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3 years ago

Find the area. please

Mathematics

1 answer:

sashaice [31]3 years ago

8 0

Answer:

16 cm²

Step-by-step explanation:

The figure is a trapezoid with two parallel sides.

The formula for the area is

A = ½(a + b)h

Data:

a = 5 cm

b = 3 cm

h = 4 cm

Calculation:

A = ½(5 + 3) × 4 = ½ × 8 × 4 = 16 cm²

The area of the trapezoid is 16 cm².

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Determine the utilization and the efficiency for each situations:A. A loan processing operation that processes an average of 7 l

Vera_Pavlovna [14]

Answer:

A) Utilization = 70%

Efficiency = 87.5%

B) Utilization = 66.67%

Efficiency = 80%

Step-by-step explanation:

Determine the utilization and the efficiency for each situations:

A. A loan processing operation that processes an average of 7 loans per day. The operation has a design capacity of 10 loans per day and an effective capacity of 8 loans per day.

Formula for Utilization = Actual Output /

Design Capacity × 100

Actual output = 7 loans per day

Design capacity = 10 loans per day

Utilization = 7/10 × 100

= 70%

Efficiency = Actual output / Effective capacity × 100

Actual output = 7 loans per day

Effective capacity = 8 loans per day

Efficiency = 7/8 × 100

= 87.5%

B. A furnace repair team that services and average of four furnaces a day if the design capacity is six furnaces a day and the effective capacity is five furnaces a day.

Formula for Utilization = Actual Output /

Design Capacity × 100

Actual output = four furnaces a day

Design capacity = six furnaces a day

Utilization = 4/6 × 100

= 66.67%

Efficiency = Actual output / Effective capacity × 100

Actual output = four furnaces a day

Effective capacity = five furnaces a day

Efficiency = 4/5 × 100

= 80%

6 0

2 years ago

The profit per acre from a grove of orange trees is given by x(190 − x) dollars, where x is the number of orange trees per acre.

nasty-shy [4]

Answer:

$P(x) = 190 x -x^2$

In order to maximize the last equation we can derivate the function in term of x and we got:

$\frac{dP}{dx} = 190 -2x$

And setting this derivate equal to 0 we got:

$\frac{dP}{dx} = 190 -2x=0$

And solving for x we got:

$x = 95$

And for this case the value that maximize the profit would be x =95 and the corresponding profit would be:

$P(x=95)= 95(190-95)= 95*95 = 9025$

Step-by-step explanation:

For this case we have the following function for the profit:

$P(x) = x(190-x)$

And we can rewrite this expression like this:

$P(x) = 190 x -x^2$

In order to maximize the last equation we can derivate the function in term of x and we got:

$\frac{dP}{dx} = 190 -2x$

And setting this derivate equal to 0 we got:

$\frac{dP}{dx} = 190 -2x=0$

And solving for x we got:

$x = 95$

And for this case the value that maximize the profit would be x =95 and the corresponding profit would be:

$P(x=95)= 95(190-95)= 95*95 = 9025$

7 0

3 years ago

Read 2 more answers

Find the area of the figure. If necessary, round to the nearest tenth.

nirvana33 [79]

Answer:9.4cm2

Step-by-step explanation:

8 0

3 years ago

77julia77 [94]

I believe the answer should b the second option :)

4 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