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3 years ago

Simplify:

Mathematics

2 answers:

Irina-Kira [14]3 years ago

7 0

Step-by-step explanation:

Let's combine the like terms first, which are x^2 & -3x^2, and 3x & -4x.

x^2 -3x^2 is -2x^2. 3x -4x is -x.

The final, simplified expression would be -2x^2 -x.

harina [27]3 years ago

6 0

Answer:

-2x²-x

Step-by-step explanation:

x² + 3x - 3x² - 4x

You need to simplify

x² -3x²=-2x²

and

$3x - 4x = - x$

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Yon buys tickets to a concert for himself and a friend. There is a tax of 6% on the price of the tickets and

Mashutka [201]

Answer:

t=1.06x+10 where x=The amount of tickets bought

Step-by-step explanation:

you multiply x by 1.06 to add the extra tax onto it.

You add the 10 to add the $10 booking fee

6 0

4 years ago

if two pounds of cheese sells for $3.18, how much will three pounds of cheese cost? SHOW YOUR WORK!! plz

OleMash [197]

If two pounds of cheese sells for $3.18, how much will three pounds of cheese cost?

If two pounds of cheese are $3.18 then we must divide it by 2.

$\frac{3.18}{2} =1.59$ <u />

Then we multiply.

$1.59*3=4.77$ <u />

So three pounds of cheese are

$ $4.77$ <u />

Hope I helped!

6 0

3 years ago

An automated egg carton loader has a 1% probability of cracking an egg, and a customer will complain if more than one egg per do

vaieri [72.5K]

Answer:

a) Binomial distribution B(n=12,p=0.01)

b) P=0.007

c) P=0.999924

d) P=0.366

Step-by-step explanation:

a) The distribution of cracked eggs per dozen should be a binomial distribution B(12,0.01), as it can model 12 independent events.

b) To calculate the probability of having a carton of dozen eggs with more than one cracked egg, we will first calculate the probabilities of having zero or one cracked egg.

$P(k=0)=\binom{12}{0}p^0(1-p)^{12}=1*1*0.99^{12}=1*0.886=0.886\\\\P(k=1)=\binom{12}{1}p^1(1-p)^{11}=12*0.01*0.99^{11}=12*0.01*0.895=0.107$

Then,

$P(k>1)=1-(P(k=0)+P(k=1))=1-(0.886+0.107)=1-0.993=0.007$

c) In this case, the distribution is B(1200,0.01)

$P(k=0)=\binom{1200}{0}p^0(1-p)^{12}=1*1*0.99^{1200}=1* 0.000006 = 0.000006 \\\\ P(k=1)=\binom{1200}{1}p^1(1-p)^{1199}=1200*0.01*0.99^{1199}=1200*0.01* 0.000006 \\\\P(k=1)= 0.00007\\\\\\P(k\leq1)=0.000006+0.000070=0.000076\\\\\\P(k>1)=1-P(k\leq 1)=1-0.000076=0.999924$