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likoan [24]
3 years ago
12

What is your estimate for the number of airline flights in a year?

Mathematics
1 answer:
Firdavs [7]3 years ago
8 0
The answer is between 45,000,000-65,000,000
You might be interested in
-1/3 -(-5/12) which shows an equivalent addition expression
dem82 [27]

Answer:

-1/3 - (-5/12) = 1/12.

Step-by-step explanation:

Simpler equation:

6/12 + (-5/12) = 1/12

Or This:

6/12 - 5/12 = 1/12.

hope this helped.

5 0
3 years ago
Find a third-degree polynomial function with real coefficients and with zeros at -4 and -3 5i
Charra [1.4K]

If -3+5i is a solution, then by the conjugate root theorem, -3-5i is also a solution. We find the polynomial by multiplying together the factors. If x = -4, then x + 4 is the factor. If x = -3+5i, then (x-(-3+5i)) is a factor, and so is (x-(-3-5i)). Simplifying those down gives us as the first factor as (x+3-5i) and the second as (x+3+5i). We can FOIL those 2 together to get their product, and then FOIL in x+4. FOILing the 2 complex factors together gives us x^2+3x+15ix+3x+9+15i-15ix-15i-25i^2. If we combine like terms and cross out things that cancel it's much easier than what it looks like there! It simplifies down to x^2+6x-25i^2. Since i^2 = -1, it simplifies further to x^2+6x-25(-1) and finally, to x^2+6x+25. Now we will FOIL in x+4. (x^2+6x+25)(x+4)=x^3+6x^2+25x+4x^2+24x+100. Our final simplified third degree polynomial is x^3+10x^2+49x+100

3 0
3 years ago
last year 950 people attended a towns annual parade. this year 1,520 people attended what was the percent increase in attendance
inessss [21]
<span>To calculate the percentage increase: First: work out the difference (increase) between the two numbers you are comparing. Then: divide the increase by the original number and multiply the answer by 100.
1520-950=570
570÷950= 0.6 × 100=60%


</span>
7 0
3 years ago
Evaluate<img src="https://tex.z-dn.net/?f=10%5E%7B6%7D" id="TexFormula1" title="10^{6}" alt="10^{6}" align="absmiddle" class="la
katovenus [111]
121,000,000,000
this is because:
10^6= 1,000,000
10^3= 1,000
121^1= 121
3 0
3 years ago
A random sample of 16 one-kilogram sugar packets is obtained and the actual weights of the packets are measured. The sample mean
elena55 [62]

Answer:

The 99% two-sided confidence interval for the average sugar packet weight is between 0.882 kg and 1.224 kg.

Step-by-step explanation:

We are in posession of the sample's standard deviation, so we use the student's t-distribution to find the confidence interval.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 16 - 1 = 15

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 35 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.99}{2} = 0.905([tex]t_{995}). So we have T = 2.9467

The margin of error is:

M = T*s = 2.9467*0.058 = 0.171

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 1.053 - 0.171 = 0.882kg

The upper end of the interval is the sample mean added to M. So it is 1.053 + 0.171 = 1.224 kg.

The 99% two-sided confidence interval for the average sugar packet weight is between 0.882 kg and 1.224 kg.

3 0
3 years ago
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