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ad-work [718]
3 years ago
12

A total of 2 yards of fabric is used to make 5 identical pillows. How much fabric is used for each pillow? Do not include units

(yards) in your answer. Modified from EngageNY ©Great Minds Disclaimer
Mathematics
1 answer:
Olenka [21]3 years ago
8 0

Answer:

2/5 yards of fabric

Step-by-step explanation:

A total of 2 yards of fabric is used to make 5 identical pillows. How much fabric is used for each pillow?

5 identical pillows = 2 yards of fabric

1 pillow = x yards

Cross Multiply

x yards × 5 pillows = 2 yards × 1 pillow

x yards = 2 yards × 1 pillow/5 pillows

x yards = 2/5 yards of fabric

Therefore, 2/5 yards of fabric was used for 1 pillow

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Plz help me out with this and do show work I will be waiting.... btw u will be marked brainliest!!!!!! Fast plz
pychu [463]

Answer:

x=16°

Step-by-step explanation:

5*x-30=2*(x+18)

For the second, I'd say it is:

5*x-30=2*x+18  

Let's solve both.  

5*x-30=2*(x+18)

5x-30=2x+36

3x=66

x=22  

5*x-30=2*x+18

3x+48

x=16°

8 0
3 years ago
How to solve for 93,94, and 95
kvasek [131]

Answer:

93- not a triangle

94- right triangle

95- I dont know sorry!


3 0
3 years ago
Tom is putting money into a savings account. He starts with $650 in the savings account, and each week he adds $60 . Let S repre
mario62 [17]
S = 60W + 650 <== ur equation

after 18 weeks...so sub in 18 for W

S = 60(18) + 650
S = 1080 + 650
S = 1730 <==
7 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
If AB=2(x+1) BC=3x+1 and AC=4(x+2) then find the value for x,AB,BC and AC
taurus [48]

Givens

AB + BC = AC

AB = 2(x + 1)

BC = 3x + 1

AC = 4(x + 2)

Substitute and Solve

AB + BC = AC

2(x + 1) + 3x + 1 = 4(x + 2)     Remove the brackets on the left

2x + 2 + 3x + 1 = 4(x + 2)      Collect the like terms on the left

5x + 3  = 4(x + 2)                  Remove the brackets on the right.

5x + 3 = 4x + 8                     Subtract 4x from both sides.

5x - 4x + 3 = 8            

x + 3 = 8                               Subtract 3 from both sides

x =8 - 3

x = 5

Answers

AB=2(5 + 1) = 2 * 6 = 12

BC = 3x + 1 = 3*5 + 1 = 15 + 1 = 16

AC = 4(5 + 2) = 4*7 = 28


7 0
3 years ago
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