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jeyben [28]
2 years ago
7

What is the difference 4 1/3-(-2/3)​

Mathematics
1 answer:
LiRa [457]2 years ago
6 0

Answer:

5

Step-by-step explanation:

remember, when you have a double negative, you can instead add so:

4 1/3-(-2/3) is the same as

4 1/3 + 2/3 = 5

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What is the range of the function represented by the table? 2/2.5 4/5.0 6/7.5 8/10.00 10/12.5
pishuonlain [190]

Answer:

{2.5, 5.0, 7.5, 10.0, 12.5}

Step-by-step explanation:

Next time, would you please type in the table vertically:

2  2.5

4  5.0

6  7.5

and so on.

The range consists of all the possible numerical outputs of the function for given inputs:  {2.5, 5.0, 7.5, 10.0, 12.5}.  Without knowing the outputs for other, unknown inputs, we cannot be more detailed in specifying the range.

3 0
3 years ago
Help asap please!!!!
Alex73 [517]

Answer:

D. 84 square feet

Step-by-step explanation:

Hope this helps and have a nice day

5 0
2 years ago
What is the square root of 120
Viefleur [7K]

\tt\sqrt{120}= \sqrt{4\cdot30}=\sqrt{4}\cdot\sqrt{30}=2\sqrt{30}\approx2\cdot5.5\approx11

5 0
3 years ago
Read 2 more answers
if your bag of M&M’s has 8 yellow out of 24 , how many of them will be yellow i’d the bag has a total of 96?
Lera25 [3.4K]

Answer: 32

Step-by-step explanation:

96 divided by 24 is 4

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7 0
3 years ago
In a survey of 300 T.V. viewers, 40% said they watch network news programs. Find the margin of error for this survey if we want
bixtya [17]

Answer:

<h2>0.05543</h2>

Step-by-step explanation:

The formula for calculating the margin of error is expressed as;

M.E = z * \sqrt{\frac{p*(1-p)}{n} } where;

z is the z-score at 95% confidence = 1.96 (This is gotten from z-table)

p is the percentage probability of those that watched network news

p = 40% = 0.4

n is the sample size = 300

Substituting this values into the formula will give;

M.E = 1.96*\sqrt{\frac{0.4(1-0.4)}{300} }\\ \\M.E = 1.96*\sqrt{\frac{0.4(0.6)}{300} }\\\\\\M.E = 1.96*\sqrt{\frac{0.24}{300} }\\\\\\M.E = 1.96*\sqrt{0.0008}\\\\\\M.E = 1.96*0.02828\\\\M.E \approx 0.05543

Hence, the margin of error for this survey if we want 95% confidence in our estimate of the percent of T.V. viewers who watch network news programs is approximately 0.05543

5 0
3 years ago
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