A park is in the shape of a rectangle 2 miles long and 1.5 miles wide. How much shorter if you walk diaganolly across the park r
2 answers:
You can answer this by using Pythagoras theorem. Okay, let me draw this first.
2 miles
V
________________________
| |
| |
| | << 1.5 miles
|_______________________|
^ Assume this is the rectangle. The long side is 2 miles while the shorter side is 1.5 miles.
So, to get the shortest distance, let's cut out a triangle from the rectangle.
/|
/ |
/ |
What we want to find is this one > / | << 2 miles
/ |
/______|
^ 1.5 miles (I know the size is not logical,but imagine it please)
So to get the diagonal length =
=
= 2.5 miles
To calculate how much shorter = ( 2 miles + 1.5 miles) - 2.5 miles
= 3.5 miles - 2.5 miles
= 1.0 miles
Hope this help !
2 miles
V
________________________
| |
| |
| | << 1.5 miles
|_______________________|
^ Assume this is the rectangle. The long side is 2 miles while the shorter side is 1.5 miles.
So, to get the shortest distance, let's cut out a triangle from the rectangle.
/|
/ |
/ |
What we want to find is this one > / | << 2 miles
/ |
/______|
^ 1.5 miles (I know the size is not logical,but imagine it please)
So to get the diagonal length =
=
= 2.5 miles
To calculate how much shorter = ( 2 miles + 1.5 miles) - 2.5 miles
= 3.5 miles - 2.5 miles
= 1.0 miles
Hope this help !
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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
where
p=probability of success
n=number of trials
x=number of successes
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.
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=probability of success
n=number of trials
x=number of successes
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.
Answer:
The height of the larger traffic signs is 51.96 inches
Step-by-step explanation:
Given that initially, A company makes traffic signs of an equilateral triangle with the perimeter of 144 inches
Now, A company makes similar traffic signs of perimeter 1.25 of original
New perimeter will be 1.25x144=180 inches
Let triangle ABC be a larger triangle with 180inches of the perimeter.
It is given that the triangle is an equilateral triangle,
Therefore, Side AB=BC=AC
The perimeter of a triangle ABC is
AB+BC+AC=180
3AB=180
AB=60 inches
Figure show that triangle ABC and AD is the height of the triangle.
Since, the triangle ABC is an equilateral triangle,
AD is the perpendicular bisector of BC
so that,
BD=DC and Angle ADB=90
In right triangle ADB,
Angle D =90 and Angle B =60
Side BD=60/2=30 inches and AB=60 inches
Using Pythagoras theorem,
AD=51.96 inches
Thus, The height of the larger traffic signs is 51.96 inches
