I need to know this answer to this question "Compare 7 in 473?"

You are downloading a song. The percent y(in decimal form) of megabytes remaining to download after x seconds is y=-0.1x+1.

igomit [66]

I would make the equation have both of your variables on the same side:

0.1x+y=1

Then I would use the cover-up method and say that

x-intercept= 10

y-intercept= 1

*Cover-Up Method: If you were finding the X-intercept, you would just ignore the y variable and divide your whole number by x.

In this case the whole number is 1. 1/0.1=10

Vice Versa with finding the Y-intercept

3 0

3 years ago

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100 points (algebra one) describe and correct the error in comparing the graphs

Pepsi [2]

Answer:

The error made is that the graph of $y=x^{2} +3$ is a vertical translation up by 3, instead of down.

Step-by-step explanation:

Exponential functions, when they are graphed, use the function $f(x)=a(x-h)^{2} +k$ .

In the equation, k expresses how much the graph shifts vertically. In this case, the function shifts vertically up by 3 from the parent function, $y=x^{2}$ . So, the translation is actually 3 units up compared to the graph of $y=x^{2}$ .

Hope this helps and have an amazing day!! ❤

3 0

3 years ago

Francine and Cheryl received equal scores on a test made up of multiple choice questions and an essay Francine got 34 multiple c

dem82 [27]

Answer: 2 points

Step-by-step explanation:

5 0

3 years ago

What

NNADVOKAT [17]

The answer is -130. hope this helps.

7 0

3 years ago

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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 more 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 expected 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