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Lapatulllka [165]
3 years ago
14

Based on the results of the simulation, if 224 groups are formed, about how many of them would you expect to contain one student

from each grade level? Round your answer to the nearest number of groups.
groups.
Mathematics
2 answers:
stich3 [128]3 years ago
6 0

Answer:

18 percent

Step-by-step explanation:

trapecia [35]3 years ago
5 0

Answer: what's the results of the stimulation?, there's not much info to answer your question.

Step-by-step explanation:

You might be interested in
What is the solution to an equation /
navik [9.2K]

Answer:

Idk what u mean?

Step-by-step explanation:

but i searched it up and here are my results

The solution of an equation is the set of all values that, when substituted for unknowns, make an equation true. For equations having one unknown, raised to a single power, two fundamental rules of algebra, including the additive property and the multiplicative property, are used to determine its solutions.Mar 20, 2020

email me if it helps

ty

6 0
3 years ago
Write an equation for an ellipse centered at the origin, which has foci at (0,\pm\sqrt{63})(0,± 63 ​ )left parenthesis, 0, comma
steposvetlana [31]

Answer:

\frac{x^{2} }{4312 } + \frac{y^{2} }{8281 }

Step-by-step explanation:

Since the foci are at(0,±c) = (0,±63) and vertices (0,±a) = (0,±91), the major axis is the y- axis. So, we have the equation in the form (with center at the origin) \frac{x^{2} }{b^{2} } + \frac{y^{2} }{a^{2} }.

We find the co-vertices b from b = ±√(a² - c²) where a = 91 and c = 63

b = ±√(a² - c²)

= ±√(91² - 63²)

= ±√(8281 - 3969)

= ±√4312

= ±14√22

So the equation is

\frac{x^{2} }{(14\sqrt{22}) ^{2} } + \frac{y^{2} }{91^{2} } = \frac{x^{2} }{4312 } + \frac{y^{2} }{8281 }

8 0
3 years ago
An instructor who taught two sections of Math 161A, the first with 20 students and the second with 30 students, gave a midterm e
uranmaximum [27]

Answer:

<em>The answers are for option (a) 0.2070  (b)0.3798  (c) 0.3938 </em>

Step-by-step explanation:

<em>Given:</em>

<em>Here Section 1 students = 20 </em>

<em> Section 2 students = 30 </em>

<em> Here there are 15 graded exam papers. </em>

<em> (a )Here Pr(10 are from second section) = ²⁰C₅ * ³⁰C₁₀/⁵⁰C₁₅= 0.2070 </em>

<em> (b) Here if x is the number of students copies of section 2 out of 15 exam papers. </em>

<em>  here the distribution is hyper-geometric one, where N = 50, K = 30 ; n = 15 </em>

<em>Then, </em>

<em> Pr( x ≥ 10 ; 15; 30 ; 50) = 0.3798 </em>

<em> (c) Here we have to find that at least 10 are from the same section that means if x ≥ 10 (at least 10 from section B) or x ≤ 5 (at least 10 from section 1) </em>

<em> so, </em>

<em> Pr(at least 10 of these are from the same section) = Pr(x ≤ 5 or x ≥ 10 ; 15 ; 30 ; 50) = Pr(x ≤ 5 ; 15 ; 30 ; 50) + Pr(x ≥ 10 ; 15 ; 30 ; 50) = 0.0140 + 0.3798 = 0.3938 </em>

<em> Note : Here the given distribution is Hyper-geometric distribution </em>

<em> where f(x) = kCₓ)(N-K)C(n-x)/ NCK in that way all these above values can be calculated.</em>

7 0
3 years ago
76.8 of what is 40.32
bearhunter [10]
Hey there

Lets solve this question together,

Convert 76.8% to a decimal.

0.768 * x=40.32

Multiply 0.768 by x to get 0.768x.

0.768x=40.32

Multiply each term by 1 / 0.768 and simplify.

Cancel the common factor of 0.768.

x=40.32 / 0.768

Divide 40.32 by 0.768 to get 52.5

x=52.5
6 0
3 years ago
What is the equation of a line that passes through the point (8, 1) and is perpendicular to the line whose equation is y=−2/3x+5
swat32

Let k:y=m_1x+b_1 and l:y=m_2x+b_2

l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}

We have y=-\dfrac{2}{3}x+5\to _1=-\dfrac{2}{3}

Therefore

m_2=-\dfrac{1}{-\frac{2}{3}}=\dfrac{3}{2}

We have the equation of a line: y=\dfrac{3}{2}x+b.

Put the coordinates of the point (8, 1) to the equation of a line:

1=\dfrac{3}{2}(8)+b

1=(3)(4)+b

1=12+b       <em>subtract 11 from both sides</em>

-11=b\to b=-11

Answer: \boxed{y=\dfrac{3}{2}x-11}

3 0
3 years ago
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