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Viefleur [7K]
3 years ago
6

Find the sum or difference of 2/3 + 5/6​

Mathematics
1 answer:
umka21 [38]3 years ago
8 0

Answer:

The sum would be

\frac{3}{2}

The difference would be

\frac{2}{3}  -  \frac{5}{6} =  -  \frac{1}{6}   \\  \\ or \\  \\  \frac{5}{6}  -  \frac{2}{3}  =  \frac{1}{6}

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Gerald buys a bag of 7,500 assorted beads online. A random sample of 150 beads contains 17 red beads. Predict the number of red
Vanyuwa [196]
850. If there is 17 red beads for every 150 beads Then you would divide 7500 by 150 and you get 50. Then multiply 50 with 17 and your answer is 850 red beads. 7500/150=50. 50 x 17= 850.
4 0
3 years ago
3_1/2 divided by 5/8<br> What is the answer
vlabodo [156]

Answer:

28/5 or 5  3/5

Step-by-step explanation:

First you need to make the mixed number into an improper fraction, like this:

3 1/2 turns into 7/2

Next you will need to do the reciprocal of division and inverse operation:

7/2 times 8/5

Then solve it:

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Last simplify:

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6 0
3 years ago
Find real or imaginary solutions of x^4-10x^2=-9
malfutka [58]
This resembles a quadratic and we can factor it using that pattern.

(x^2)^2-(10x^2)+9=0

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6 0
3 years ago
When mrs. myles gave a test, the scores were normally distributed with a mean of 72 and a standard deviation of 8. this means th
irina [24]
The empirical rule states that in a normal distribution,
68% of data is within 1 std deviation of the mean
95%  of data is within 2 std deviation of the mean
99.7% of data is within 3 std deviation of the mean

In this case 95% of the cases would be within two std deviations of the mean 
    mean - 8         and    mean + 8
     72   - 8 = 64   and    72 + 8 = 80 

then 95% of the scores are between  64% and 80% on the test. 
5 0
3 years ago
The perimeter of a rectangle downtown courtyard is 174 feet. The width of the rectangle is twice the length. Find the length and
kirill [66]

Answer:

Step-by-step explanation:

L = Length

W = Width

W = 2L

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2(2L) + 2L = 174

            6L = 174

              L = 29 ft

W = 2(29) = 58 ft

5 0
3 years ago
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