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coldgirl [10]
1 year ago
11

The food service manager conducted a random survey of students to determine their preference for new lunch menu items. Which of

these
inferences can be drawn from the manager's data? Select all that apply.
Item
Number of
Students
Pizza
Salad Bar
Pita Sandwiches
Burritos
Mathematics
1 answer:
gogolik [260]1 year ago
4 0

Answer:

<u>Inferences A, B and C are correct.</u>

Complete question, data and inferences:

Item                         Number of students

Pizza                                   6

Salad Bar                           27

Pita Sandwiches                13

Burritos                             24

The food service manager conducted a random survey of students to determine their preference for new lunch menu items. Which of these inferences can be drawn from the manager’s data? Select all that apply.  

A. About twice as many students prefer salad bar as compared to pita sandwiches.  

B. About three-fourths of students prefer either salad bar or burritos.

C. About half as many students prefer pita sandwiches as compared to burritos.

D. Burritos are the most preferred item.

E. Twice as many students prefer pizza as compared to pita sandwiches.

Source:

Question that can be found at quizizz dot com. We can't copy and paste the exact URL because brainly will block it.

Step-by-step explanation:

A. About twice as many students prefer salad bar as compared to pita sandwiches.   True. Two times 13 is 26 and the number of students that prefer Salad Bar is 27.

B. About three-fourths of students prefer either salad bar or burritos.  True. 51 students are the 73% (about three-fourths of the students surveyed) of the total students surveyed, 70.

C. About half as many students prefer pita sandwiches as compared to burritos. True. 13 is almost the half of 24.

D. Burritos are the most preferred item.  False. Salad Bar is the most preferred item.

E. Twice as many students prefer pizza as compared to pita sandwiches.  False. It's the opposite actually; twice as many students prefer pita sandwiches as compared to pizza.

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Ksenya-84 [330]
The four options are attached below

<u><em>Answer:</em></u>
Second attachment is the correct choice

<u><em>Explanation:</em></u>
ASA (angle-side-angle) means that two angles and the included side between them in the first triangle are congruent to the corresponding two angles and the included side between them in the second triangle

<u>Now, let's check the choices:</u>
<u>First attachment:</u>
It shows that two sides and the included angle between them in the first triangle is congruent to the corresponding two sides and the included angle between them in the second one. This is congruency by SAS. Therefore, this option is incorrect

<u>Second attachment:</u>
It shows that two angles and the included side  between them in the first triangle is congruent to the corresponding two sides and the included angle between them in the second triangle. This is congruency by ASA. Therefore, this option is correct

<u>Third attachment:</u>
It shows that the three angles in the first triangle are congruent to the corresponding three angles in the second one. This is not enough to prove congruency. Therefore, this option is incorrect

<u>Fourth attachment:</u>
It shows that the three sides in the first triangle are congruent to the corresponding three sides in the second one. This is congruency by SSS. Therefore, this option is incorrect.

Based on the above, the second attachment is the only correct one

Hope this helps :)

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Step-by-step explanation:

Given: ΔABC is equilateral and AB = a

The diagram is given below :

AM is a median , PB ⊥ AB , PM = b

Now, by using properties of equilateral triangle, median is perpendicular bisector and each angle is of 60°.

We get, ∠AMB = 90°. So, by linear pair ∠AMB + ∠PMB = 180° ⇒ ∠PMB = 90°. Also, ∠ABC = 60° and ∠ABP = 90° (given) So, ∠PBM = 30°

Since, AM is perpendicular bisector of BC. So,

MB = \frac{a}{2}

Now in ΔAMB , By using Pythagoras theorem

AB^{2}=AM^{2}+MB^{2}\\AM^{2}=AB^{2}-MB^{2}\\AM^{2}=a^{2}-(\frac{a}{2})^{2}\\AM=\frac{\sqrt{3}\cdot a}{2}

Now, in ΔBMP :

sin\thinspace 30^{o}=\frac{\text{Perpendicular}}{\text{Hypotenuse}}\\\\sin\thinspace 30^{o}=\frac{\text{MB}}{\text{PB}}\\\\PB=\frac{\text{MB}}{\text{sin 30}}\\\\PB=\frac{\frac{a}{2}}{\frac{1}{2}}\implies PB = a\\\\tan\thinspace 30^{o}=\frac{\text{Perpendicular}}{\text{Base}}\\\\tan\thinspace 30^{o}=\frac{\text{MB}}{\text{PM}}\\\\PM=\frac{\text{MB}}{\text{tan 30}}\\\\PM=\frac{\frac{a}{2}}{\frac{1}{\sqrt3}}\implies PM=b= \frac{\sqrt{3}\cdot a}{2}

Perimeter of ABM = AB + PB + PM + AM

\text{Perimeter = }a+a+b+ \frac{\sqrt{3}\cdot a}{2}\\\\=2\cdot a + \frac{\sqrt{3}\cdot a}{2} +\frac{\sqrt{3}\cdot a}{2}\\\\=2\cdot a +\sqrt{3}\cdot a\\\\=(2+\sqrt3})\cdot a

Hence, Perimeter of ΔABP = (2 + √3)·a units

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where 3 - 1 = 2 is the distance from <em>y</em> = 3 to the axis of revolution, and similarly (1 + sec(<em>x</em>)) - 1 = sec(<em>x</em>) is the distance from <em>y</em> = 1 + sec(<em>x</em>) to the axis. The integrand is symmetric about <em>x</em> = 0, so the integral "folds" in on itself, and the integral from -π/3 to π/3 is twice the integral from 0 to π/3.

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