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3 years ago

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Jennifer scores a 77 on her algebra 2 test. The mean on the test was 73. The standard deviation on the test was 5. What was her

Z-score

1 answer:

MA_775_DIABLO [31]3 years ago

7 0

Answer:

0.80

Step-by-step explanation:

The z-score is found from the formula ...

Z = (X -μ)/σ

Z = (77 -73)/5 = 4/5 = 0.80

Jennifer's Z-score was 0.80.

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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

If f(x) = x3 and g(x) = (x + 1)3, which is the graph of g(x)?

ale4655 [162]

As you can see below, in red is <span>g(x) = (x + 1)^3 and in blue is f(x)=x^3. Adding 1 to x moves the graph to the left by 1.

Hope this was the answer you were looking for and I hope you have a great day!
</span>

6 0

2 years ago

Read 2 more answers

Mark records his science scores in each monthly assessment over a period of 5 months. In the first assessment he scores 76%. In

Trava [24]

This is the complete question:

Mark records his science scores in each monthly assessment over a period of 5 months. In the first assessment he scores 76%. In the second assessment he scores 73%. After that, his scores keep increasing by 2% in every assessment. Assume that x represents the number of assessments since he starts recording and y represents the scores in each assessment, which of the following describes the situation?

a. a relation only b.

b. neither a function nor a relation

c. a function only

d. both a relation and a function

Answer:

<u><em>d. both a relation and a function</em></u>

Step-by-step explanation:

<em>A relation</em> is any statement, either verbal or mathematical, that connects, interrelates or associates, two or more elements (objects, persons, variables).

The text describes a situation where the scores in the assesments are related to the number of statements since Mark starts recording, hence this is a relation.

<h2>Is this relation a function or not?</h2>

In order for a relation be a function, the association has to be unambiguos. This means that for a given input only one output can exist. If an input can have two or more outputs, then you can not determine which is the output for that input. This is what defines whether a relation is a function or not.

In the situation described in the text, x is the input, i.e. the number of assessments since Mark starts recording the scores. Definetely, the number of assessment is not repeated: there is only one first assessment, only one second assessment, only one third assessment, ... the number of assessment cannot not get repeated. Of course, if the input is not repeated, there is only one output associated to the input (each assessment has just one associated score) and the relation is a function.

3 0

3 years ago

Round to nearest thousandth!! (3 decimal places) photo attached

aliina [53]

Answer:

<em>Probability point lies on shaded region: ( About ) 0.294</em>

Step-by-step explanation:

<em>~ If we are to find the probability, we must find the area of the shaded figure, over that of the total area of this trapezoid ( whole figure ) ~</em>

Let us get the general dimensions. Firstly, the base of the trapezoid is composed of parts 5, 12, and 5. This is an isoceles trapezoid, meaning that the triangles ( shaded region ) are congruent to one another, including the altitudes. That would mean the non - shaded region is a parallelogram, provided altitudes are congruent and by definition of a trapezoid the bases are congruent.

1. Given the information above, the smaller base is congruent to the opposite side of the parallelogram created, so it is 12 units in length, the larger base being 12 + 5 + 5 ⇒ 22 units

2. By Pythagorean Theorem, if x ⇒ altitude of one of the triangles ( shaded region), 5^2 + x^2 = 13^2 ⇒ 25 + x^2 = 169 ⇒ x^2 = 144 ⇒ altitude = 12 units

3. Now we can find the area of this trapezoid by: ( base + base/2 ) * altitude ⇒ ( 12 + 22 / 2 ) * 12 ⇒ ( 34 / 2 ) * 12 ⇒ ( 17 ) * 12 ⇒ 204 units^2

5. The shaded region is composed of two triangles, and knowing the triangles are ≅, let us solve for the area of one triangle and multiply that by 2 to find the total area of the shaded region. Area of triangle: 1/2 * base * altitude ⇒ 1/2 * 5 * 12 ⇒ 30 units^2. Area of shaded region: 30 * 2 = 60 units^2

<em>Probability point lies on shaded region: 60/204 ⇒ ( About ) 0.294</em>

8 0

3 years ago

Para embalar una caja se emplea 4,2 m de cinta adhesiva. ?Cuántas cajas se podrán embalar con tres rollos que tienen 3 hm, 7 dam

kompoz [17]

For this case, we perform the conversions:

First roll:

$1 \ Hectometer --------> 100 \ meters\\3 \ Hectometer --------> x$

$x = \frac {3 * 100} {1}\\x = 300 \ meters.$

We make a rule of three to determine the number of "c" boxes that can be packed with 300 meters of adhesive tape.

1 -----------> 4.2

c -----------> 300

$c = \frac {300 * 1} {4.2}\\c = 71.42857143\\c = 71$

You can pack 71 boxes.

Second roll:

$1 \ Decametro --------> 10 \ meters\\7 \ Decameter --------> x\\x = \frac {7 * 10} {1}\\x = 70 \ meters.$

We make a rule of three to determine the number of "c" boxes that can be packed with 70 meters of adhesive tape.

1 -----------> 4.2

c -----------> 70

$c = \frac {70 * 1} {4.2}\\c = 16.66666667\\c = 16$

You can pack 16 boxes.

Third roll:

1 -----------> 4.2

c -----------> 50

$c = \frac {50 * 1} {4.2}\\c = 11.9047619\\c = 11$

You can pack 11 boxes.

Thus, in total you can pack $11 + 16 + 71 = 98 \ boxes$

Answer:

98 boxes

4 0

3 years ago