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alexandr1967 [171]
3 years ago
11

Customers of a satellite dish company must pay for installation of the dish, and a monthly fee for service. After 6 months, Ryan

had paid $594. After 11 months, he had paid $989. Write and solve a linear equation to find the cost for installation
Mathematics
1 answer:
FromTheMoon [43]3 years ago
7 0

Cost of installation fee is $120.

Step-by-step explanation:

Let,

Monthly fee = x

Installation fee = y

According to given statement;

6x+y=594   Eqn 1

11x+y=989   Eqn 2

Subtracting Eqn 1 from Eqn 2

(11x+y)-(6x+y)=989-594\\11x+y-6x-y=395\\5x=395\\Dividing\ both\ sides\ by\ 5\\\frac{5x}{5}=\frac{395}{5}\\x=79\\

Putting x=79 in Eqn 1

6(79)+y=594\\474+y=594\\y=594-474\\y=120

Cost of installation fee is $120.

Keywords: Linear equations, subtraction

Learn more about linear equations at:

  • brainly.com/question/5282516
  • brainly.com/question/5424148

#LearnwithBrainly

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Answer:

I think it's 476 students have gadgets

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3 years ago
Jamar's car used 1/40 ​ of a gallon to travel 3/8 ​ of a mile. How many miles can the car go on one gallon of gas?
dybincka [34]

Answer:

15 miles.

Step-by-step explanation:

3/8 x 40 = 120/8

now divide,

120/8 = 15

6 0
3 years ago
The radius of a circle is 5 centimeters. What is the length of a 90 arc?
Marta_Voda [28]

Answer:

The length of arc is 7.85cm

Step-by-step explanation:

You can find the length of arc by using the formula :

\frac{θ}{360}  \times 2 \times π \times  {r}

θ = 90°

r = 5cm

Arc = (90/360)×2×π×5

= (1/4)×2×π×5

= (1/4)×10×π

= (5/2)π

= 7.85 cm (3s.f)

5 0
3 years ago
Read 2 more answers
PLEASE HELP! BRAINLIEST to correct answer!!!
irinina [24]

Answer:

Vertex: (4,9)

X int: (1,0),(7,0) = x=-1, x=-7

y iny: (0,-7)

4 0
3 years ago
A university found that 20% of its students withdraw without completing the introductory statistics course. Assume that 20 stude
EleoNora [17]

Answer:

a) P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)

And we can use the probability mass function and we got:

P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115  

P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576  

P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369  

And adding we got:

P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061

b) P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182  

c) P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]

P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115  

P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576  

P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369

P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054

And replacing we got:

P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886

d) E(X) = 20*0.2= 4

Step-by-step explanation:

Previous concepts  

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".  

Solution to the problem  

Let X the random variable of interest, on this case we now that:  

X \sim Binom(n=20, p=0.2)  

The probability mass function for the Binomial distribution is given as:  

P(X)=(nCx)(p)^x (1-p)^{n-x}  

Where (nCx) means combinatory and it's given by this formula:  

nCx=\frac{n!}{(n-x)! x!}  

Part a

We want this probability:

P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)

And we can use the probability mass function and we got:

P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115  

P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576  

P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369  

And adding we got:

P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061

Part b

We want this probability:

P(X=4)

And using the probability mass function we got:

P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182  

Part c

We want this probability:

P(X>3)

We can use the complement rule and we got:

P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]

P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115  

P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576  

P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369

P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054

And replacing we got:

P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886

Part d

The expected value is given by:

E(X) = np

And replacing we got:

E(X) = 20*0.2= 4

3 0
3 years ago
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