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3 years ago

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The results of a recent survey of the students at Rockdale Middle School show that 17 out of every 22 students enjoy the cafeter

ia’s cheeseburgers, "5" /"6" of the students enjoy the cafeteria’s pizza, and 80.5% enjoy the cafeteria’s salad bar. Based upon this survey, which of these three foods is most enjoyed?

1 answer:

madam [21]3 years ago

5 0

Answer:

Pizza is the most enjoyed food with 83.3%

Step-by-step explanation:

Given

Cheeseburgers = 17 out of 22 students

Pizza = 5/6

Salad Bar = 80.5%

Required

Which of the food is most enjoyed

First, we have to convert all goven parameters to percentage

$Cheeseburgers = \frac{17}{22}$

$Cheeseburgers = 77.3\%$

$Pizza = \frac{5}{6}$

$Pizza = 83.3\%$

Comparing the percentages;

Pizza is the most enjoyed food with 83.3%

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Because gambling is a big business, calculating the odds of a gambler winning or losing in every game is crucial to the financi

IRISSAK [1]

Answer:

a) 0.1165

b) 0.0983

c) 0.000455

d) 0.787

e) 0.767

Step-by-step explanation:

5 bars, 4 lemons, 3 cherries, and a bell

Total = 5+4+3+1 = 13

The probability of getting a bar on a slot, P(Ba) = 5/13 = 0.385

A lemon, P(L) = 4/13 = 0.308

A cherry, P(C) = 3/13 = 0.231

A bell, P(Be) = 1/13 = 0.0769

a) Probability of getting 3 lemons = (4/13) × (4/13) × (4/13) = 256/2197 = 0.1165

b) Probability of getting no fruit symbol

On each slot, there are 4+3 = 7 fruit symbols.

Probability of getting a fruit symbol On a slot = 7/13

Probability of not getting a fruit symbol = 1 - (7/13) = 6/13 = 0.462

Probability of not getting a fruit symbol On the three slots = 0.462 × 0.462 × 0.462 = 0.0983

c) Probability of getting 3 bells, the jackpot = (1/13) × (1/13) × (1/13) = 1/2197 = 0.000455

d) Probability of not getting a bell on the 3 slots

Probability of not getting a bell on one slot = 1 - (1/13) = 12/13 = 0.923

Probability of not getting a bell on the 3 slots = (12/13) × (12/13) × (12/13) = 1728/2197 = 0.787

e) Probability of at least one bar is a sum of probabilities

Note that Probability of getting a bar = 5/13 and probability of not getting a bar = 8/13

1) Probability of getting 1 bar and other stuff on the 2 other slots (this can happen in 3 different orders) = 3 × (5/13)×(8/13)×(8/13) = 960/2197 = 0.437

2) Probability of getting 2 bars and other stuff on the remaining slot (this can also occur in 3 different orders) = 3 × (5/13)×(5/13)×(8/13) = 600/2197 = 0.273

3) Probability of getting 3 bars on the slots machine = (5/13) × (5/13) × (5/13) = 125/2197 = 0.0569

Probability of at least one bar = 0.437 + 0.273 + 0.0569 = 0.7669 = 0.767

5 0

3 years ago

Help ASAP can u explain a little too? (pls help if u don’t know it don’t answer pls) Thank You:)

larisa [96]

Since they are similar, the dimensions are in the same ratio. L1 = 5, L2 = 15, so they are in a 3:1 ratio. So if V1 = 60, then W1×H1 = 60/5 = 12
W2 must also be 3×W1 and H2 3×H1, and
3×3 = 9. So take 12×9 (W×H1×9) ×15 (L2) = V2
V2 = 12×9×15 = 1620 cm^3

Let me know the right answer when you find out!

8 0

3 years ago

igomit [66]

If the base is square, and the perimeter is 14.5, that means that each side of the base is 14.5/4=3.625. Since we now know the length and the width, as well as the height which is 16.8, we plug into the formula $\frac{lwh}{3}$ . So, plug in the details $\frac{3.625*3.625*16.8}{3}$ Your answer will be 73.6 cm^3

4 0

3 years ago

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Daniel [21]

Answer:

Step-by-step explanation:

As the statement is ‘‘if and only if’’ we need to prove two implications

$f : X \rightarrow Y$ is surjective implies there exists a function $h : Y \rightarrow X$ such that $f\circ h = 1_Y$ .
If there exists a function $h : Y \rightarrow X$ such that $f\circ h = 1_Y$ , then $f : X \rightarrow Y$ is surjective

Let us start by the first implication.

Our hypothesis is that the function $f : X \rightarrow Y$ is surjective. From this we know that for every $y\in Y$ there exist, at least, one $x\in X$ such that $y=f(x)$ .

Now, define the sets $X_y = \{x\in X: y=f(x)\}$ . Notice that the set $X_y$ is the pre-image of the element $y$ . Also, from the fact that $f$ is a function we deduce that $X_{y_1}\cap X_{y_2}=\emptyset$ , and because $f$ the sets $X_y$ are no empty.

From each set $X_y$ choose only one element $x_y$ , and notice that $f(x_y)=y$ .

So, we can define the function $h:Y\rightarrow X$ as $h(y)=x_y$ . It is no difficult to conclude that $f\circ h(y) = f(x_y)=y$ . With this we have that $f\circ h=1_Y$ , and the prove is complete.

Now, let us prove the second implication.

We have that there exists a function $h:Y\rightarrow X$ such that $f\circ h=1_Y$ .

Take an element $y\in Y$ , then $f\circ h(y)=y$ . Now, write $x=h(y)$ and notice that $x\in X$ . Also, with this we have that $f(x)=y$ .

So, for every element $y\in Y$ we have found that an element $x\in X$ (recall that $x=h(y)$ ) such that $y=f(x)$ , which is equivalent to the fact that $f$ is surjective. Therefore, the prove is complete.

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3 years ago

Which statement about light is FALSE? a Light travels about 300,000 km/second. b Light travels in the form of a wave. c Light fr

Travka [436]

Answer: I believe it’s c

Step-by-step explanation:

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3 years ago

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