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7nadin3 [17]
3 years ago
12

Claudia is making a gelatin dessert for a party. She plans on making 12 servings for every 5 people. If each pan Claudia uses to

make the dessert holds 8 servings, what is the minimum number of these pans that she needs in order to make enough to feed 10 people?
Mathematics
1 answer:
Oksana_A [137]3 years ago
6 0

Answer:

She would need a minimum of 15 pans for 10 people

Step-by-step explanation:

12(5) = 60 12(10) = 120

8(?) = 120

120/8 = 15

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WILL MARK BRAINLIST PLS HELP:))
dusya [7]

Answer:

Vertex: (3,0)

Max/min: min

axis of symmetry: x=3

Domain: (-∞,∞)

Range: [3,∞)

zeroes: (3,0)

Step-by-step explanation:

Vertex is where the graph changes directions (so in this case it's the point where it changes from decreasing to increasing) which I think is (3,0)

It's a minimum because the coefficent for the degree is positive

The axis of symmetry is just the x value of the vertex (which is x= 3)

the domain is all possible x values (-∞,∞)

The range is all possible y values [3,∞)

The zeroes is where the line hits the x axis, which is (3,0)

6 0
3 years ago
Customers arrive at a service facility according to a Poisson process of rate λ customers/hour. Let X(t) be the number of custom
mash [69]

Answer:

Step-by-step explanation:

Given that:

X(t) = be the number of customers that have arrived up to time t.

W_1,W_2... = the successive arrival times of the customers.

(a)

Then; we can Determine the conditional mean E[W1|X(t)=2] as follows;

E(W_!|X(t)=2) = \int\limits^t_0 {X} ( \dfrac{d}{dx}P(X(s) \geq 1 |X(t) =2))

= 1- P (X(s) \leq 0|X(t) = 2) \\ \\ = 1 - \dfrac{P(X(s) \leq 0 , X(t) =2) }{P(X(t) =2)}

=  1 - \dfrac{P(X(s) \leq 0 , 1 \leq X(t)) - X(s) \leq 5 ) }{P(X(t) = 2)}

=  1 - \dfrac{P(X(s) \leq 0 ,P((3 \eq X(t)) - X(s) \leq 5 ) }{P(X(t) = 2)}

Now P(X(s) \leq 0) = P(X(s) = 0)

(b)  We can Determine the conditional mean E[W3|X(t)=5] as follows;

E(W_1|X(t) =2 ) = \int\limits^t_0 X (\dfrac{d}{dx}P(X(s) \geq 3 |X(t) =5 )) \\ \\  = 1- P (X(s) \leq 2 | X (t) = 5 )  \\ \\ = 1 - \dfrac{P (X(s) \leq 2, X(t) = 5 }{P(X(t) = 5)} \\ \\ = 1 - \dfrac{P (X(s) \LEQ 2, 3 (t) - X(s) \leq 5 )}{P(X(t) = 2)}

Now; P (X(s) \leq 2 ) = P(X(s) = 0 ) + P(X(s) = 1) + P(X(s) = 2)

(c) Determine the conditional probability density function for W2, given that X(t)=5.

So ; the conditional probability density function of W_2 given that  X(t)=5 is:

f_{W_2|X(t)=5}}= (W_2|X(t) = 5) \\ \\ =\dfrac{d}{ds}P(W_2 \leq s | X(t) =5 )  \\ \\  = \dfrac{d}{ds}P(X(s) \geq 2 | X(t) = 5)

7 0
4 years ago
Adult tickets cost 8 dollars students cost 4 of they sold 30more adult tickets as made 840 dollars how many adult tickets were s
g100num [7]
I do not quite understant
6 0
3 years ago
1/3t + 3/4 = 2/4 -t solve
olga55 [171]
Okay so I am going to summarize the work out process because its a lot to

Here we go

1/3 (t) + 3/4 - 2/4 - t = ?

1/2 (simplify )

(1/3 (T)+3/4 - 1/2 - (t) = ?

t (2) / 2

1 - 2(t) / 2 = ?

3/4 (simplify this )

1/3(t)+ 3/4 - [1 - 2(t) / 2 = ?

1/3 (this is re last one you have to simplify)

L (Denominator): 3

R (Denominator): 4

L: [L.C.M] : 4

R: [L.C.M] : 3

Basically , we just switched the dominators around

So, Therefore The of t is -3/16


T = -3/16
4 0
3 years ago
To rent a certain meeting room a college charges a reservation fee of $33 and an additional fee of $9.80 per hour the club wants
sammy [17]

Answer:

They must rent for less than 7 hours

Step-by-step explanation:

The charge for the room is the fee plus the hourly rate times the hour

f = 33+9.80t

This must be less than 101.60

101.60> 33+9.80t

Subtract 33 from each side

101.60-33> 33-33+9.80t

68.6> 9.80t

Divide each side by 9.8

68.8/9.8 > t

7>t

They must rent for less than 7 hours

7 0
3 years ago
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