Find the perimeter of each polygon 4cm

if you were to solve the following system by substitution what would be the best variable to solve and from what equation? 2x+8y

Nina [5.8K]

All the numbers in the first equation have a common factor of 2. Removing that gives
.. x +4y = 6
making it easy to solve for x
.. x = 6 -4y

My choice would be to solve for x using the first equation.

_____
On second thought, it might actually be easier to solve either equation for 8y. That term then directly substitutes into the other equation (equivalent to adding the two equations).
.. 8y = 3x -11 . . . . . from the second equation
.. 2x +(3x -11) = 12 . . . substituting into the first equation
.. 5x = 23 . . . . . . . . . . collect terms, add 11 (what you would get by adding the equations in the first place)
.. x = 4.6
.. y = (3*4.6 -11)/8 = 0.35

4 0

3 years ago

Hi uwu pls help me?? i really need it :'( dont make me cry

densk [106]

Answer:

4z-(3z) z

Step-by-step explanation:

I hope that helps

8 0

3 years ago

I reallyyy just need the answer fast!

Mars2501 [29]

Answer:

68×32= 2176

Step-by-step explanation:

5 0

2 years ago

Cecily’s teacher held a raffle. To win the raffle, a student has to pick a paper scroll with an integer written on it. The chart

Luda [366]

Answer:

b

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

A telemarketer is successful at getting people to donate money for her organization in 55% of all calls she makes. She must get

Tems11 [23]

An interesting twist to a binomial distribution problem.

Given:
p=55%=0.55 for probability of success in solicitation
x=4=number of successful solicitations
n=number of calls to be made
P(x,n,p)>=89.9%=0.899 (from context, it is >= and not =, which is almost impossible)

From context of question, all calls are assumed independent, with constant probability of success, so binomial distribution is applicable.

The number of successes, x, is then given by
$P(x)=C(n,x)p^x(1-p)^{n-x}$ where
p=probability of success
n=number of trials
x=number of successes $C(n,x)=\frac{n!}{x!(n-x)!}$

Here we need n such that
P(x,n,p)>=0.899
given
x>=4, p=0.55, which means we need to find

Method 1: if a cumulative binomial distribution table is available, we can look up n=9,10,11 and find
P(x>=4,9,0.55)=0.834
P(x>=4,10,0.55)=0.898
P(x>=4,11,0.55)=0.939
So she must make (at least) 11 calls to make sure the probability of meeting her quota is 89.9% or more.

Method 2: using technology.
Similar to method 1, we can look up the probabilities directly, for n=9,10,11
P(x>=4,9,0.55)=0.834178
P(x>=4,10,0.55)=0.8980051
P(x>=4,11,0.55)=0.9390368

Method 3: using simple calculator
Here we need to calculate the probabilities for each value of n=10,11 and sum the probabilities of FAILURE S=P(0,n,0.55)+P(1,n,0.55)+P(2,n,0.55)+P(3,n,0.55)
so that the probability of success is 1-S.
For n=10,
P(0,10,0.55)=0.000341
P(1,10,0.55)=0.004162
P(2,10,0.55)=0.022890
P(3,10,0.55)=0.074603
So that
S=0.000341+0.004162+0.022890+0.074603
=0.101995
and Probability of getting 4 successes (or more)
=1-S
=0.898005, missing target by 0.1%

So she will have to make 11 phone calls, bring up the probability to 93.9%. The work is similar to that of n=10.

8 0

3 years ago