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BartSMP [9]
3 years ago
11

theo had a balance of -$4 in his savings account.after making a deposit, he was $25 in his account. what is the overall change t

o his account?
Mathematics
2 answers:
Anika [276]3 years ago
8 0

Answer:

+$29

Step-by-step explanation:

Depositing is putting money in the bank. To get to zero, Theo first added 4. The add 25 to get $25.

Licemer1 [7]3 years ago
7 0

Answer:

$29.00

Step-by-step explanation:

To get out of debt

-4+4=0

To get to $25

0+25=25

Put them together

4+25=29

Hope this helps!

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grigory [225]

Answer:

y=-3/2x+6

Step-by-step explanation:

You can find the slope by taking two points on the graph, and making the one that occurs earlier in the graph (from left to right) the first point (x1, y1) and the one that occurs later in the graph the second point (x2, y2). The equation is m (or slope)=(y2-y1)/(x2-x1). I took the first two points in the table for this. m=(12-18)/-4-(-8)

the double negative on the bottom becomes an addition=> (12-18)/(-4+8)

the top simplifies to be -6 and the bottom simplifies to 4=>-6/4

this fraction can be reduced to -3/2, which is the slope of the graph.

Now, use point slope form (y-y1=m(x-x1)) to find the equation of the graph. Plug any coordinate on the graph in for x1 and y1 here. It should be correct as long as it is a point on the graph, but I am using the point (-8, 18) here.

=>y-18=-3/2(x-(-8))

the double negative in the parentheses becomes a positive=> y-18=-3/2(x+8)

distribute the -3/2 to every term in the parentheses=> y-18=-3/2x-12

add 18 to both sides, cancelling out the -18 on the left side of the equation=>y=-3/2x+6 (-12+18=6 to get 6 for b).

Therefore, the equation is y=-3/2x+6

6 0
2 years ago
A store selling newspapers orders only n = 4 of a certain newspaper because the manager does not get many calls for that publica
umka2103 [35]

Answer:

a) The expected value is 2.680642

b) The minimun number of newspapers the manager should order is 6.

Step-by-step explanation:

a) Lets call X the demanded amount of newspapers demanded, and Y the amount of newspapers sold. Note that 4 newspapers are sold when at least four newspaper are demanded, but it can be <em>more</em> than that.

X is a random variable of Poisson distribution with mean \mu = 3 , and Y is a random variable with range {0, 1, 2, 3, 4}, with the following values

  • PY(k) = PX(k) = ε^(-3)*(3^k)/k! for k in {0,1,2,3}
  • PY(4) = 1 -PX(0) - PX(1) - PX(2) - PX (3)

we obtain:

PY(0) = ε^(-3) = 0.04978..

PY(1) = ε^(-3)*3^1/1! = 3*ε^(-3) = 0.14936

PY(2) = ε^(-3)*3^2/2! = 4.5*ε^(-3) = 0.22404

PY(3) = ε^(-3)*3^3/3! = 4.5*ε^(-3) = 0.22404

PY(4) = 1- (ε^(-3)*(1+3+4.5+4.5)) = 0.352768

E(Y) = 0*PY(0)+1*PY(1)+2*PY(2)+3*PY(3)+4*PY(4) =  0.14936 + 2*0.22404 + 3*0.22404+4*0.352768 = 2.680642

The store is <em>expected</em> to sell 2.680642 newspapers

b) The minimun number can be obtained by applying the cummulative distribution function of X until it reaches a value higher than 0.95. If we order that many newspapers, the probability to have a number of requests not higher than that value is more 0.95, therefore the probability to have more than that amount will be less than 0.05

we know that FX(3) = PX(0)+PX(1)+PX(2)+PX(3) = 0.04978+0.14936+0.22404+0.22404 = 0.647231

FX(4) = FX(3) + PX(4) = 0.647231+ε^(-3)*3^4/4! = 0.815262

FX(5) = 0.815262+ε^(-3)*3^5/5! = 0.91608

FX(6) = 0.91608+ε^(-3)*3^6/6! = 0.966489

So, if we ask for 6 newspapers, the probability of receiving at least 6 calls is 0.966489, and the probability to receive more calls than available newspapers will be less than 0.05.

I hope this helped you!

8 0
3 years ago
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