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laila [671]
2 years ago
6

Marked price = Rs 650 and Discount percentage = 2.5%. find selling price​

Mathematics
1 answer:
levacccp [35]2 years ago
3 0

Answer:

The selling price is <u>Rs.633.75</u>.

<h3><u>Solution</u> :</h3>

Here we have given that :

  • \pink\star Marked Price = Rs.650
  • \pink\star Discount % = 2.5 %

We need to find the selling price.

Firstly, finding the discount by substituting the values in the formula :

{\longrightarrow{\sf{Discount = Marked \:  Price \times  Discount \%}}}

{\longrightarrow{\sf{Discount =650\times 2.5\%}}}

{\longrightarrow{\sf{Discount =650\times  \dfrac{2.5}{100}}}}

{\longrightarrow{\sf{Discount = \dfrac{650 \times 2.5}{100}}}}

{\longrightarrow{\sf{Discount = \dfrac{1625}{100}}}}

{\longrightarrow{\sf{Discount = 16.25}}}

Hence, the discount is Rs.16.25.

\rule{200}2

Now, finding the selling price by substituting the values in the formula :

{\longrightarrow{\sf{Selling \:  Price = Marked \:  Price -  Discount}}}

{\longrightarrow{\sf{Selling \:  Price = 650 - 16.25}}}

{\longrightarrow{\sf{Selling \:  Price =633.75}}}

Hence, the selling price is Rs.633.75.

\rule{300}{2.5}

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a) P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211

b) P(X> 2)=1-P(X\leq 2)=1-[0.0211+0.0995+0.211]=0.668

c) P(2 \leq x \leq 5)=0.211+0.264+0.218+0.123=0.816

Step-by-step explanation:

1) 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".

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

X \sim Binom(n=10, p=0.32)

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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Part a

P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211

Part b

P(X> 2)=1-P(X\leq 2)=1-[P(X=0)+P(X=1)+P(X=2)]

P(X=0)=(10C0)(0.32)^0 (1-0.32)^{10-0}=0.0211  

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P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211

P(X> 2)=1-P(X\leq 2)=1-[0.0211+0.0995+0.211]=0.668

Part c

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P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211

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