Estimate the sum of 196 and 482
2 answers:
we know that
A quick way to estimate the sum of two numbers is to round each number and then add the rounded numbers
case a) We round the number to the nearest hundred
so
Round up is equal to
Round up is equal to
Find the estimate sum
case b) We round the number to the nearest tens
so
Round up is equal to
Round down is equal to
Find the estimate sum
You might be interested in
Answer:
0.3874 = 38.74% probability that exactly 9 of the 10 consumers recognize her brand name.
0.6126 = 61.26% probability that the number who recognize her brand name is not nine.
Step-by-step explanation:
For each consumer, there are only two possible outcomes. Either they recognize the name, or they do not. The probability of a customer recognizing the name is independent of anu other customer. This means that the binomial probability distribution is used to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
The brand name of Mrs. Fields (cookies) has a 90% recognition rate.
This means that .
Sample of 10
This means that
Find the probability that exactly 9 of the 10 consumers recognize her brand name.
This is P(X = 9). So
0.3874 = 38.74% probability that exactly 9 of the 10 consumers recognize her brand name.
Also, find the probability that the number who recognize her brand name is not nine.
1(100%) subtracted by those who recognize. So
1 - 0.3874 = 0.6126
0.6126 = 61.26% probability that the number who recognize her brand name is not nine.
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.