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sineoko [7]
3 years ago
7

What is the slope of the line between (?4, 4 and (?1, ?2?

Mathematics
1 answer:
mrs_skeptik [129]3 years ago
7 0
The slope is calculated through the formula:
dy/dx.
If you apply this to your question you get:
(4-2)/(4-1) = 2/3 or 0.667
Hope this helps!
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Một lô hàng chứa rất nhiều sản phẩm, trong đó tỷ lệ sản phẩm loại tốt là 70%. chọn ngẫu nhiên từ lô hàng ra 5 sản phẩm. gọi X là
den301095 [7]

trả lời : lâp bảng

X=0 thì xác suất lấy ra được sản phẩm tốt là 5*30%=1,5%

X =1 thì xác suất lấy ra được là 1* 70%*4*30%=0,84%

X=2 thì xác suất lấy ra được sản phẩm tốt là 2*70%*3*30%=1,26%

X=3 thì xác suất lấy ra được sản phẩm tốt là 3*70%*2*30%=1,26%

X=4 thì xác suất lấy ra được sản phẩm tốt là 4*70%*1*30%=0,84%

X=5 thì xác suất lấy ra được sản phẩm tốt là 5*70%=3,5%

X 0 1 2 3 4 5

P 1,5 0,84 1,26 1,26 0,84 3,5

b,

E(x)= 1,5*0+ 0,84*1+1,26*2+1,26*3+0,84*4+3,5*5=28

E(x^2)=1,5*0^2+0,84*1^2+1,26*2^2+1,26*3^2+0,84*4^2+3,5*5^2=118,16

===> V(x)= (E(x))^2-E(x^2)=28^2-118,16=665,84

8 0
3 years ago
Solve for t. 4=2+2/5t
maxonik [38]
T=5

4=2+2/5t
Subtract 2 from both sides.
2=2/5t
Multiply both sides by 5/2.
t=5
8 0
2 years ago
Read 2 more answers
A website is offering a promotion, during which customers can buy up to 100 photos for a flat fee. The cost per photo varies inv
Alecsey [184]
The complete question in the attached figure

we have that
 for c=5 and n=20------------> n*c=20*5=100
 for c=2.5 and n=40------------> n*c=40*2.5=100
 for c=2 and n=50------------> n*c=50*2=100
 for c=1.25 and n=80------------> n*c=80*1.25=100

therefore
<span>the function that models the data is n*c=100
</span>
the answer is nc=100





6 0
3 years ago
What is a partial quotient 318 divided by 53
Montano1993 [528]
It is six, or 318/53 is that wat u r asking
7 0
3 years ago
Read 2 more answers
Answer the questions about congruent triangles.
Brrunno [24]

Answer:

<B = <G

Step-by-step explanation:

ABC ≅FGH

<A = <F

<B = <G

<C = <H

because corresponding parts of congruent triangles are congruent

3 0
3 years ago
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