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Kruka [31]
3 years ago
8

PLEASEEEEEE HELP ME OUT

Mathematics
1 answer:
Nat2105 [25]3 years ago
5 0
We can divide the shape into a rectangle and a triangle:

(1) Area of rectangle = length*width
From the graph we can see that the length of the rectangle is 7 units and the width is 5 units
Area = 7*5 = 35 units^2

(2) Area of triangle = 1/2*bh
From the graph we can see that the base of the triangle is 4 units and its height is 5 units
Area = 1/2*4*5 = 10 units^2

(3) Total area = area of triangle + area of rectangle
= 10 + 35
= 45 units^2
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You can see that your function is y-wise going all the way down to negative infinity but then stops and continues the path along 2 on the upper bound.

The inequality describing the interval is thusly B,

-\infty\lt y\lt2

Hope this helps :)

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3 years ago
2/3 devided by 3/4 plsss
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8/9 or 0.889

Step-by-step explanation:

Used a calculator

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If you in k12 how many point does a test bring up your grade if u make 100
Masja [62]

It depends on many different things. What grade you have on that class. what are your grades on other assignments. And how much of your grade the test is worth.

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3 years ago
The average weight of the largest bone in the inner ear is approximately 0.000109 ounces.drag the tiles to show 0.000109 in scie
VladimirAG [237]

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it would be 1.09x10^4

Step-by-step explanation:

3 0
4 years ago
Many smartphones, especially those of the LTE-enabled persuasion, have earned a bad rap for exceptionally bad battery life. Batt
BartSMP [9]

Answer:

Null hypothesis: the variance in hours of usage for talking is not greater than the the variance in hours of usage for internet.

H_o: \sigma _1 ^2 \leq \sigma _2^2

Alternative hypothesis: the variance  in hours of usage for talking is  greater than the the variance in hours of usage for internet.

H_a : \sigma_1^2 > \sigma_2^2

\mathbf{s_ 1 =16.11}

\mathbf{s_2 = 7.98}

Step-by-step explanation:

Let x_1 and x_2 be the two variables that represents the battery life in hours for talking usage and battery life in hours for internet usage respectively.

The hypothesis can be formulated as:

Null hypothesis: the variance in hours of usage for talking is not greater than the the variance in hours of usage for internet.

H_o: \sigma _1 ^2 \leq \sigma _2^2

Alternative hypothesis: the variance  in hours of usage for talking is  greater than the the variance in hours of usage for internet.

H_a : \sigma_1^2 > \sigma_2^2

The standard deviation for the battery usage for talking is :

\bar x_1 = \dfrac{1}{n_1} \sum x_i  \\ \\ \bar x_1  = \dfrac{1}{12}(35.8 +22.4+...+35.5) \\ \\ \bar x_1 = \dfrac{241.2}{12}  \\ \\ \bar x_1 =20.1

The standard deviation Is:

s_ 1 = \sqrt{\dfrac{1}{n_1-1}\sum (x{_1i}-\bar x_i)^2}

s_ 1 = \sqrt{\dfrac{1}{12-1}\sum (35.8- 20.1)^2+ (35.5-20.1)^2}

s_ 1 = \sqrt{259.568}

\mathbf{s_ 1 =16.11}

The standard deviation for the battery life usage for the internet is :

\bar x_2 = \dfrac{1}{n_2} \sum x_{2i}

\bar x_2 = \dfrac{1}{10} (24.0+12.5+36.4+...+4.7})

\bar x_2 = \dfrac{115}{10}

\bar x_2 = 11.5

Thus; the standard deviation is:

s_2 = \sqrt{\dfrac{1}{n_2-1}(x_{2i}- \bar x_2)^2}

s_2 = \sqrt{\dfrac{1}{10-1}(24-11.5)^2+(4.7-11.5)^2}

s_2 = \sqrt{63.60}

\mathbf{s_2 = 7.98}

4 0
3 years ago
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