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3 years ago

What is the fraction for 1261.4523

Mathematics

2 answers:

pickupchik [31]3 years ago

7 0

The answer might be this so try 12614523/10000

ioda3 years ago

4 0

I got 12614523/10000

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a rectangle has a width of 4 cm and a length of 8cm is scaled by a factor of 6.what are the side lengths of the scaled copy

vichka [17]

Answer:

$\huge\boxed{\text{Width: } 24 \ \ \ \text{Length: } 32}$

Step-by-step explanation:

When we scale an object by a scale factor, the side lengths become the scale factor times bigger than the previous sides. Therefore, with a scale factor of 6, the side lengths would be multiplied by 6.

$4 \cdot 6 = 24\\\\4 \cdot 8 = 32$

Hope this helped!

6 0

3 years ago

Read 2 more answers

Need help. asap. also need an explanation for the problem. :)

Maurinko [17]

Answer:

124858

Step-by-step explanation:

The first 4 digits are simple, you multiply the first digit of the equation by the 2nd digit and then for the other 2 you multiply the first digit of the equation by the 3rd digit.

6 + 2 + 8

6 * 2 = 12

6 * 8 = 48

Then the last 2 digits are the sum of the products of the 1 and 2 and 1 and 3 and subtract it by the 2

12 + 48 = 60

60 - 2 = 58

Put them together

7 0

2 years ago

. If m 4 = 105, then what is m 8?

polet [3.4K]

I believe the answer is 210 because 1m= 26.25 and that x's 8= 210

7 0

3 years ago

Read 2 more answers

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

If x=1 and y=2, what is the value of the expression 2x^2-3xy

hjlf

2x² - 3xy
2(1)² - 3(1)(2)
2(1) - 3(2)
2 - 6
-4

7 0

3 years ago