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dusya [7]
3 years ago
14

Addison used 5/6yard of ribbon to decorate a photo frame. she used 1/3 yards of ribbon to decorate her scrapbook. which fraction

strip should Addison trade for the 1/3 strip in order to find how many yards of ribbon she used in all
Mathematics
1 answer:
zavuch27 [327]3 years ago
6 0
1/3 will become 2/6.

Steps in adding fraction.
Step 1. Make sure that the denominators are the same:
5/6 + 1/3      * 6 and 3 are the denominators. they are not the same. however, 6 is twice the number of 3. Thus, multiply 2 to the numerator and denominator of 1/3

1/3 * 2/2 = 2/6

5/6 + 2/6

Step 2. Add the numerators. Put the sum above the common denominator.

(5+2) / 6 = 7/6

Step 3. Simplify the fraction.

7/6 = 1 1/6 

Addison used 1 1/6 yards of ribbon in all.
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Kendra is participating in a fundraiser walk-a-thon she walks 4 miles in 60 minutes how many minutes will it take to walk 7 mile
Lemur [1.5K]

Answer:

105 min

Step-by-step explanation:

60 minutes ÷ 4 miles= 15 minutes per mile

15min per mile x 7 miles = 105 minutes

8 0
3 years ago
What is the slope of the line that passes through the points (2,1) and (-4,-5)
cestrela7 [59]
\text{Slope =} \dfrac{1+5}{2+4} =   \dfrac{6}{6} = 1
6 0
3 years ago
Read 2 more answers
The following data gives the speeds (in mph), as measured by radar, of 10 cars traveling north on I-15. 76 72 80 68 76 74 71 78
____ [38]

Answer:

71.123 mph ≤ μ ≤ 77.277 mph

Step-by-step explanation:

Taking into account that the speed of all cars traveling on this highway have a normal distribution and we can only know the mean and the standard deviation of the sample, the confidence interval for the mean is calculated as:

m-t_{a/2,n-1}\frac{s}{\sqrt{n} } ≤ μ ≤ m+t_{a/2,n-1}\frac{s}{\sqrt{n} }

Where m is the mean of the sample, s is the standard deviation of the sample, n is the size of the sample, μ is the mean speed of all cars, and t_{a/2,n-1} is the number for t-student distribution where a/2 is the amount of area in one tail and n-1 are the degrees of freedom.

the mean and the standard deviation of the sample are equal to 74.2 and 5.3083 respectively, the size of the sample is 10, the distribution t- student has 9 degrees of freedom and the value of a is 10%.

So, if we replace m by 74.2, s by 5.3083, n by 10 and t_{0.05,9} by 1.8331, we get that the 90% confidence interval for the mean speed is:

74.2-(1.8331)\frac{5.3083}{\sqrt{10} } ≤ μ ≤ 74.2+(1.8331)\frac{5.3083}{\sqrt{10} }

74.2 - 3.077 ≤ μ ≤ 74.2 + 3.077

71.123 ≤ μ ≤ 77.277

8 0
3 years ago
Which of the following is NOT a reason fitness testing is useful in schools?
EastWind [94]
The most relevant answer is B
6 0
2 years ago
Juan ramirez sells suits in a major department store on weekends. He earns a commission of 5 percent on the first 10 suits, and
Allisa [31]
 Juan Ramirez made $185.00

For the first 10 suits he made $12.50 each suit, then he made $20 for the additional 3. 

250*.05=12.50

250*.08=20

12.50*10=125

3*20=60

125+60=185.
3 0
3 years ago
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