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mart [117]
3 years ago
13

Henry and reiko both used 1 yard of ribbon to make bows. write two different fractions to show that Henry and reiko use the same

amount of ribbon
Mathematics
1 answer:
Natalka [10]3 years ago
4 0
To show that they used the same amount of ribbon you could take the 1 yard and divided into different sized equal groups.

1 yard divided by four equals 1/4 yard for each of four ribbons.

1 yard divided by six equals 1/6 yard for each of six ribbons.

You could have ribbons the size of 1/4 our yard and 1/6 yard, but you would have different amounts of each.
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Silas had 3 times as much money as Raja. Silas spent $84 of his money to buy a birthday gift for his grandma while Raja was awar
Tamiku [17]

Answer:

Total amount Silas and Raja has at first = $252

Step-by-step explanation:

Let,

Amount Silas had = x

Amount Raja had = y

According to given statement;

x = 3y       Eqn 1

x-84 = y+42     Eqn 2

Putting value of x from Eqn 1 in Eqn 2

3y - 84 = y+42

3y-y = 42+84

2y = 126

Dividing both sides by 2

\frac{2y}{2}=\frac{126}{2}\\y=63

Putting y = 63 in Eqn 1

x = 63(3)

x = 189

Total amount they had = 189 + 63 = $252

Hence,

Total amount Silas and Raja has at first = $252

8 0
2 years ago
A regression equation was estimated as = –100 + 0.5x. if x = 20,the predicted value of y is _____.
ratelena [41]
Any answer things? :|
5 0
3 years ago
Full-time ph.d. students receive an average of $12,837 per year with a standard deviation of $3000. find the probability that th
Ainat [17]
It is about 41.4% according to my TI-83/84 calculator.

7 0
2 years ago
Solve for g:<br> 3/16=(-5/4)+g
Musya8 [376]

Answer:

g = 23/16

Step-by-step explanation:

3/16 = (-5/4) + g

g = 5/4 + 3/16

g = 23/16

7 0
2 years ago
uppose germination periods, in days, for grass seed are normally distributed and have a known population standard deviation of 2
natka813 [3]

Answer: 0.701

Step-by-step explanation:

Formula :  EBM =z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}} , where \alpha= significance level , \sigma = Population standard deviation, n= sample size.

As per given,  n= 22

\sigma = 2

Critical z- value for 90% confidence level : z_{\alpha/2}=1.645

Then,

 EBM =(1.645)\dfrac{2}{\sqrt{22}}\\\\=(1.645)\dfrac{2}{4.690416}\\\\\approx0.701

Hence ,  error bound (EBM) of the confidence interval with a 90% confidence level= ± 0.701

5 0
3 years ago
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