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Dimas [21]
4 years ago
11

What is the mean absolute deviation (MAD) of the data set?

Mathematics
1 answer:
erastova [34]4 years ago
7 0
The answer is 4 i think
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I need help with number 1
rjkz [21]

Answer:∠1=55° ∠2=55° ∠3=55° ∠4=70° ∠5=55°

Step-by-step explanation:

We know that opposite angles of rhombus are congruent and the diagonals of rhombus cut the vertex angles into half. So ∠4=70° and ∠2=∠3=∠1=∠5.

Let ∠2+∠3=x

the sum of four angles of rhombus= 360°

70+70+ x +x = 360°

140+2x = 360°

2x=220°

x=110°

∠2=110°/2=55°=∠3=∠1=∠5

3 0
3 years ago
Please help<br> Picture inserted<br> Multiple choice <br> Will appreciate if u help
lukranit [14]

Answer:

Uhm I can not see the pic sorry :c

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Uring lunch hour at a local restaurant, 90% of the customers order a meat entrée and 10% order a vegetarian entrée. Of the custo
Furkat [3]

Answer:

Total percent of customers who order a drink with their entrée = 76%

Step-by-step explanation:

Let

Total customers = 100%

customers who order a meat entrée = 90%

customers who order a vegetarian entrée = 10%

Of the customers who order a meat entrée, 80% order a drink

% of customers who order a meat entrée and drink = 80% of 90%

= (80/100 * 90/100) * 100

= (0.8 * 0.9) * 100

= 0.72 * 100

= 72%

% of customers who order a meat entrée and drink = 72%

Of the customers who order a vegetarian entrée, 40% order a drink.

% of customers who order a vegetarian entrée and drink = 40% of 10%

= (40/100 * 10/100) * 100

= (0.4 * 0.1) * 100

= 0.04 * 100

= 4%

% of customers who order a vegetarian entrée and drink = 4%

What is the percent of customers who order a drink with their entrée?

Total percent of customers who order a drink with their entrée = % of customers who order a meat entrée and drink + % of customers who order a vegetarian entrée and drink

= 72% + 4%

= 76%

Total percent of customers who order a drink with their entrée = 76%

3 0
3 years ago
Help thankss <br>answer is <br><img src="https://tex.z-dn.net/?f=%20%28%20-%2014.0%29" id="TexFormula1" title=" ( - 14.0)" alt="
lyudmila [28]
We have point P=(x_0,y_0)=(2,-4), so first calculate f'(x_0). There will be:

y=f(x)=2x^3-5x^2\\\\\\\\f'(x)=\big(2x^3-5x^2\big)'=\big(2x^3\big)'-\big(5x^2\big)'=2\big(x^3\big)'-5\big(x^2\big)'=\\\\=2\cdot3x^2-5\cdot2x=\boxed{6x^2-10x}\\\\\\\\f'(x_0)=f'(2)=6\cdot2^2-10\cdot2=6\cdot4-20=24-20=\boxed{4}

Now, we can write the equation of the normal line as:

y-y_0=-\dfrac{1}{f'(x_0)}(x-x_0)\\\\\\ y-(-4)=-\dfrac{1}{4}(x-2)\\\\\\y+4=-\dfrac{1}{4}x+\dfrac{1}{2}\\\\\\y=-\dfrac{1}{4}x+\dfrac{1}{2}-4\\\\\\\boxed{y=-\dfrac{1}{4}x-\dfrac{7}{2}}

Normal line (and every line) crosses x-axis when y = 0, so coordinates of A:

\boxed{y=0}\qquad\qquad\text{and}\\\\\\y=-\dfrac{1}{4}x-\dfrac{7}{2}\\\\\\&#10;0=-\dfrac{1}{4}x-\dfrac{7}{2}\\\\\\\dfrac{1}{4}x=-\dfrac{7}{2}\quad|\cdot4\\\\\\x=-\dfrac{7\cdot4}{2}\\\\\\x=-7\cdot2\\\\\\\boxed{x=-14}\\\\\\\boxed{A=(-14,0)}
8 0
3 years ago
The auditorium holds 200 people. Tickets cost $12 at the door and $8.50 if purchased in advance. The Honor Society has a goal of
arsen [322]

Answer:

they have to sell 575 tickets so they could reach their goal.

4 0
3 years ago
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