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Nadusha1986 [10]
3 years ago
6

Please anyone that can help

Mathematics
1 answer:
ivann1987 [24]3 years ago
8 0
Triangular Prism- SA= 1/2bh
                          = 1/2 × 7 × 11
                          = 38 1/2 × 4 = 154
                             7 × 7 = 49
                          = 154 + 49
                          = 204in³

Rectangular Prism- SA= 2lw + 2lh + 2wh
                             = 2 × 11 × 7 + 2 × 11 × 7 + 2 × 7 × 7
                             = 154 + 154 + 98
                             = 406in³

Therefore, the rectangular prism has the larger surface area.
You might be interested in
Which of the following functions best represents the graph?
marshall27 [118]

Answer:

f(x)=x^3+x^2-4x-4

Step-by-step explanation:

From the graph, the x-intercepts are;

x=-2

x=-1

x=2

These are root of the polynomial function represented by the given graph.

By the remainder theorem;

f(-2)=0,f(-1)=0,f(2)=0

According to the factor theorem, if (x-a) is a factor of f(x), then f(a)=0

This implies that;

(x+2),(x+1),(x-2) are factors of the required function.

Hence; f(x)=(x+1)(x-2)(x+2)

We expand using difference of two squares to obtain;

f(x)=(x+1)(x^2-4)

We expand using the distributive property to get;

f(x)=x^3-4x+x^2-4

Rewrite in standard form to obtain;

f(x)=x^3+x^2-4x-4

6 0
3 years ago
Read 2 more answers
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 number of entrees purchased in a single order at a Noodles &amp; Company restaurant has had an historical average of 1.35 en
trasher [3.6K]

Answer:

a) 0.3571

b) The p-value is 0.362007.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 1.35

Sample mean, \bar{x} = 1.4

Sample size, n = 26

Alpha, α = 0.01

Sample standard deviation, s = 0.7

First, we design the null and the alternate hypothesis

H_{0}: \mu = 1.35\text{ entrees per order}\\H_A: \mu > 1.35\text{ entrees per order}

We use One-tailed t test to perform this hypothesis.

a) Formula:

t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n-1}} }

Putting all the values, we have

t_{stat} = \displaystyle\frac{1.4 - 1.35}{\frac{0.7}{\sqrt{25}} } = 0.3571

b) The p-value at t-statistic 0.3571 and degree of freedom 25 is 0.362007.

4 0
3 years ago
Help me solve this equation
klio [65]

Answer:

9 is the value of a in the equation

7 0
3 years ago
How do you write 4.092 in expanded notation?
DanielleElmas [232]
4+ 0.0 + 0.09+ 0.002. This should be the expanded notation form.
6 0
3 years ago
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