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xxMikexx [17]
3 years ago
15

A shopkeeper allows discount of 12% on the marked price of a certain article and makes a profit of 10%. if the article costs the

shopkeeper Rs.420. what price should be the marked price?​
Mathematics
1 answer:
romanna [79]3 years ago
8 0

Answer:

rs288

Step-by-step explanation:

MP of the article = 210×120100×10087.5= Rs.288210×120100×10087.5= Rs.288

Part of solved Discount questions and answers : >> Aptitude >> Discount

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M Enter the number that belongs in<br> the green box
gladu [14]

Answer:

m<C = 180 - (90+32)

= 180 - 122

= 58°

4 0
2 years ago
The life of a red bulb used in a traffic signal can be modeled using an exponential distribution with an average life of 24 mont
BartSMP [9]

Answer:

See steps below

Step-by-step explanation:

Let X be the random variable that measures the lifespan of a bulb.

If the random variable X is exponentially distributed and X has an average value of 24 month, then its probability density function is

\bf f(x)=\frac{1}{24}e^{-x/24}\;(x\geq 0)

and its cumulative distribution function (CDF) is

\bf P(X\leq t)=\int_{0}^{t} f(x)dx=1-e^{-t/24}

• What is probability that the red bulb will need to be replaced at the first inspection?

The probability that the bulb fails the first year is

\bf P(X\leq 12)=1-e^{-12/24}=1-e^{-0.5}=0.39347

• If the bulb is in good condition at the end of 18 months, what is the probability that the bulb will be in good condition at the end of 24 months?

Let A and B be the events,

A = “The bulb will last at least 24 months”

B = “The bulb will last at least 18 months”

We want to find P(A | B).

By definition P(A | B) = P(A∩B)P(B)

but B⊂A, so  A∩B = B and  

\bf P(A | B) = P(B)P(B) = (P(B))^2

We have  

\bf P(B)=P(X>18)=1-P(X\leq 18)=1-(1-e^{-18/24})=e^{-3/4}=0.47237

hence,

\bf P(A | B)=(P(B))^2=(0.47237)^2=0.22313

• If the signal has six red bulbs, what is the probability that at least one of them needs replacement at the first inspection? Assume distribution of lifetime of each bulb is independent

If the distribution of lifetime of each bulb is independent, then we have here a binomial distribution of six trials with probability of “success” (one bulb needs replacement at the first inspection) p = 0.39347

Now the probability that exactly k bulbs need replacement is

\bf \binom{6}{k}(0.39347)^k(1-0.39347)^{6-k}

<em>Probability that at least one of them needs replacement at the first inspection = 1- probability that none of them needs replacement at the first inspection. </em>

This means that,

<em>Probability that at least one of them needs replacement at the first inspection =  </em>

\bf 1-\binom{6}{0}(0.39347)^0(1-0.39347)^{6}=1-(0.60653)^6=0.95021

5 0
3 years ago
How will fit in the box
r-ruslan [8.4K]
12 will fit in the box for number 1

6 0
3 years ago
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A sleep time of 15.9 hours per day for a newborn baby is at the 10h percentile of the distribution of sleep times for all newbor
Nuetrik [128]

Answer:

option (D) 16.5

Step-by-step explanation:

Data provided:

Sleep time of , X = 15.9 is at the 10th percentile

Standard deviation, σ = 0.5  hour

also,

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or

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Now,

\\Z=\frac{X-\mu}{\sigma}

or

X = μ + ( σ × Z )

or

μ = X - ( σ × Z )

on substituting the respective values, we get

μ = 15.9 - 0.5 × (-1.28)

or

μ = 16.5

Hence, the correct answer is option (D) 16.5

8 0
3 years ago
How do I solve x-11/2&lt;-7
Mekhanik [1.2K]

Answer:

x < -3

Step-by-step explanation:

1. multiply both sides by 2

2. isolate the x by adding 11 to both sides

3. simplify

(i have included an attachment showing the work for clarification)

4 0
3 years ago
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