For each value of y, determine whether it is a solution to 7=y÷4.

Mathematics

2 answers:

Simora [160]3 years ago

5 0

Answer:

this brainly place is terrible and it doesn't work and none of the answers are right

Step-by-step explanation:

so yah like no one should use this place its for people with money and stuff ughhhhhhh *stupidness*

rjkz [21]3 years ago

4 0

Answer:

Step-by-step explanation:

7 = y/4

7*4 = y

y = 28

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The daily demand for ice cream cones at a price of $1.20 per cone is 50 cones. At a price of $2.20 per cone, the demand is 30 co

Lapatulllka [165]

Answer:

<h2>The demand at a price of $1.50 per cone is 44 cones</h2>

Step-by-step explanation:

let us apply the interpolation formula to solve this problem

we have

$\frac{y-y1}{x-x1} =\frac{y2-y1}{x2-x1}$

Now let us set the values to the notation above

x= $1.20

x1= $2.20

x2= $1.50

y = 50 cones

y1= 30 cones

y2= ?

$y2= \frac{(y-y1)}{(x-x1)} (x2-x1)+y1$

Substituting our given data we have

$y2= \frac{(50-30)}{(1.2-2.2)} (1.5-2.2)+30\\\\y2= \frac{20}{-1} (-0.7)+30\\\\y2= (20*0.7)+30\\y2= 14+30\\y2= 44$

the demand at a price of $1.50 per cone is 44 cones

4 0

2 years ago

If two standard six-sided dice are tossed, what is the probability that a 5 is rolled on at least one of the two dice? express y

zaharov [31]

We define the probability of a particular event occurring as:
$\frac{number\ of \ desired\ outcomes}{number\ of\ possible\ outcomes}$

What are the total number of possible outcomes for the rolling of two dice? The rolls - though performed at the same time - are <em>independent</em>, which means one roll has no effect on the other. There are six possible outcomes for the first die, and for <em>each </em>of those, there are six possible outcomes for the second, for a total of 6 x 6 = 36 possible rolls.

Now that we've found the number of possible outcomes, we need to find the number of <em>desired</em> outcomes. What are our desired outcomes in this problem? They are asking for all outcomes where there is <em>at least one 5 rolled</em>. It turns out, there are only 3:

(1) D1 - 5, D2 - Anything else, (2), D1 - Anything else, D2 - 5, and (3) D1 - 5, D2 - 5

So, we have $\frac{3}{36} = \frac{1}{12}$ probability of rolling at least one 5.