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gregori [183]
3 years ago
15

I need help dudessss

Mathematics
1 answer:
matrenka [14]3 years ago
6 0

Answer:

3.618

Step-by-step explanation:

10^4=$10,000

36180/10000= 3.618

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2. Find the perimeter of a rectangle of<br> Length 13m with an area of 65m
I am Lyosha [343]

Answer:

Perimeter = 36m

Step-by-step explanation:

Area = l x w

65m = 13m x w

5m = w

Perimeter = 2(l + w)

Perimeter = 2(13m + 5m)

Perimeter = 36m

3 0
3 years ago
Read 2 more answers
I need question 3 only
dsp73

Answer:

  • table: 14, 16, 18
  • equation: P = 2n +12

Step-by-step explanation:

Perimeter values will be ...

rectangles . . . perimeter

  1 . . . 14

  2 . . . 16

  3 . . . 18

__

The perimeter of a figure is twice the sum of the length and width. Here, the length is a constant 6. The width is n, the number of rectangles. So, the perimeter is ...

  P = 2(6 +n) = 12 +2n

Your equation is ...

  P = 2n +12 . . . . . . . . perimeter P of figure with n rectangles.

_____

<em>Additional comment</em>

You may be expected to write the equation using y and x for the perimeter and the number of rectangles. That would be ...

  y = 2x +12 . . . . . . . . . perimeter y of figure with x rectangles

4 0
3 years ago
Find the value of x.
Paha777 [63]

Answer:

125 minus 81 will give me 34°

5 0
2 years ago
Read 2 more answers
Which formula can be used to find the nth term of a geometric sequence where the fifth term is 1/16 and the common ratio is 1/4
stepladder [879]
A₅ = 1/16  and r= 1/4

Let see how to build up this formula that is going to give that term of rank n

1st term =a₁ = To be calculated

1st a₁   = a₁  x  r°
2nd a₂  = a₁  x  r¹
3rd a₃  =  a₁  x  r²
4th a₄   = a₁  x  r³
5th a₅   = a₁  x  r⁴
.......................
.......................
nth : a(n) = a₁  x r⁽ⁿ-¹)
Note when that the subscript of a is the same as the exponent mines 1

We know the ratio r =1/4 & the fifth term, a₅ =1/16 (given). Now let's apply the formula to calculate the unknown a₁.

a(n) = a₁  x r⁽ⁿ-¹) ==>a₅ = a₁ x (1/4)⁽⁵⁺¹⁾ ===> 1/16 = a₁ x (1/4)⁴
1/167 = a₁ (1/256) ==> a₁ =16 & the formula becomes
a₅ = 16(1/4)⁴

4 0
3 years ago
Read 2 more answers
A shop sells a particular of video recorder. Assuming that the weekly demand for the video recorder is a Poisson variable with t
julia-pushkina [17]

Answer:

a) 0.5768 = 57.68% probability that the shop sells at least 3 in a week.

b) 0.988 = 98.8% probability that the shop sells at most 7 in a week.

c) 0.0104 = 1.04% probability that the shop sells more than 20 in a month.

Step-by-step explanation:

For questions a and b, the Poisson distribution is used, while for question c, the normal approximation is used.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}

In which

x is the number of successes

e = 2.71828 is the Euler number

\lambda is the mean in the given interval.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

The Poisson distribution can be approximated to the normal with \mu = \lambda, \sigma = \sqrt{\lambda}, if \lambda>10.

Poisson variable with the mean 3

This means that \lambda= 3.

(a) At least 3 in a week.

This is P(X \geq 3). So

P(X \geq 3) = 1 - P(X < 3)

In which:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

Then

P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}

P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498

P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494

P(X = 2) = \frac{e^{-3}*3^{2}}{(2)!} = 0.2240

So

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0498 + 0.1494 + 0.2240 = 0.4232

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 1 - 0.4232 = 0.5768

0.5768 = 57.68% probability that the shop sells at least 3 in a week.

(b) At most 7 in a week.

This is:

P(X \leq 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)

In which

P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}

P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498

P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494

P(X = 2) = \frac{e^{-3}*3^{2}}{(2)!} = 0.2240

P(X = 3) = \frac{e^{-3}*3^{3}}{(3)!} = 0.2240

P(X = 4) = \frac{e^{-3}*3^{4}}{(4)!} = 0.1680

P(X = 5) = \frac{e^{-3}*3^{5}}{(5)!} = 0.1008

P(X = 6) = \frac{e^{-3}*3^{6}}{(6)!} = 0.0504

P(X = 7) = \frac{e^{-3}*3^{7}}{(7)!} = 0.0216

Then

P(X \leq 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 0.0498 + 0.1494 + 0.2240 + 0.2240 + 0.1680 + 0.1008 + 0.0504 + 0.0216 = 0.988

0.988 = 98.8% probability that the shop sells at most 7 in a week.

(c) More than 20 in a month (4 weeks).

4 weeks, so:

\mu = \lambda = 4(3) = 12

\sigma = \sqrt{\lambda} = \sqrt{12}

The probability, using continuity correction, is P(X > 20 + 0.5) = P(X > 20.5), which is 1 subtracted by the p-value of Z when X = 20.5.

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

Z = \frac{20 - 12}{\sqrt{12}}

Z = 2.31

Z = 2.31 has a p-value of 0.9896.

1 - 0.9896 = 0.0104

0.0104 = 1.04% probability that the shop sells more than 20 in a month.

5 0
2 years ago
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