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Kamila [148]
3 years ago
10

x%5E%7B2%7D" id="TexFormula1" title="\frac{1}{3}x^{2} +5x+\frac{4}{3}=2x-\frac{1}{3} x^{2}" alt="\frac{1}{3}x^{2} +5x+\frac{4}{3}=2x-\frac{1}{3} x^{2}" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Ber [7]3 years ago
5 0
X1= -4, x2= -1/2 is the correct answer
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Mentalmath grade 3 $1.00-$0.37
Mkey [24]

Answer:

$0.63

Step-by-step explanation:

6 0
3 years ago
Fifty-three percent of employees make judgements about their co-workers based on the cleanliness of their desk. You randomly sel
GarryVolchara [31]

Answer:

The unusual X values ​​for this model are: X = 0, 1, 2, 7, 8

Step-by-step explanation:

A binomial random variable X represents the number of successes obtained in a repetition of n Bernoulli-type trials with probability of success p. In this particular case, n = 8, and p = 0.53, therefore, the model is {8 \choose x} (0.53) ^ {x} (0.47)^{(8-x)}. So, you have:

P (X = 0) = {8 \choose 0} (0.53) ^ {0} (0.47) ^ {8} = 0.0024

P (X = 1) = {8 \choose 1} (0.53) ^ {1} (0.47) ^ {7} = 0.0215

P (X = 2) = {8 \choose 2} (0.53)^2 (0.47)^6 = 0.0848

P (X = 3) = {8 \choose 3} (0.53) ^ {3} (0.47)^5 = 0.1912

P (X = 4) = {8 \choose 4} (0.53) ^ {4} (0.47)^4} = 0.2695

P (X = 5) = {8 \choose 5} (0.53) ^ {5} (0.47)^3 = 0.2431

P (X = 6) = {8 \choose 6} (0.53) ^ {6} (0.47)^2 = 0.1371

P (X = 7) = {8 \choose 7} (0.53) ^ {7} (0.47)^ {1} = 0.0442

P (X = 8) = {8 \choose 8} (0.53)^{8} (0.47)^{0} = 0.0062

The unusual X values ​​for this model are: X = 0, 1, 7, 8

6 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 successesC(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
PLEASE HELP <br> GIVEN THE GRAPH OF A FUNCTION, IDENTIFY ALL ITS FEATURES
Novosadov [1.4K]

Answer:

Once again, it’s a nonlinear function.

Graph is not in a straight line.

It has arrows on it’s ends.

Step-by-step explanation:

This line is squiggly. It is bent at the origin and some others. It also has arrows at the line’s ends.

8 0
2 years ago
The State Board of Education has $2,183 to buy new calculators. If each calculator costs $37, how many calculators can the board
Tema [17]

Answer:

59

Step-by-step explanation:

They could buy 59

You solve this by dividing the total amount of money ($2183) by the other amount ($37) which is 59

5 0
3 years ago
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