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morpeh [17]
3 years ago
8

N a second grade class containing 12 girls and 11 boys, 2 students are selected at random to give out the math papers. what is t

he probability that both are girls?
Mathematics
1 answer:
ira [324]3 years ago
7 0
\displaystyle
|\Omega|=\binom{23}{2}=\dfrac{23!}{2!21!}=\dfrac{22\cdot23}{2}=253\\
|A|=\binom{12}{2}=\dfrac{12!}{2!10!}=\dfrac{11\cdot12}{2}=66\\\\
P(A)=\dfrac{66}{253}=\dfrac{6}{23}
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Determine if the sequence below is arithmetic or geometric and determine the common difference/ratio in simplest form.
Nana76 [90]

Step-by-step explanation:

this is clearly not a linear sequence (the terms don't have the same difference).

so, it has to be a geometric sequence.

the common ratio is r.

s2 = s1 × r

16 = 64 × r

r = 16/64 = 1/4

control :

s3 = s2×r

4 = 16 × 1/4 = 4

correct.

6 0
2 years ago
The line passes through the point (3,0) and has a slope of -3
Lesechka [4]

Answer:

the y intercept would be -3 if thats what youre asking

Step-by-step explanation:

6 0
2 years ago
Drag the tiles to the correct boxes to complete the pairs not all tiles will be used match each quadratic graph to its respectiv
ch4aika [34]

Answer:

Part 1) The function of the First graph is f(x)=(x-3)(x+1)

Part 2) The function of the Second graph is f(x)=-2(x-1)(x+3)

Part 3) The function of the Third graph is f(x)=0.5(x-6)(x+2)

See the attached figure

Step-by-step explanation:

we know that

The quadratic equation in factored form is equal to

f(x)=a(x-c)(x-d)

where

a is the leading coefficient

c and d are the roots or zeros of the function

Part 1) First graph

we know that

The solutions or zeros of the first graph are

x=-1 and x=3

The parabola open up, so the leading coefficient a is positive

The function is equal to

f(x)=a(x-3)(x+1)

Find the value of the coefficient a

The vertex is equal to the point (1,-4)

substitute and solve for a

-4=a(1-3)(1+1)

-4=a(-2)(2)

a=1

therefore

The function is equal to

f(x)=(x-3)(x+1)

Part 2) Second graph

we know that

The solutions or zeros of the first graph are

x=-3 and x=1

The parabola open down, so the leading coefficient a is negative

The function is equal to

f(x)=a(x-1)(x+3)

Find the value of the coefficient a

The vertex is equal to the point (-1,8)

substitute and solve for a

8=a(-1-1)(-1+3)

8=a(-2)(2)

a=-2

therefore

The function is equal to

f(x)=-2(x-1)(x+3)

Part 3) Third graph

we know that

The solutions or zeros of the first graph are

x=-2 and x=6

The parabola open up, so the leading coefficient a is positive

The function is equal to

f(x)=a(x-6)(x+2)

Find the value of the coefficient a

The vertex is equal to the point (2,-8)

substitute and solve for a

-8=a(2-6)(2+2)

-8=a(-4)(4)

a=0.5

therefore

The function is equal to

f(x)=0.5(x-6)(x+2)

3 0
4 years ago
Can someone help me wirh this ASAP
madreJ [45]
Heyoo. I used the Pythagorean Theorem to solve it since that theorem can be used to find the missing side of any right angle triangle. Hope this helps!

5 0
3 years ago
A shipment of 288 reams of paper was delivered each of the 30 classroom received an equal share of the paper any extra reams of
Katen [24]
Each classroom would get 9.6
3 0
3 years ago
Read 2 more answers
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