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igomit [66]
2 years ago
12

I need help with this​

Mathematics
1 answer:
Schach [20]2 years ago
7 0
8- 1/3 quarts
4- 2/3 quarts
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Solve quadratic equations by factoring<br> x cubed +3x squared - 4x= 0
Papessa [141]
<span>x cubed +3x squared - 4x= 0
= x^3 + 3x^2 -4x = 0
=x(x^2 +3x -4) = 0
= x (x+4)(x-1) = 0

x = 0
x+ 4 = 0 then x = -4
x - 1 = 0 then x = 1
answer: x = 0, x = -4 and x = 1</span>
8 0
3 years ago
Read 2 more answers
Phoebe wants to get atleast a grade of 90 in the math class. Her grade will be the average of two tests. She scored a 94 on the
marin [14]
First things first, you will have 94 + x. Place those on top of a fraction bar and on the bottom put the two numbers over 2. So you have 94 + x over 2. Then, write the symbol ‘greater than or equal to’ and on the other side place your 90. So, 90 + x over 2 is greater than or equal to 90. This is a difficult thing to write out and explain so I hope this makes sense. :)
7 0
3 years ago
Read 2 more answers
Which of the following is correct?
Andru [333]

Answer:

none

Step-by-step explanation:

7 0
2 years ago
I need help wit one &amp; two please help
seropon [69]

23 is less than 25 which means it's closer to 20, so you would round it down to 20. 1.75 is over 0.50, 1/2, so you would round it up to 2. If you multiply 2 and 20 you would get 40.

Number two gets the same explanation just with different numbers.

3 0
2 years ago
A recent survey based on a random sample of n=470 voters, predicted that independent candidate for the mayoral election will get
julsineya [31]

I have an expression

\sigma = \sqrt{p(1-p)/n}

floating around in my head; let's see if it makes sense.

The variance of binary valued random variable b that comes up 1 with probability p (so has mean p) is

E( (b-p)^2 ) =  (-p)^2(1-p) + (1-p)^2p = p(1-p)

That's for an individual sample. For the observed average we divide by n, and for the standard deviation we take the square root:

\sigma = \sqrt{p(1-p)/n}

Plugging in the numbers,

\sigma = \sqrt{.24(1-.24)/490} = 0.019 = 1.9\%

One standard deviation of the average is almost 2% so a 27% outcome was 3/1.9 = 1.6 standard deviations from the mean, corresponding to a two sided probability of a bit bigger than 10% of happening by chance.

So this is borderline suspect; most surveys will include a two sigma margin of error, say plus or minus 4 percent here, and the results were within those bounds.

7 0
3 years ago
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