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MissTica
2 years ago
12

Student Council is selling t-shirts and hats for $21 and $14, respectively. The club needs to raise at least $2730. They have re

ceived pre-sale orders of 80 t-shirts and 40 hats. Graph a system of linear inequalities that shows the number of t-shirts and hats they must sell in order to reach their goal. Use x for the number of t-shirts and y for the number of hats.
Choose the graph that shows all the possible combinations. List two possibilities.
Mathematics
1 answer:
sesenic [268]2 years ago
3 0

The required inequality is 3x + 2y > 390 and two possibilities is given by (100,45) and (80,75).

Student Council is selling t-shirts and hats for $21 and $14, respectively. The club needs to raise at least $2730.

<h3>What is inequality?</h3>

Inequality can be define as the relation of equation contains the symbol of ( ≤, ≥, <, >) instead of equal sign in an equation.

Let x for the number of t-shirts and y for the number of hats.

Now,
equation for to raise at least $2730.
21x + 14y > 2730
3x + 2y > 390

Two possibility can be caroled out from the graph i,e.  (100,45) and (80,75).

Thus, the required inequality is 3x + 2y > 390 and two possibilities is given by (100,45) and (80,75).  

Learn more about inequality here:

brainly.com/question/14098842

#SPJ1

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Please help me i will give Brainly
Degger [83]

Answer:

Step-by-step explanation:

\frac{3x + 5}{2x + 7} = 5

do cross multiplication

3x + 5 = 5(2x + 7)

3x + 5 = 10x + 35

5 - 35 = 10x - 3x

-30 = 7x

-30/7 = x

20 A

since there are 7 angles given it means that the polygon is heptagon as heptagon has 7 sides.

sum of interior angle of heptagon = (n-2)*180

(7-2)*180

5*180

900

Now ,

110 + 90 + 150 + 102 + 110 + 170 + x = 900

732 + x = 900

x = 900 - 732

x = 168 degree

20 B

since 5 angles are given it means that the polygon is pentagon as pentagon has 5 sides.

sum of interior angles of a pentagon = (n-2)*180

(5-2)*180

3*180

540 degree

Now ,

110 + 95 + 120 + 114 + x = 540

465 + x = 540

x = 540 - 465

x = 75 degree

21

let one rational number be x

according to the question,

1/7 * x = 2

x/7 = 2

do cross multiplication

x = 14

6 0
3 years ago
How do you solve this 2x +5 - 7x = 15
Serga [27]
1. combine like terms (2x and -7x) to create -5x
2. subtract 5 from each side to create: -5x=10
3. divide both sides by -5
4. x=-2
7 0
3 years ago
7. Stan is serving lemonade at the school's dance out of no
Lilit [14]

Answer: 540 cups

Step-by-step explanation:

Diameter = 20

Radius = 20/2=10

Volume of a cylinder: pi(r^2)h = 3600pi

Cone = (pi(r^2)h)/3 = (20pi)/3

3600pi/(20pi/3) = 540 cups

7 0
3 years ago
Read 2 more answers
__Na2O + __ H2O → __NaOH
Andrews [41]

Answer:

Na20+H20 -> 2NaOH

Step-by-step explanation:

8 0
3 years ago
One of the earliest applications of the Poisson distribution was in analyzing incoming calls to a telephone switchboard. Analyst
grandymaker [24]

Answer:

(a) P (X = 0) = 0.0498.

(b) P (X > 5) = 0.084.

(c) P (X = 3) = 0.09.

(d) P (X ≤ 1) = 0.5578

Step-by-step explanation:

Let <em>X</em> = number of telephone calls.

The average number of calls per minute is, <em>λ</em> = 3.0.

The random variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em> = 3.0.

The probability mass function of a Poisson distribution is:

P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0,1,2,3...

(a)

Compute the probability of <em>X</em> = 0 as follows:

P(X=0)=\frac{e^{-3}3^{0}}{0!}=\frac{0.0498\times1}{1}=0.0498

Thus, the  probability that there will be no calls during a one-minute interval is 0.0498.

(b)

If the operator is unable to handle the calls in any given minute, then this implies that the operator receives more than 5 calls in a minute.

Compute the probability of <em>X</em> > 5  as follows:

P (X > 5) = 1 - P (X ≤ 5)

              =1-\sum\limits^{5}_{x=0} { \frac{e^{-3}3^{x}}{x!}} \,\\=1-(0.0498+0.1494+0.2240+0.2240+0.1680+0.1008)\\=1-0.9160\\=0.084

Thus, the probability that the operator will be unable to handle the calls in any one-minute period is 0.084.

(c)

The average number of calls in two minutes is, 2 × 3 = 6.

Compute the value of <em>X</em> = 3 as follows:

<em> </em>P(X=3)=\frac{e^{-6}6^{3}}{3!}=\frac{0.0025\times216}{6}=0.09<em />

Thus, the probability that exactly three calls will arrive in a two-minute interval is 0.09.

(d)

The average number of calls in 30 seconds is, 3 ÷ 2 = 1.5.

Compute the probability of <em>X</em> ≤ 1 as follows:

P (X ≤ 1 ) = P (X = 0) + P (X = 1)

             =\frac{e^{-1.5}1.5^{0}}{0!}+\frac{e^{-1.5}1.5^{1}}{1!}\\=0.2231+0.3347\\=0.5578

Thus, the probability that one or fewer calls will arrive in a 30-second interval is 0.5578.

5 0
3 years ago
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