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Feliz [49]
3 years ago
13

Help me fast please lskrwe

Mathematics
2 answers:
Elanso [62]3 years ago
8 0

Answer: c = 1.59 * a

Step-by-step explanation:

You got the answer correct.

Sever21 [200]3 years ago
5 0

Answer:

You are correct, it is c = 1.59 * a

Step-by-step explanation:

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Find measure &lt; BPD <br> A. 9 degrees<br> B. 146 degrees <br> C. 155 degrees <br> D. 14 degrees
Aleks04 [339]

Answer:

The answer is B

Step-by-step explanation:

As the picture has already given you the answer, ∠BPD = 146°.

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A high school will graduate 100 students. of these, 52 students plan to attend college.
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A total of 48 students will not go to college
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HURRY PLEASE! Which expression is equivalent to ^4sqrt6/^3sqrt2?<br><br> (options in the image)
melamori03 [73]

Answer:

According to the given problems the one that gets closer is the first option. ^12sqrt27/2

Step-by-step explanation:

<em>Simplify the radical by breaking the radicand up into a product of known factors.</em>

The answer for the first question is 12 √ 27 / 2


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3 years ago
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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
3 years ago
Given x-intercepts of -3 and 2 find and equation for a graph that goes through both points. (You do NOT have to graph the equati
lesya [120]

Answer:

1? lllllllllllll the answer is 1 I think

6 0
3 years ago
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