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max2010maxim [7]
2 years ago
5

As the weight of purchased items increases, the shipping charges increase, as shown in the table below. Weight, in oz Total Ship

ping Cost not more than 5 $9.50 more than 5, not more than 10 $13.25 more than 10, not more than 15 $17.00 more than 15, not more than 20 $20.75 Assuming only positive domain values, which statement is true of the graph that represents the data in the table? Beginning at 5 ounces, the graph is discontinuous at every fifth integer of the domain. The range values graphed for the set of data are $10, $14, $17, and $21. For every 1 ounce increase in weight, the total shipping cost increases by $3.75. The left side of each horizontal interval is a closed circle, and the right side is an open circle.
Mathematics
2 answers:
ankoles [38]2 years ago
6 0
The correct answer is beginning at 5 ounces, the graph is discontinuous every 5th integer of the domain.

It is a step function; there is not a continuous increase between values.  The shipping level is fixed for each price, then jumps to the next tier.  This will be a discontinuous graph.
chubhunter [2.5K]2 years ago
3 0

Answer:

The answer is A for yall who probably didnt understand their explanantion

Step-by-step explanation:

every 5, 5, 10, 15

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4. The data shows the number of siblings each student in a 9th-grade class has.
antiseptic1488 [7]

Answer:

The correct option is Mean and standard deviation.

Step-by-step explanation:

The data set provided, arranged in ascending order is:

S = {0 , 0 , 0 , 1 , 1 , 1 , 1 , 1 , 1 , 2 , 2 , 2 , 2 , 2 , 2 , 2 , 3 , 3 , 3 , 4 , 4}

Now, it is provided that Macy was absent when the data was collected and the class guessed she had 4 siblings. When  she returned, they found out she actually has 9 siblings.

So, the new data set is:

X = {0 , 0 , 0 , 1 , 1 , 1 , 1 , 1 , 1 , 2 , 2 , 2 , 2 , 2 , 2 , 2 , 3 , 3 , 3 , 4 , 9}

The mean of a data set is the average value representing the entire data.

\bar X=\frac{1}{n}\sum X_{i}

The median is the middle value of the data set.

The inter-quartile range is the difference between the first and the third quartile.

The standard deviation is the value that represents how dispersed the data values are.

SD=\sqrt{\frac{1}{n}\sum (X_{i}-\bar X)^{2}}

From all the above statistic, the mean and standard deviation are the only ones that uses all the observations in the data set to compute their value.

So, on changing one of the 4s by a 9, the set of measures that will be affected are the mean and standard deviation.

Thus, the correct option is Mean and standard deviation.

3 0
1 year ago
Miranda and Savannah are excited that a new store just opened in town! They go together the first day it opens!
pogonyaev

Answer:

Miranda has been to the store 9 times.

Savannah has gone to the store 7 times.

Total each spent = $ 63 (approx.)

7 0
2 years ago
Read 2 more answers
A set of eight plates cost $112 what is the cost of each plate
-BARSIC- [3]
14 dollars . Just divide.
7 0
2 years ago
Factor the polynomial 12c9 + 28c7. Find the GCF of 12c9 and 28c7. 4c7 Write each term as a product, where one factor is the GCF.
saul85 [17]
We are asked in the problem to determine the factored form of a polynomial given the expression 12c9 + 28c7 by finding the GCF. GCF or the greatest common factor is the greatest number (including variable if applicable) that is divisible to the terms included in the polynomial. In this case, the GCF of 12 and 28 is 4 while the GCF of c9 and c7 is c7. We multiply both GCFs from the variable and numerical side.
Hence the complete GCF is 4c7. In this case, the factored form of the polynomial using the GCF is 4c7 ( 3c2 + 7). The answer to this problem is B. 
6 0
1 year ago
Read 2 more answers
The harmonic motion of a particle is given by f(t) = 2 cos(3t) + 3 sin(2t), 0 ≤ t ≤ 8. (a) When is the position function decreas
iren [92.7K]

For the last part, you have to find where f'(t) attains its maximum over 0\le t\le8. We have

f'(t)=-6\sin3t+6\cos2t

so that

f''(t)=-18\cos3t-12\sin2t

with critical points at t such that

-18\cos3t-12\sin2t=0

3\cos3t+2\sin2t=0

3(\cos^3t-3\cos t\sin^2t)+4\sin t\cos t=0

\cos t(3\cos^2t-9\sin^2t+4\sin t)=0

\cos t(12\sin^2t-4\sin t-3)=0

So either

\cos t=0\implies t=\dfrac{(2n+1)\pi}2

or

12\sin^2t-4\sin t-3=0\implies\sin t=\dfrac{1\pm\sqrt{10}}6\implies t=\sin^{-1}\dfrac{1\pm\sqrt{10}}6+2n\pi

where n is any integer. We get 8 solutions over the given interval with n=0,1,2 from the first set of solutions, n=0,1 from the set of solutions where \sin t=\dfrac{1+\sqrt{10}}6, and n=1 from the set of solutions where \sin t=\dfrac{1-\sqrt{10}}6. They are approximately

\dfrac\pi2\approx2

\dfrac{3\pi}2\approx5

\dfrac{5\pi}2\approx8

\sin^{-1}\dfrac{1+\sqrt{10}}6\approx1

2\pi+\sin^{-1}\dfrac{1+\sqrt{10}}6\approx7

2\pi+\sin^{-1}\dfrac{1-\sqrt{10}}6\approx6

4 0
1 year ago
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