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2 years ago

The manufacturer of widgets spends $5 to make each widget and sells them for $8. The manufacturer also has fixed costs

Mathematics

2 answers:

Bezzdna [24]2 years ago

5 0

Answer:c(x)=5x+1200

Step-by-step explanation:

S_A_V [24]2 years ago

4 0

Answer:

$C(x) = 5x + 1200$

Step-by-step explanation:

The cost function is given by:

C(x)=Fixed cost + cost of x widgets.

The fixed cost is $1200

The cost of producing x widgets is 5x.

Therefore the total cost when x widgets are manufactured is

$C(x) = 5x + 1200$

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Shinto is serving pizza. He serves 1/5 of the pizza to one friend and 5/12 to another friend. If he had 9/10 of a pizza to start

never [62]

Answer: 17/60

Step-by-step explanation:

Change all of the fractions to have a common denominator:

LCM of 10, 12, 15 is 60

You get 12/60, 25/60, and 54/60.

54/60 is what he had at the start, and so you can just subtract what he gave away

54/60-12/60=42/60

42/60-25/60=17/60

17 is a prime number so you cannot simplify the fraction

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2 years ago

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Sladkaya [172]

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Estimate 9.586+3.097 then give me answer

zzz [600]

Answer is: its 12.683

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2 years ago

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Let represent the number of tires with low air pressure on a randomly chosen car. The probability distribution of is as follows.

Sindrei [870]

Answer:

a) $P(X=3) = 0.1$

b) $P(X\geq 3) =1-P(X$

And replacing we got:

$P(X \geq 3) = 1- [0.2+0.3+0.1]= 0.4$

c) $P(X=4) = 0.3$

d) $P(X=0) = 0.2$

e) $E(X) =0*0.2 +1*0.3+2*0.1 +3*0.1 +4*0.3= 2$

f) $E(X^2)= \sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2) =0^2*0.2 +1^2*0.3+2^2*0.1 +3^2*0.1 +4^2*0.3= 6.4$

And the variance would be:

$Var(X0 =E(X^2)- [E(X)]^2 = 6.4 -(2^2)= 2.4$

And the deviation:

$\sigma =\sqrt{2.4} = 1.549$

Step-by-step explanation:

We have the following distribution

x 0 1 2 3 4

P(x) 0.2 0.3 0.1 0.1 0.3

Part a

For this case:

$P(X=3) = 0.1$

Part b

We want this probability:

$P(X\geq 3) =1-P(X$

And replacing we got:

$P(X \geq 3) = 1- [0.2+0.3+0.1]= 0.4$

Part c

For this case we want this probability:

$P(X=4) = 0.3$

Part d

$P(X=0) = 0.2$

Part e

We can find the mean with this formula:

$E(X)= \sum_{i=1}^n X_i P(X_i)$

And replacing we got:

$E(X) =0*0.2 +1*0.3+2*0.1 +3*0.1 +4*0.3= 2$

Part f

We can find the second moment with this formula

$E(X^2)= \sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2) =0^2*0.2 +1^2*0.3+2^2*0.1 +3^2*0.1 +4^2*0.3= 6.4$

And the variance would be:

$Var(X0 =E(X^2)- [E(X)]^2 = 6.4 -(2^2)= 2.4$

And the deviation:

$\sigma =\sqrt{2.4} = 1.549$

4 0

3 years ago

At which root does the graph of f(x) = (x + 4)(x + 7,5) cross the x-axis?<br> -7<br> -4<br> 4<br> 7

Alona [7]

Answer:

Answer: -7

Step-by-step explanation:

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1 year ago