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3 years ago

Tessellations that use only one type of regular polygon are called semi-regular tessellations. A. True B. False

1 answer:

5 0

The answer is B... False. Tansellations that use only one type of regular polygon are not called semi-regular tessellations

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At noon on tuesday in Chicago the temperature in degrees Fahrenheit was -2 Fahrenheit at noon on Wednesday the temperature was 8

bagirrra123 [75]

The temperature was 6 degrees Fahrenheit

4 0

3 years ago

Read 2 more answers

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

How many tickets were sold after June 15th?

MatroZZZ [7]

A) about 45

If we sold around 90, it would NOT be shown in the graph. And starting on June 15, we sold (per say 50). On June 19 we end with approximately 100. So 45 would be the most probable answer.

Hope this helps!

3 0

2 years ago

Read 2 more answers

Find the eqution of a line that passes through point a and b a=(4,7) b=(2,3)

Lena [83]

Answer:

y=2x-1

Step-by-step explanation:

straight line equation

y=mx + c

find gradient(m) first,use formula;

m=3-7/2-4

m= -4/-2

m=2

use one the coordinates

(2,3)

3=2(2)+c

3=4+c

c=3-4

c = -1

y = 2x - 1

4 0

3 years ago

I dont know how to do it like how to graph the -x or the y

geniusboy [140]

Answer:

see explanation and attachment

Step-by-step explanation:

(1) y = -x + 2

so slope is -1 and y-intercept is (0,2)

(2) 2x - y = 7

y = 2x - 7

so slope is 2 and y-intercept is (0,-7)

7 0

2 years ago