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3 years ago

Who is represented equally under Virginia? Who is represented equally under New Jersey?

Mathematics

1 answer:

Ainat [17]3 years ago

6 0

Answer:The Virginia Plan proposed

instead a legislative branch

consisting of two chambers

(bicameral legislature), in each of

which the states would be

represented in proportion to

their “Quotas of contribution, or

to the number of free

inhabitants.”

Step-by-step explanation:

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Please help! I need help with part B if 1000 students were surveyed what would the probability be of how many plays sports?

nevsk [136]

Answer:

188 students would play sports, out of 1,000.

Step-by-step explanation:

If the probability that a random high school student plays a sport is 0.188 (150/800), which means for every 1 high school student 0.188 of them play a sport. Multiply 0.188x1,000 to get 188 students who play a sport.

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3 years ago

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Please help me with numbers 1 & 2. Please show work, thank you.

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1.) Jarvis lives 3 miles away from Cassidy and Theo lives 6 miles away.
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2.) Jorge's watering can can hold 8 cups of water.
(Older brother's can hold 16 cups; 8+8+16= 32; there are 32 cups in 2 gallons)

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4 years ago

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Given the regular price of an item and the discount rate, how do you find the discount?

Allisa [31]

Answer:

discount divided by the normal/actual

Step-by-step explanation:

5 0

3 years ago

Find the limit

Lana71 [14]

Step-by-step explanation:

<h3>Appropriate Question :-</h3>

Find the limit

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

On substituting directly x = 1, we get,

$\rm \: = \: \sf \dfrac{1-2}{1 - 1}-\dfrac{1}{1 - 3 + 2}$

$\rm \: = \sf \: \: - \infty \: - \: \infty$

which is indeterminant form.

Consider again,

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

can be rewritten as

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 3x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 2x - x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( x(x - 2) - 1(x - 2))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ {(x - 2)}^{2} - 1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 2 - 1)(x - 2 + 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)(x - 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)}{x(x - 2)}\right]$

$\rm \: = \: \sf \: \dfrac{1 - 3}{1 \times (1 - 2)}$

$\rm \: = \: \sf \: \dfrac{ - 2}{ - 1}$

$\rm \: = \: \sf \boxed{2}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right] = 2 \: }}$

$\rule{190pt}{2pt}$

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3 years ago

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A 2-column table with 7 rows. The first column is labeled x with entries negative 3, negative 2, negative 1, 0, 1, 2, 3. The sec

DochEvi [55]

The statements that are true about the intervals of the continuous function are Options 2, 4 and 5

f(x) ≤ 0 over the interval [0, 2].
f(x) > 0 over the interval (–2, 0).
f(x) ≥ 0 over the interval [2, ).

<h3>What is the statement about?</h3>

Looking at the values given, the intervals which satisfies the condition are known to be:

f(x)<=0 over the interval [0,2]

f(x)>0 over the interval (-2,0)

f(x)>=0 over the interval [2,∞)

Because:

Since the table with x and f(x) values, we have to examine analyze the table and see the each option that is in line with f(x) or not .

Examine the values of x that is from -3 to 3, the f(x) values are both positive and negative . hence f(x)>0 is false over the interval (-∞,3)

Looking at the the interval from 0 to 2, the f(x) values are 0 and negative. Hence, f(x)<=0 over the interval [0,2]

When you look over the interval (-1,1), the f(x) values are said to be both positive and negative and as such, f(x)<0 is false over the interval (-1,1)

When you look at the interval (-2,0) , the f(x) is positive and as such, f(x)>0 over the interval (-2,0)

Looking at the interval [2,∞), f(x) is positive and as such, f(x)>=0 over the interval [2,∞)

Therefore, Option 2, 4 and 5 are correct.

Learn more about interval from

brainly.com/question/14454639

#SPJ1

6 0

2 years ago