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vaieri [72.5K]
3 years ago
8

If the simple interest on ​$2000 for 10 years is ​$​,1000 then what is the interest​ rate?

Mathematics
1 answer:
astra-53 [7]3 years ago
3 0

\bf \underline{Given :}

\sf{•  \: Principal  \: (P) =  \$  \: 2000}

\sf{•  \: Simple \:  Interest  \: (I) =  \$  \: 1000}

\sf{•  \: Time  \: (T) = 10  \: years}

\sf{• \:  Interest  \: Rate \:  (R) =  \: ?}

\bf \underline{Solution:-}

\sf {We  \: know \:  that, }

\bf \red{\bigstar{ \: I = PRT}}

\sf{⟹1000 = 2000 \times  \frac{R}{100}  \times 10 }

\sf{⟹1000 = \frac{R}{100}   \times 20000}

\sf{⟹ \frac{R}{100}  =  \frac{1000}{20000} }

\sf{⟹R =  \frac{1}{20} \times 100 }

\sf⟹ R  = 5

\pink{\bf{Hence,  \: the  \: interest  \: rate \:  is  \: 5   \: \%. }}

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Some parts of California are particularly earthquake-prone. Suppose that in one metropolitan area, 33% of all homeowners are ins
Lina20 [59]

Answer:

(a) The probability mass function of <em>X</em> is:

P(X=x)={4\choose x}\ (0.33)^{x}\ (1-0.33)^{4-x};\ x=0,1,2,3...

(b) The most likely value for <em>X</em> is 1.32.

(c) The probability that at least two of the four selected have earthquake insurance is 0.4015.

Step-by-step explanation:

The random variable <em>X</em> is defined as the number among the four homeowners  who have earthquake insurance.

The probability that a homeowner has earthquake insurance is, <em>p</em> = 0.33.

The random sample of homeowners selected is, <em>n</em> = 4.

The event of a homeowner having an earthquake insurance is independent of the other three homeowners.

(a)

All the statements above clearly indicate that the random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 4 and <em>p</em> = 0.33.

The probability mass function of <em>X</em> is:

P(X=x)={4\choose x}\ (0.33)^{x}\ (1-0.33)^{4-x};\ x=0,1,2,3...

(b)

The most likely value of a random variable is the expected value.

The expected value of a Binomial random variable is:

E(X)=np

Compute the expected value of <em>X</em> as follows:

E(X)=np

         =4\times 0.33\\=1.32

Thus, the most likely value for <em>X</em> is 1.32.

(c)

Compute the probability that at least two of the four selected have earthquake insurance as follows:

P (X ≥ 2) = 1 - P (X < 2)

              = 1 - P (X = 0) - P (X = 1)

              =1-{4\choose 0}\ (0.33)^{0}\ (1-0.33)^{4-0}-{4\choose 1}\ (0.33)^{1}\ (1-0.33)^{4-1}\\\\=1-0.20151121-0.39700716\\\\=0.40148163\\\\\approx 0.4015

Thus, the probability that at least two of the four selected have earthquake insurance is 0.4015.

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The solution is k=19/5
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