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skad [1K]
3 years ago
8

Please guys say me all answer correctly​

Mathematics
1 answer:
Ratling [72]3 years ago
3 0

Answer:

9222521251

Step-by-step explanation:

+6261113051+

151232065+3

156

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Crude oil imports to one country from another for 2009–2013 could be approximated by the following model where t is time in year
DIA [1.3K]

Answer:

Import in 2012 was the greatest

Step-by-step explanation:

Given

l(t) = -33t² + 800t - 3000

9 ≤ t ≤ 13

Required

Determine the year with highest import.

To do this we simply substitute the values of t from 9 to 13 in the given function.

When t = 9

l(t) = -33t² + 800t - 3000

l(9) = -33 * 9² + 800 * 9 - 3000

l(9) = -33 * 81 + 800 * 9 - 3000

l(9) = -2673 + 7200 -3000

l(9) = 1527

When t = 10

l(10) = -33 * 10² + 800 * 10 - 3000

l(10) = -33 * 100 + 800 * 10 - 3000

l(10) = -3300 + 8000 - 3000

l(10) = 1700

When t = 11

l(11) = -33 * 11² + 800 * 11 - 3000

l(11) = -33 * 121 + 800 * 11 - 3000

l(11) = -3993 + 8800 - 3000

l(11) = 1807

When t = 12

l(12) = -33 * 12² + 800 * 12 - 3000

l(12) = -33 * 144 + 800 * 12 - 3000

l(12) = -4752 + 9600 - 3000

l(12) = 1848

When t = 13

l(13) = -33 * 13² + 800 * 13 - 3000

l(13) = -33 * 169 + 800 * 13 - 3000

l(13) = -5577 + 10400 - 3000

l(13) = 1823

Comparing the values of l(t) for the range of t = 9 to 13,

The highest value of l(t) is:

l(12) = 1848

Hence, the year with the highest import is 2012

6 0
3 years ago
GUEST, GUEST An airline charges an additional fee for luggage that exceeds the 50-pound weight limit. Drag an arrow to the numbe
Evgesh-ka [11]
To create this inequality graph, you will have an open circle at 50 and an arrow pointing to the right.

Since the bag must be over 50, we draw an open circle at fifty.
Since the fee goes towards the bags over 50, we draw an arrow pointing to the right.
5 0
3 years ago
Enter your answer and show all the steps that you use to solve this problem in the space provided.
SOVA2 [1]

Answer:

The value of the expression at b = -2 is thus -6

Step-by-step explanation:

We have been given the expression;

|b|+b^{3}

and we are required to evaluate the expression for b = -2

We simply substitute b with -2 in the given expression and simplify;

|-2|+(-2)^{3}\\\\|-2|=2\\\\(-2)^{3}=-2*-2*-2=-8\\ \\|-2|+(-2)^3=2+(-8)=2-8=-6

3 0
3 years ago
Which is greater, 5/6, 7/8 or 5/8?
Mashutka [201]
5/6 i think it’s that
7 0
3 years ago
Read 2 more answers
The weights in pounds of 23 dogs were used to construct the following stem-and-leaf display using the first digit as the stem an
Norma-Jean [14]

Answer:

The median is 52 pounds. The lower quartile is 45 pounds. The upper quartile is 65 pounds. We can identify that one value is suspicious because one whisker is out of the range.

Step-by-step explanation:

The median is the value separating the higher half from the lower half of a data sample. According to the diagram, we can find the median of weight of 23 dogs:

3/ 2 4

4/ 0 3 4 5 7 8 9

5/ 0 1 2 3 4 5

6/ 1 2 5 6 7

7/ 0 1

8 /

9/ 8

There are 11 values before and after the median. Median is 52 pounds.

The lower quartile is the value separating from the lowest value and the median and the upper quartile is the value separating from the highest value and the median. These values are underline in the diagram:

3/ 2 4

4/ 0 3 4 <u>5</u> 7 8 9

5/ 0 1 2 3 4 5

6/ 1 2 <u>5</u> 6 7

7/ 0 1

8 /

9/ 8

So, the lower quartile Q1= 45 pounds and the upper quartile Q3= 65 pounds.

The Interquartile Range (IQR) is given by:

IQR=Q3-Q2=65-45=20

In the following box diagram, we can identify the median, the lower quartile and the upper quartile. We calculate the lower and the upper whisker:

The lower 1.5*IQR whisker is given by: Q1 - 1.5 * IQR= 45-1.5*20=15

The upper 1.5*IQR whisker is given by: Q3 + 1.5 * IQR=65+1.5*20=95

Therefore, in the diagram there is un value out of the range.

The weight 98 pounds is out of the range.

5 0
3 years ago
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