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agasfer [191]
3 years ago
6

The maximum weight allowed per car on The Twister carnival ride is 263 pounds. Your friend weighs 86 pounds. To be able to ride

in a car together, how much can you weigh? Write and solve the inequality. (1 point) x + 177 ≤ 263; at most 86 pounds x – 86 ≤ 263; at most 177 pounds x + 86 ≤ 263; at most 177 pounds x + 86 ≤ 177; at most 263 pound
Mathematics
1 answer:
Gelneren [198K]3 years ago
8 0
Answer is 177
 sovle the ret
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write an equation to represent"three consecutive integers is 12 less than 4 times the middle integer'
Sever21 [200]

Consider that the three consecutive integers are:

least integer = n

middle integer = n + 1

greatest integer = n + 2

THe expression "three consecutive integers is 12 less than 4 times the middle integer" can be written as follow:

n + (n + 1) + (n + 2) = 4(n +1) - 12

In order to find the numbers, proceed as follow:

n + (n + 1) + (n + 2) = 4(n +1) - 12 cancel parenthesis

n + n + 1 + n + 2 = 4n + 4 - 12 simplify like terms

3n + 3 = 4n - 8 subtract 4n and 3 both sides

3n - 4n = - 8 - 3

-n = -11

n = 11

Hence, the three consecutive integers are:

n = 11

n + 1 = 12

n + 2 = 13

5 0
1 year ago
Quadrilateral abcd is a parallelogram if both pairs of opposite angles are congruent. prove that quadrilateral abcd is parallelo
Sati [7]

Answer:

you how are doing please study and work hard

8 0
2 years ago
In what table does y vary directly with x?
timurjin [86]

Answer:

The phrase “ y varies directly as x” or “ y is directly proportional to x” means that as x gets bigger, so does y, and as x gets smaller, so does y. That concept can be translated in two ways. for some constant k. The k is called the constant of proportionality. Therefore a chance of probability is a binary table.

Step-by-step explanation:

8 0
2 years ago
Rework problems 13 and 14 from section 2.4 of your textbook (page 81) about the bucket containing orange tennis balls and yellow
DaniilM [7]

The formula C(n, r)= \frac{n!}{r!(n-r)!}, where r! is 1*2*3*...r

is the formula which gives us the total number of ways of forming groups of r objects, out of n objects.

for example, given 10 objects, there are C(10,6) ways of forming groups of 6, out of the 10 objects.

----------------------------------------------------------------------------------------------------

The bucket contains 6 orange balls and 7 yellow balls, that is a total of 13 balls.

consider the balls of the same color to be different objects (as they indeed are, don't think of them as identical)

The total number of groups of 5 balls out of the 13, that we can form is :

\displaystyle{C(13, 5)= \frac{13!}{5!8!}= \frac{13\cdot12\cdot11\cdot10\cdot9\cdot8!}{5!8!}

\frac{13\cdot12\cdot11\cdot10\cdot9}{5\cdot4\cdot3\cdot2\cdot1}=13\cdot11\cdot9= 1287


<span>1) What is the probability that, of the 5 balls selected at random, at least one is orange and at least one is yellow? 

the event can happen as follows:
 
{1 orange, 4 yellow} or </span>{2 orange, 3 yellow} or {3 orange, 2 yellow} or {4 orange, 1 yellow}


that is, the only cases we are not considering are {5 orange} or {5 yellow}


5 orange balls can be selected in C(6, 5) = 6!/(5!1!)=6 different ways, 

and 5 yellow balls can be selected in C(7, 5)=7!/(5!2!)=(7*6)/2=21 different ways.

so 5 orange balls or 5 yellow balls can be selected in 6+21=27 different ways, 


this means that P(5 orange balls or 5 yellow balls)=27/1287=0.02


P({1 orange, 4 yellow} or {2 orange, 3 yellow} or {3 orange, 2 yellow} or {4 orange, 1 yellow}) = 1-P(5 orange balls or 5 yellow balls)=1-0.02=0.98



Part 2

The event "at least two are orange and at least two are yellow " can happen in the following ways:

{2 orange, 3 yellow} or {3 orange, 2 yellow}

2 orange, 3 yellow balls can be selected in C(6,2)*C(7,3) because 

any selected 2 of the 6 orange balls, can be combined with any group of 3, out of the 7 yellow balls.

so we calculate:


\displaystyle{ C(6,2)\cdot C(7,3)= \frac{6!}{2!4!} \cdot \frac{7!}{4!3!}=15\cdot \frac{7\cdot6\cdot5}{3\cdot2}=525&#10;

\displaystyle{ C(6,3)\cdot C(7,2)= \frac{6!}{3!3!} \cdot \frac{7!}{5!2!}= \frac{6\cdot5\cdot4}{3\cdot2} \cdot21=420&#10;


thus, 

P ({2 orange, 3 yellow} or {3 orange, 2 yellow})=


 P (2 orange, 3 yellow) + P (3 orange, 2 yellow)=


\displaystyle{\frac{C(6,2)\cdot C(7,3)+C(6,3)\cdot C(7,2)}{1287}= \frac{525+420}{1287}= 0.734


8 0
3 years ago
Consider the function f(x)=x^2-2x+6. find the value of f(x) when x=3
Delvig [45]
We plug in x=3 into the function, 3^2-2(3)+6, simplifies to 9-6+6, 3+6 = 9
8 0
3 years ago
Read 2 more answers
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