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prisoha [69]
2 years ago
11

Solve the equation 40a^2+4a=0

Mathematics
1 answer:
sergiy2304 [10]2 years ago
7 0
40a^2+4a=0\ \ \ \ |divide\ both\ sides\ by\ 4\\\\10a^2+a=0\\\\a(10a+1)=0\iff a=0\ or\ 10a+1=0\\\\a=0\ or\ 10a=-1\\\\\boxed{a=0\ or\ a=-0.1}
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4cmx4cmx4cm pls SOMEONE 25 POINTS
STALIN [3.7K]

Answer: 4cm x 4cm x 4cm = 64cm

Step-by-step explanation:

4x4 = 16x4 = 64

7 0
2 years ago
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A box of 15 cookies costs $9.<br> MY<br> What is the cost for 1 cookie?
Nitella [24]

Answer: $0.6

Step-by-step explanation:

15cookies= $9

1 cookie = x

Cross multiply

15x= 9

Divide both sides by 15

x= 9/15= 3/5= $0.6

4 0
3 years ago
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Barry grows apples, blueberries, grapes, and zucchini on his farm. He sells his grapes for $3.50 per pound. How much money did h
vaieri [72.5K]

Answer:

ok so this might look confusing what you have to do is just ignore part of the problem in this one just ignore apples, blueberries and zucchini their not important

on saturday it looks like he sold 14 pounds on sunday 15 pounds

14+15=29

and each pound is 3.50 so

29*3.5= 29*3+29*0.5

29*3=87

29*0.5=14.5

87+14.5=101.5

c

Hope This Helps!!!

7 0
1 year ago
Determine if the two triangles are congruent.If they are state how you know
Finger [1]

Answer:

1. yes, both triangles sides are congruent

2. yes, they have congruent sides and they have a congruent angle

3. yes, congruent sides and angle

4. yes, 2 congruent angles

5. no, only one congruent angle not enough proof

6. yes, 2 congruent angles

7. no, only one congruent angle not enough proof

8. no, only one congruent side

6 0
2 years ago
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What is the formula for the expected number of successes in a binomial experiment with n trials and probability of success​ p? C
charle [14.2K]

Answer:

(D)E[ X ] =np.

Step-by-step explanation:

Given a binomial experiment with n trials and probability of success​ p,

f(x)=\left(\begin{array}{c}n\\k\end{array}\right)p^x(1-p)^{n-x}, 0\leq  x\leq n

E(X)=\sum_{x=0}^{n}xf(x)= \sum_{x=0}^{n}x\left(\begin{array}{c}n\\k\end{array}\right)p^x(1-p)^{n-x}

Since each term of the summation is multiplied by x, the value of the term corresponding to x = 0 will be 0. Therefore the expected value becomes:

E(X)=\sum_{x=1}^{n}x\left(\begin{array}{c}n\\x\end{array}\right)p^x(1-p)^{n-x}

Now,

x\left(\begin{array}{c}n\\x\end{array}\right)= \frac{xn!}{x!(n-x)!}=\frac{n!}{(x-)!(n-x)!}=\frac{n(n-1)!}{(x-1)!((n-1)-(x-1))!}=n\left(\begin{array}{c}n-1\\x-1\end{array}\right)

Substituting,

E(X)=\sum_{x=1}^{n}n\left(\begin{array}{c}n-1\\x-1\end{array}\right)p^x(1-p)^{n-x}

Factoring out the n and one p from the above expression:

E(X)=np\sum_{x=1}^{n}n\left(\begin{array}{c}n-1\\x-1\end{array}\right)p^{x-1}(1-p)^{(n-1)-(x-1)}

Representing k=x-1 in the above gives us:

E(X)=np\sum_{k=0}^{n}n\left(\begin{array}{c}n-1\\k\end{array}\right)p^{k}(1-p)^{(n-1)-k}

This can then be written by the Binomial Formula as:

E[ X ] = (np) (p +(1 - p))^{n -1 }= np.

5 0
3 years ago
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