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choli [55]
3 years ago
14

30 POINTS!!!!1 PLEASE HELP!!!!!

Mathematics
1 answer:
Ahat [919]3 years ago
4 0

Answer:

From what I could tell, the answer is 16.

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Which equation represents a function that is nonlinear?
tekilochka [14]

Answer:

I dont get itt

Step-by-step explanation:

8 0
3 years ago
Five consecutive positive integers are chosen at random. If the average of the five integers is odd, what is the remainder when
kupik [55]

Answer:

The remainder would be 1 or 3

Step-by-step explanation:

Consider the 5 consecutive positive integers are,

x,  x + 1,  x + 2,  x + 3,  x + 4,

Average =\frac{\text{Sum of all observation}}{\text{Number of observations}}

Since, the average of these 5 numbers = \frac{x+x+1+x+2+x+3+x+4}{5}

=\frac{5x+10}{5}

= x + 2

If x + 2 = odd,

⇒ x = odd - 2 = odd - even = odd

⇒ x + 4 = odd + even = odd

∵ an odd number is represented by '2n + 1'

Where, n = 0, 1, 2, 3, ........

Now, 2n + 1 = 1( mod 4) if n = even

While, 2n + 1 = 3( mod 4) if n = odd,

Hence, when the largest of the five integers is divided by 4 remainder would be 1 or 3.

4 0
2 years ago
An airline experiences a no-show rate of 6%. What is the maximum number of reservations that it could accept for a flight with a
LekaFEV [45]

Let Xb be the number of reservations that are accommodated. Xb has the binomial distribution with n trials and success probability p = 0.94

In general, if X has the binomial distribution with n trials and a success probability of p then
P[Xb = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)
for values of x = 0, 1, 2, ..., n
P[Xb = x] = 0 for any other value of x.

To use the normal approximation to the binomial you must first validate that you have more than 10 expected successes and 10 expected failures. In other words, you need to have n * p > 10 and n * (1-p) > 10.

Some authors will say you only need 5 expected successes and 5 expected failures to use this approximation. If you are working towards the center of the distribution then this condition should be sufficient. However, the approximations in the tails of the distribution will be weaker espeically if the success probability is low or high. Using 10 expected successes and 10 expected failures is a more conservative approach but will allow for better approximations especially when p is small or p is large.

If Xb ~ Binomial(n, p) then we can approximate probabilities using the normal distribution where Xn is normal with mean μ = n * p, variance σ² = n * p * (1-p), and standard deviation σ

I have noted two different notations for the Normal distribution, one using the variance and one using the standard deviation. In most textbooks and in most of the literature, the parameters used to denote the Normal distribution are the mean and the variance. In most software programs, the standard notation is to use the mean and the standard deviation.

The probabilities are approximated using a continuity correction. We need to use a continuity correction because we are estimating discrete probabilities with a continuous distribution. The best way to make sure you use the correct continuity correction is to draw out a small histogram of the binomial distribution and shade in the values you need. The continuity correction accounts for the area of the boxes that would be missing or would be extra under the normal curve.

P( Xb < x) ≈ P( Xn < (x - 0.5) )
P( Xb > x) ≈ P( Xn > (x + 0.5) )
P( Xb ≤ x) ≈ P( Xn ≤ (x + 0.5) )
P( Xb ≥ x) ≈ P( Xn ≥ (x - 0.5) )
P( Xb = x) ≈ P( (x - 0.5) < Xn < (x + 0.5) )
P( a ≤ Xb ≤ b ) ≈ P( (a - 0.5) < Xn < (b + 0.5) )
P( a ≤ Xb < b ) ≈ P( (a - 0.5) < Xn < (b - 0.5) )
P( a < Xb ≤ b ) ≈ P( (a + 0.5) < Xn < (b + 0.5) )
P( a < Xb < b ) ≈ P( (a + 0.5) < Xn < (b - 0.5) )

In the work that follows X has the binomial distribution, Xn has the normal distribution and Z has the standard normal distribution.

Remember that for any normal random variable Xn, you can transform it into standard units via: Z = (Xn - μ ) / σ

In this question Xn ~ Normal(μ = 0.94 , σ = sqrt(0.94 * n * 0.06) )

Find n such that:

P(Xb ≤ 160) ≥ 0.95

approximate using the Normal distribution
P(Xn ≤ 160.5) ≥ 0.95

P( Z ≤ (160.5 - 0.94 * n) / sqrt(0.94 * n * 0.06)) ≥ 0.95

P( Z < 1.96 ) ≥ 0.95

so solve this equation for n

(160.5 - 0.94 * n) / sqrt(0.94 * n * 0.06) = 1.96

n = 164.396

n must be integer valued so take the ceiling and you have:

n = 165.

The air line can sell 165 tickets for the flight and accommodate all reservates at least 95% of the time

If you can understand that...

7 0
3 years ago
Increase £84 by 23%<br> TIA X
zmey [24]

Answer:

103.32

Step-by-step explanation:

May way to answer a question like this

84(1+23/100)

Lets break it down:

Step 1.     23/100=0.23

Step2.     0.23+1=1.23

Step3.      1.23 x 84=103.32

I hope this helps You

4 0
2 years ago
To find the volume of a rectangular prism you multiply the 3 dimensions, length x width x height. What combination of numbers wi
aleksklad [387]
1, 1 and 17. Since 17 is a prime number, there isn’t exactly numbers that are multiplied by each other that give 17.
8 0
1 year ago
Read 2 more answers
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