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Komok [63]
3 years ago
8

Jill bought oranges and bananas. she bought 12 pieces of fruit and spent $5. Oranges cost $0.50 each and bananas cost $0.25 each

. write a system of equations to model the problem. Then solve the system algebraically. How many oranges and how many bananas did Jill but?
Mathematics
1 answer:
inysia [295]3 years ago
3 0
The first part is 0.5x+0.25y= $12.00
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7(9k + 6m) <br><br><br>(50 points)
vladimir1956 [14]

21 • (3k + 2m)

Step by step solution :

Step  1  :

Step  2  :

Pulling out like terms :

2.1     Pull out like factors :

  9k + 6m  =   3 • (3k + 2m)  

Final result :

 21 • (3k + 2m)

6 0
3 years ago
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A fair coin is flipped 5 times
wolverine [178]

Answer:

2/5

Step-by-step explanation:

The answer is common sense.

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3 years ago
the length of a rectangular patio is 8 feet less than twice its width. the area of the patio is 280 square feet. find the dimens
jolli1 [7]

Answer:

The length of the rectangle 'l' = 20

The width of the rectangle 'w' = 14

Step-by-step explanation:

<u>Explanation</u>:-

Let 'x' be the width

Given data the length of a rectangular patio is 8 feet less than twice its width

2x-8 = length

The area of rectangle = length X width

Given area of rectangle = 280 square feet

x(2x-8) = 280

2(x)(x-4) =280

x(x-4) =140

x^2 -4x -140=0

x^2-14x+10x-140=0

x(x-14)+10(x-14)=0

(x+10)(x-14) =0

x = -10 and x = 14

we can choose only x =14

The width of the rectangle 14

The length of the rectangle 2x-8 = 2(14)-8 = 28 -8 =20

The length of the rectangle 'l' = 20

The width of the rectangle 'w' = 14

6 0
3 years ago
Lifetime of $1 Bills The average lifetime of circulated $1 bills is 18 months. A researcher believes that the average lifetime i
OLEGan [10]
<h2>Answer with explanation:</h2>

Let \mu be the population mean lifetime of circulated $1 bills.

By considering the given information , we have :-

H_0:\mu=18\\\\H_a:\mu\neq18

Since the alternative hypotheses is two tailed so the test is a two tailed test.

We assume that the lifetime of circulated $1 bills is normally distributed.

Given : Sample size :  n=50 , which is greater than 30 .

It means the sample is large so we use z-test.

Sample mean : \overline{x}=18.8

Standard deviation : \sigma=2.8

Test statistic for population mean :-

z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}

\Rightarrow\ z=\dfrac{18.8-18}{\dfrac{2.8}{\sqrt{50}}}\approx2.02

The p-value= 2P(z>2.02)=0.0433834

Since the p-value (0.0433834) is greater than the significance level (0.02) , so we do not reject the null hypothesis.

Hence, we conclude that we do not have enough evidence to support the alternative hypothesis that the average lifetime of a circulated $1 bill differs from 18 months.

6 0
3 years ago
I have no idea help please
Sliva [168]

Answer:

C

Step-by-step explanation:

3 0
3 years ago
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