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Jlenok [28]
3 years ago
15

The box plots below show student grades on the most recent exam compared to overall grades in the class: Two box plots are shown

. The top one is labeled Class. Minimum at 68, Q1 at 71, median at 84, Q3 at 89, maximum at 100. The bottom box plot is labeled Exam. Minimum at 55, Q1 at 76, median at 85, Q3 at 94, maximum at 100. Which of the following best describes the information about the medians? A) The class and exam medians are approximately equal. B) The class median is much higher than the exam median. C) The first quartile for class and exam is the same, but the exam has the lowest median. D) The low outlier on exams affects the median significantly.
Mathematics
1 answer:
juin [17]3 years ago
7 0

Answer:

the answer should be C please give brainliest

Step-by-step explanation:

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Plz help this is complicated
abruzzese [7]

Answer:

Graph attached.

Step-by-step explanation:

In the question the given parameters are as followed.

y=7cos(\frac{x}{2})+1

1). Amplitude a = 7

2). Period = \frac{2\pi }{\frac{1}{2}}=2\pi \times 2=4\pi

3). Vertical shift = 1 means mid line of the graph will be y =1 and maximum of the graph is 8 and minimum is -6.

7 0
3 years ago
{1, -2),(-2, 0),(-1, 2),(1, 3)}<br> Function
Over [174]

Answer:

this is a function

Step-by-step explanation:

the x does not repeat itself

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3 years ago
X(15-x)=50 quadratic equations
meriva

Answer:

15x -  {x}^{2}  =50 \\  {x }^{2}  - 15x  + 50 = 0 \\  {x}^{2} - 10x - 5x + 50 = 0 \\ x(x - 10) - 2(x - 10) = 0 \\ (x - 2)(x - 10) = 0 \\ x = 2 \: and \: x = 10

3 0
2 years ago
What doe decimals have that fractions don't have?
chubhunter [2.5K]
A decimal point................................................................

4 0
3 years ago
How many different strings of length 12 containing exactly five a's can be chosen over the following alphabets? (a) The alphabet
N76 [4]

Answer:

Part (A) The required ways are 792.

Part (B) The required ways are 101376.

Step-by-step explanation:

Consider the provided information.

Part (A) The alphabet {a, b}

The length of strings is 12 that containing exactly five a's.

The number of ways are: \frac{12!}{5!7!}

After filling "a" we have now 7 places.

For 7 places we have "a" and "b" alphabet but we already select a's so now the remaining place have to fill by "b" only.

Thus, the required ways are: \frac{12!}{5!7!}\times 1=792

Part (B) The alphabet {a, b, c}

We have selected five a's now we have now 7 places.

For 7 places we have "b" and "c".

Thus, there are 2 choices for each 7 place that is 2^7

Therefore the total number of ways are: 792\times 2^7=101376

Thus, the required ways are 101376.

7 0
3 years ago
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