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Studentka2010 [4]
3 years ago
9

If f(x)=x-3/x and g(x) = 5x – 4, what is the domain of (f o g) (x)?

Mathematics
1 answer:
8_murik_8 [283]3 years ago
7 0

Answer:

Step-by-step explanation:

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What additional info do you need
Novay_Z [31]

Answer:

  ∠C ≅ ∠M  or  ∠B ≅ ∠L

Step-by-step explanation:

You are given an angle and its opposite side as being congruent. AAS requires two congruent angles and one side, so you need another set of congruent angles (one in each triangle). It does not matter which they are. The above-listed pairs are appropriate.*

_____

* Since the figure cannot be assumed to be drawn to scale, either of angles B or C could be declared congruent to either of angles L or M. However, it appears that angles B and L are opposite the longest side of the triangle, so it makes good sense to declare that pair congruent. The same congruence statement (ΔBCD≅ΔLMN) would result from declaring angles C and M congruent. So, either declaration will work (matches the last answer choice.)

__

AAS requires two angles and a side. One side is already marked, so we do not need any more information about sides. (The second and third answer choices can be rejected as irrelevant.)

7 0
3 years ago
Brittany is making trays of fudge. Her recipe calls for 2/3 cup of walnuts for each batch of fudge . How many cups of walnut doe
aleksley [76]

Answer:

maybe 18

Step-by-step explanation:

3 0
3 years ago
Given that (-1, a) is the midpoint of the line segment joining (0, -2) and (b, 8) then find the values of a and b
Nikitich [7]

Answer: a=3 & b=-2


Step-by-step explanation:


6 0
2 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
3 years ago
The mean breaking strength of yarn used in manufacturing drapery material is required to be more than 100 psi. Past experience h
Ganezh [65]

Answer:

H0: μ = 100 ; H1: μ > 100 ;

Test statistic = 0.75 ;

We fail to reject null ;

Kindly check explanation for the rest

Step-by-step explanation:

Given that :

H0: μ = 100

H1: μ > 100

Sample mean, x = 100.6 ; n = 9 ; sample Standard deviation, s = 2.4

Test statistic :

(x - μ) ÷ s/sqrt(n)

(100.6 - 100) ÷ 2.4 / sqrt(9)

0.6 ÷ 0.8 = 0.75

Probability of not rejecting the null :

P = P(Z > 0.75) = 0.22663

α = 0.05

Since, P > α ; we fail to reject the Null ; there is no sufficient evidence to accept the claim that mean strength is > 100 psi

μ + Zcritical*s/sqrt(n)

Zcritical at (1 - α) = 1.645

100 + 1.645*(2.4/3)

100 + 1.316 = 101.316

(x - μ) ÷ s/sqrt(n)

(101.316 - 102) ÷ 2.4 / sqrt(9)

-0.684 ÷ 0.8 = 0.75

= - 0.855

P(Z < - 0.855) = 0.1963 (Z probability calculator).

95% lower confidence interval :

x - error margin, E

E = Zcritical * s/sqrt(n)

E = 1.645 * 0.8 = 1.316

100.6 - 1.316 = 99.284

From part c :

Interval becomes ;

(99.284, 101.316) ; this interval contains the hypothesized mean value of 100. Hence we fail to reject the null.

8 0
2 years ago
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