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2 years ago

9

Three less than four times a number is 5

1 answer:

Margarita [4]2 years ago

5 0

Answer: 2

Step-by-step explanation:

5+3=8

8/4=2

‍♂️ I’m not the best at explaining

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What is the maximum value of P = 4x + 2y, given the constraints on x and y listed below?

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When you graph the constraints the four points you can use are (0,0), (10,0), (0,2), and (6,2).

(10,0) is the point that gives you the maximum value which is 40.

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Write words to match the expression 3+(4×12)

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13 cans of paint with 5/6 liters of paint in each were used to paint 4 identical cars. How can you calculate the amount of paint

brilliants [131]

Answer:

(B) 5/6 ·13÷4

(C) 5/6 ÷4·13

(D) 5/6 · 13/4

Step-by-step explanation:

Amount of paint per can = $\frac{5}{6}$ liters.

Number of cans used to paint 4 cars = 13

Total amount of paint used for 4 cars = $13 * \frac{5}{6}$ liters.

= $\frac{13 * 5}{6}$ liters.

= $\frac{65}{6}$ liters.

Total amount of paint used for 1 car = $\frac{65}{6} \div 4$ liters.

= $\frac{65}{24}$ liters.

Of the given options, option B,C and D correspond to the same value.

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2 years ago

There are three power plants [X, Y, Z] that at any given time each one either generates electricity or idles. Event A is that pl

insens350 [35]

We're told that

$P(A\cap B)=0.15$

$P(A\cup B)^C=0.06\implies P(A\cup B)=0.94$

$P(B\mid A)=P(B^C\mid A)=0.5$

where the last fact is due to the law of total probability:

$P(A)=P(A\cap B)+P(A\cap B^C)$

$\implies P(A)=P(B\mid A)P(A)+P(B^C\mid A)P(A)$

$\implies 1=P(B\mid A)+P(B^C\mid A)$

so that $B\mid A$ and $B^C\mid A$ are complementary.

By definition of conditional probability, we have

$P(B\mid A)=P(B^C\mid A)$

$\implies\dfrac{P(A\cap B)}{P(A)}=\dfrac{P(A\cap B^C)}{P(A)}$

$\implies P(A\cap B)=P(A\cap B^C)$

We make use of the addition rule and complementary probabilities to rewrite this as

$P(A\cap B)=P(A\cap B^C)$

$\implies P(A)+P(B)-P(A\cup B)=P(A)+P(B^C)-P(A\cup B^C)$

$\implies P(B)-[1-P(A\cup B)^C]=[1-P(B)]-P(A\cup B^C)$

$\implies2P(B)=2-[P(A\cup B)^C+P(A\cup B^C)]$

$\implies2P(B)=[1-P(A\cup B)^C]+[1-P(A\cup B^C)]$

$\implies2P(B)=P(A\cup B)+P(A\cup B^C)^C$

$\implies2P(B)=P(A\cup B)+P(A^C\cap B)\quad(*)$

By the law of total probability,

$P(B)=P(A\cap B)+P(A^C\cap B)$

$\implies P(A^C\cap B)=P(B)-P(A\cap B)$

and substituting this into $(*)$ gives

$2P(B)=P(A\cup B)+[P(B)-P(A\cap B)]$

$\implies P(B)=P(A\cup B)-P(A\cap B)$

$\implies P(B)=0.94-0.15=\boxed{0.79}$

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2 years ago