helps!

7 0
2 years ago
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