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frosja888 [35]
1 year ago
10

Katherine is landscaping her home with juniper trees and pansies. She wants to arrange 15 pansies around each of 8 trees. Each t

ree costs $20.75 and a six-pack of pansies costs $2.50. Explain how to write an expression to find Katherine’s final cost.
Mathematics
2 answers:
Doss [256]1 year ago
5 0
Let
X-----------------> number of pansies
y-----------------> number of trees

we know that
x=15*8----------> x=120 pansies
y=8 trees

cost of each trees is----------> $<span>20.75
</span>cost of each pansies is------> $2.50/6------> $5/12

[<span>expression to find Katherine’s final cost]=[cost trees]+[cost pansies]
</span>[cost trees]=y*$20.75
[cost pansies]=x*($5/12)   

[expression to find Katherine’s final cost]=y*($20.75)+x*($5/12)
[expression to find Katherine’s final cost]=8*($20.75)+120*($5/12)

[expression to find Katherine’s final cost]=$166+$50
[expression to find Katherine’s final cost]=$216

the answer is 
[expression to find Katherine’s final cost]=y*($20.75)+x*($5/12)
[expression to find Katherine’s final cost]=8*($20.75)+120*($5/12)

Katherine’s final cost is $216
Arlecino [84]1 year ago
4 0

Answer:

Add the cost of pansies to 15 and the cost of trees to 8. Then add them together.

Step-by-step explanation:

That's what I said and I got it correct so there you go, hope this helps.

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Answer:

(C)Determine the principal square root of both sides of the equation.

Step-by-step explanation:

Given: Isosceles right triangle XYZ (45°–45°–90° triangle)

To Prove: In a 45°–45°–90° triangle, the hypotenuse is \sqrt{2} times the length of each leg.

Proof:

\angle XYZ$ is 90^\circ$ and the other 2 angles are 45^\circ. \\\overline{XY} = a, \overline{YZ}= a, \overline{XZ}= c.\\

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Since a=b in an isosceles triangle:

a^2 + a^2 = c^2\\$Combining like terms\\2a^2 = c^2\\$Determine the principal square root of both sides of the equation.\\\sqrt{c^2}=\sqrt{2a^2}  \\\\c=a\sqrt{2} \\

Therefore, the next step is to Determine the principal square root of both sides of the equation.

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Answer:

a) 91.33% probability that at most 6 will come to a complete stop

b) 10.91% probability that exactly 6 will come to a complete stop.

c) 19.58% probability that at least 6 will come to a complete stop

d) 4 of the next 20 drivers do you expect to come to a complete stop

Step-by-step explanation:

For each driver, there are only two possible outcomes. Either they will come to a complete stop, or they will not. The probability of a driver coming to a complete stop is independent of other drivers. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

20% of all drivers come to a complete stop at an intersection having flashing red lights in all directions when no other cars are visible.

This means that p = 0.2

20 drivers

This means that n = 20

a. at most 6 will come to a complete stop?

P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{20,0}.(0.2)^{0}.(0.8)^{20} = 0.0115

P(X = 1) = C_{20,1}.(0.2)^{1}.(0.8)^{19} = 0.0576

P(X = 2) = C_{20,2}.(0.2)^{2}.(0.8)^{18} = 0.1369

P(X = 3) = C_{20,3}.(0.2)^{3}.(0.8)^{17} = 0.2054

P(X = 4) = C_{20,4}.(0.2)^{4}.(0.8)^{16} = 0.2182

P(X = 5) = C_{20,5}.(0.2)^{5}.(0.8)^{15} = 0.1746

P(X = 6) = C_{20,6}.(0.2)^{6}.(0.8)^{14} = 0.1091

P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.0115 + 0.0576 + 0.1369 + 0.2054 + 0.2182 + 0.1746 + 0.1091 = 0.9133

91.33% probability that at most 6 will come to a complete stop

b. Exactly 6 will come to a complete stop?

P(X = 6) = C_{20,6}.(0.2)^{6}.(0.8)^{14} = 0.1091

10.91% probability that exactly 6 will come to a complete stop.

c. At least 6 will come to a complete stop?

Either less than 6 will come to a complete stop, or at least 6 will. The sum of the probabilities of these events is decimal 1. So

P(X < 6) + P(X \geq 6) = 1

We want P(X \geq 6). So

P(X \geq 6) = 1 - P(X < 6)

In which

P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0115 + 0.0576 + 0.1369 + 0.2054 + 0.2182 + 0.1746 = 0.8042

P(X \geq 6) = 1 - P(X < 6) = 1 - 0.8042 = 0.1958

19.58% probability that at least 6 will come to a complete stop

d. How many of the next 20 drivers do you expect to come to a complete stop?

The expected value of the binomial distribution is

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

4 of the next 20 drivers do you expect to come to a complete stop

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