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aleksandr82 [10.1K]
3 years ago
10

A grocery store sells tangerines in 2/5 kg. bags. A customer bought 4 kg of tangerines for a school party. How many bags did he

buy? ASAP
Mathematics
2 answers:
zhenek [66]3 years ago
6 0

Answer:

There are 10 in 4kg

Step-by-step explanation:

Given

1\ Tangerine = \frac{2}{5}kg

Required

Determine the number in 4kg

Represent this with x.

To calculate this, we make use of the following:

x * \frac{2}{5}kg = 4kg

Multiply through by \frac{5}{2}

x * \frac{2}{5}kg *\frac{5}{2}= 4kg *\frac{5}{2}

x  = 4*\frac{5}{2}

x  =10

<em>Hence, there are 10 in 4kg</em>

grigory [225]3 years ago
4 0

Answer: the answer in that there are 10 in 4kg

Step-by-step explanation:

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Answer:

a) P(X=0)=(2C0)(0.11)^0 (1-0.11)^{2-0}=0.7921  

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And then we have our probability distribution like this:

X     |       0    |      1     |     2    |

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Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=20, p=0.48)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

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Solution for the problem

Part a

On this case since we select a sample size of n =2 we have the following values for the number of left handed X=0,1,2. We can find the probabilities for each case since we know that p=0.11.

P(X=0)=(2C0)(0.11)^0 (1-0.11)^{2-0}=0.7921  

P(X=1)=(2C1)(0.11)^1 (1-0.11)^{2-1}=0.1958  

P(X=2)=(2C2)(0.11)^2 (1-0.11)^{2-2}=0.0121  

And then we have our probability distribution like this:

X     |       0    |      1     |     2    |

P(X) |  0.7921 | 0.1958 | 0.0121|

Part b

For this case we want this probability:

P(X \geq 1) = P(X=1)+P(X=2) = 0.1958+0.0121=0.2079

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