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3 years ago

What is the value of x2 + 2y ÷ 2w + 3z for w = 2, x = 5, y = 8, and z = 3?

Mathematics

1 answer:

pav-90 [236]3 years ago

5 0

The answer to this question is 23

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Malik estimated that 150 people would attend a concert. The exact attendance was 215 What is his percent error, to the nearest p

Afina-wow [57]

Answer:

I think around 43.3%

Step-by-step explanation:

150

150 x 100

=43.3333333333333

3 0

3 years ago

A company weighs 1,096 boxes of cereal and finds that 12.5% of them weighed less than one pound. Which of

Mice21 [21]

Answer:

137 boxes

Step-by-step explanation:

1096 boxes * 0.125(12.5% of boxes weighing less than a pound) = 137 boxes

8 0

3 years ago

Alice searches for her term paper in her filing cabinet, which has several drawers. She knows thatshe left her term paper in dra

katen-ka-za [31]

You made a mistake with the probability $p_{j}$ , which should be $p_{i}$ in the last expression, so to be clear I will state the expression again.

So we want to solve the following:

Conditioned on this event, show that the probability that her paper is in drawer $j$ , is given by:

(1) $\frac{p_{j} }{1-d_{i}p_{i} } , if j \neq i,$ and

(2) $\frac{p_{i} (1-d_{i} )}{1-d_{i}p_{i} } , if j = i.$

so we can say:

$A$ is the event that you search drawer $i$ and find nothing,

$B$ is the event that you search drawer $i$ and find the paper,

$C_{k}$ is the event that the paper is in drawer $k, k = 1, ..., n.$

this gives us:

$P(B) = P(B \cap C_{i} ) = P(C_{i})P(B | C_{i} ) = d_{i} p_{i}$

$P(A) = 1 - P(B) = 1 - d_{i} p_{i}$

Solution to Part (1):

if $j \neq i$ , then $P(A \cap C_{j} ) = P(C_{j} )$ ,

this means that

$P(C_{j} |A) = \frac{P(A \cap C_{j})}{P(A)} = \frac{P(C_{j} )}{P(A)} = \frac{p_{j} }{1-d_{i}p_{i} }$

as needed so part one is solved.

Solution to Part(2):

so we have now that if $j$ = $i$ , we get that:

$P(C_{j}|A ) = \frac{P(A \cap C_{j})}{P(A)}$

remember that:

$P(A|C_{j} ) = \frac{P(A \cap C_{j})}{P(C_{j})}$

this implies that:

$P(A \cap C_{j}) = P(C_{j}) \cdot P(A|C_{j}) = p_{i} (1-d_{i} )$

so we just need to combine the above relations to get:

$P(C_{j}|A) = \frac{p_{i} (1-d_{i} )}{1-d_{i}p_{i} }$

as needed so part two is solved.

8 0

3 years ago

Solve the equation x^13-2x^12-x^11+2x^10=0 in the real number system

ASHA 777 [7]

Answer:

x ={0; - 1 ; 2 ; 1}

Step-by-step explanation:

x¹³-2x¹²-x¹¹+2x¹⁰=0

x¹⁰(x³-2x²-x+2)=0

x¹⁰(x+1)(x-2)(x-1)=0

x¹⁰=0 ∨ x+1=0 ∨ x-2=0 ∨ x-1=0

x=0 x= - 1 x= 2 x= 1

3 0

4 years ago

HCF of (a^3+b^3 ) & (a^4+a^2b^2+b^4)?

VashaNatasha [74]

Answer:

a^2-ab+b^2

Step-by-step explanation:

factorize both of them.

8 0

3 years ago

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