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2 years ago

11

Mark the absolute maximum point of the graph

1 answer:

Elodia [21]2 years ago

8 0

(7,7) is the maximum point in the graph. This is because it is the highest shown point that is above 0.

Brainliest???

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Jeff wants to buy tickets for a football game for his friends. The stadium is selling tickets for $7 each. The stadium requires

aniked [119]

Your answer is D){28, 42, 56, 70, 84}

To answer this question you just have to make sure the amount he pays is a multiple of 7.If you know you times table then this should be easy

Times table of 7:

7,14,21,28,35,42,49,55,63,70,77,84.

Now you just need to find an answer choice who has these numbers.Lets check it out!

A) 11 isn't a multiple of 7

B) 56 isn't a multiple of 7

C) None of these are multiples of 7 except 7 so this is also incorrect

D) This is correct because they are all multiples of 7

Hope this makes sense.

7 0

2 years ago

Slide the green dot Use the number line to plot the numbers. Then arrange

lbvjy [14]

I don't see any green dot, sorry.

6 0

2 years ago

Below are survival times (in days) of 13 guinea pigs that were injected with a bacterial infection in a medical study:

tresset_1 [31]

Outliers are data that are relatively far from other data elements.

The dataset has an outlier and the outlier is 120

The dataset is given as:

91 83 84 79 91 93 95 97 97 120 101 105 98

Sort the dataset in ascending order

79 83 84 91 91 93 95 97 97 98 101 105 120

<h3>The lower quartile (Q1)</h3>

The Q1 is then calculated as:

$Q1 = \frac{N +1}{4}th$

So, we have:

$Q1 = \frac{13 +1}{4}th$

$Q1 = \frac{14}{4}th$

$Q1 = 3.5th$

This is the average of the 3rd and the 4th element

$Q1 = \frac{1}{2} \times (84 + 91)$

$Q1 = 87.5$

<h3>The upper quartile (Q3)</h3>

The Q3 is then calculated as:

$Q3 = 3 \times \frac{N +1}{4}th$

So, we have:

$Q3 = 3 \times \frac{13 +1}{4}th$

$Q3 = 3 \times 3.5th$

$Q3 = 10.5th$

This is the average of the 10th and the 11th element.

$Q_3 =\frac12 \times (98 + 101)$

$Q_3 =99.5$

<h3>The interquartile range (IQR)</h3>

The IQR is then calculated as:

$IQR = Q_3 -Q_1$

$IQR = 99.5 - 87.5$

$IQR = 12$

Also, we have:

$IQR(1.5) = 12 \times 1.5$

$IQR(1.5) = 18$

<h3>The outlier range</h3>

The lower and the upper outlier range are calculated as follows:

$Lower = Q_1 - IQR(1.5)$

$Lower = 87.5- 18$

$Lower = 69.5$

$Upper = Q_3 + IQR(1.5)$

$Upper = 99.5 + 18$

$Upper = 117.5$

120 is greater than 117.5.

Hence, the dataset has an outlier and the outlier is 120

Read more about outliers at:

brainly.com/question/9933184

6 0

2 years ago

Factor completely 5ab + 3ay + 5b + 3y. (5b + 3y)(a + 1) (5b − 3y)(a + 1) (5b + 3y)(a − 1) Prime

Katen [24]

<span>5ab + 3ay + 5b + 3y
a(5b + 3y) + 1(5b + 3y)
(a+1)(5b + 3y)
(5b + 3y)(a+1)</span>

7 0

3 years ago

Read 2 more answers

Select the correct answer.

Afina-wow [57]

Answer:

$d)\ \frac{81}{m^{7}}$

Step-by-step explanation:

$(3m^-4)^3(3m^5)$

First, simplify $(3m^{-4})^3$ :

$(3m^{-4})^3\\\\3^3*m^{-4*3}\\\\27m^{-12}$

$27m^{-12}*3m^5\\\\27*3 *m^{-12+5}\\\\81m^{-7}\\\\81*\frac{1}{m^{7}} \\\\\frac{81}{m^{7}}$

Hope this helps!

5 0

2 years ago