Arestaurant sells 330 sandwiches each day. for each $0.25 decrease in price, the restaurant sells about 15 more sandwiches. how much should the restaurant charge to maximize daily revenue? what is the maximum daily revenue? each sandwich is $6

The maximum daily revenue is with 1 decrease in price.

Step-by-step explanation:

Let x be the number of $0.25 decreases in price and R to be the daily revenue in dollar.

Since,

R(x)=

=

= 1980+7.5x-3.75x^2

∴       ....(1)

For a quadratic equation in standard form , the axis of symmetry is a vertical line

Here, a= -3.75 b=7.5

then,

Substitute the value of x=1 in equation (1), to find the maximum revenue;

=

Therefore, the maximum daily revenue is, with x=1 decrease in price.

85.560%

step-by-step explanation:

conditional probability is defined as:

p(a|b) = p(a∩b) / p(b)

in other words, the probability that a occurs, given that b has occurred = the probability that both a and b occur / the probability that b occurs.

in this example:

p(turns 64 | turns 44) = p(turns 64 & turns 44) / p(turns 44).

if someone turns 64, they must have already turned 44, so:

p(turns 64 | turns 44) = p(turns 64) / p(turns 44)

on page 42, we see that of the original 100,000, 94944 turn 44, and 81234 turn 64.   therefore:

p(turns 64 | turns 44) = 81234 / 94944

p(turns 64 | turns 44) = 0.85560

96 39 cause 135 -96=39 im smart (i used a calculater)