Harley begins solve this problem by combining like terms 3x + 5x =10

The Carolina panthers play 8 home games each year.Tickets cost an average of $35. The stadium seats 74,000 people. Revenue is a

tatiyna

Answer: the range for one game is {$0, $2,590,000}

Step-by-step explanation:

The revenue, on average, can be written as:

Y = $35*x

where y is the revenue and x is the number of tickets sold.

The domain of this function is equal to {0, 74000} this means that they can sell any whole number of tickets between 0 and 74000.

To find the range we need to evaluate y in bot extremes of the domain (because we have a linear relation)

minimum revenue

y(0) = $35*0 = $0

maximum revenue

y(74000) = $35*74000 = $2,590,000

So the range for one game is {$0, $2,590,000}

4 0

4 years ago

There are 10 marbles in a bag. 3 are red, 4 are blue, 2 are green, and 1 is yellow. You plan to draw 2 marbles from the bag with

schepotkina [342]

<h2>Probability that the second marble will be blue is 0.44</h2>

Step-by-step explanation:

There are 10 marbles in a bag. 3 are red, 4 are blue, 2 are green, and 1 is yellow.

Total number of marbles = 10

Given that first marble is yellow.

New total number of marbles = 10 - 1 = 9

Now we have 3 are red, 4 are blue, 2 are green.

Number of blue marbles = 4

We need to find drawing probability of blue marble from 9 marbles.

Let the probability be, p

$p=\frac{\texttt{Total blue marbles}}{\texttt{Total number of marbles}}\\\\p=\frac{4}{9}\\\\p=0.44$

Probability that the second marble will be blue is 0.44

4 0

3 years ago

Trangle s and tringle l are scaled copys of one another.

KATRIN_1 [288]

A) find the length of the bottom of both
S=2 units L=5 units

So S/L = scale factor

=2/5

B) L/S
=5/2

C) S/M=3/2
S=2
2/m=3/2
3m=2*2
3m=4
M=4/3

M/S=the opposite of s to m
So it is 2/3

7 0

4 years ago

Let the universal set, s, have 66 elements. a and b are subsets of s. set a contains 19 elements and set b contains 34 elements.

Otrada [13]

The cardinality of the union between two sets is given by

$N(A\cup B) = N(A) + N(B) - N(A \cap B)$

In fact, if an element is in both A and B, it is counted in both N(A) and N(B), so we're counting it twice. By subtracting the cardinality of the interserction, we're taking back one of these "extra-counted" element, and the count is correct. In your case, the numbers are

$N(A \cup B) = 19+34-10 = 43$

This means that the two sets cover, without repetitions, 43 of the 66 elements in the sets. So, there are

$66-43 = 23$

elements which are in neither A nor B

5 0

3 years ago

Find the value of x. Round to the nearest tenth.

Yuki888 [10]

Answer:

D

Step-by-step explanation:

SOH

sinФ = opposite/hypentenous

sin35 = x/20

0.574 = x/20

cross multiply

x = 20(0.574)

x = 11.48

Round to the nearest tenth.

x = 11.5

3 0

3 years ago