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2 years ago

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 & 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

I need help with this​

I need help with this