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ehidna [41]
3 years ago
6

Hey:) can you please help me posted picture of question

Mathematics
2 answers:
Keith_Richards [23]3 years ago
4 0
We have the following expression:
 (8x5 + 9) (8x3 + 9)
 Rewriting we have:
 64x ^ (5 + 3) + 72x ^ 5 + 72x ^ 3 + 81
 Rewriting we have:
 64x ^ 8 + 72x ^ 5 + 72x ^ 3 + 81
 Answer:
 
The solution for this case is given by:
 
64x ^ 8 + 72x ^ 5 + 72x ^ 3 + 81
elixir [45]3 years ago
3 0
Option B gives the correct answer.

The step by step solution using the FOIL method of multiplication is given below:

(8 x^{5}+9)(8 x^{3}+9) \\  \\ 
  =8 x^{5} (8 x^{3}+9) +9(8 x^{3}+9) \\  \\ 
=64 x^{8} +72 x^{5}+72 x^{3}+81
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Answer:

BGR=2

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Slope: -5, y-intercept:3
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Because of safety considerations, in May 2003 the Federal Aviation Administration (FAA) changed its guidelines for how small com
Rasek [7]

Answer:

a) The 95% confidence interval for the mean summer weight (including carry-on luggage) of Frontier Airlines passengers is between 179 and 187 pounds. This means that we are 95% sure that the mean summer weight of all Frontier Airlines passengers is between these two values.

b)

The 95% confidence interval for the mean winter weight (including carry-on luggage) of Frontier Airlines passengers is between 185.4 pounds and 194.6 pounds. This means that we are 95% sure that the mean winter weight of all Frontier Airlines passengers is between these two values.

c) They are respected, as the upper bound of both intervals is below the new FAA recommendations.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve these questions.

Question a:

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 = 100 - 1 = 99

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 99 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.95}{2} = 0.975. So we have T = 1.9842

The margin of error is:

M = T\frac{s}{\sqrt{n}} = 1.9842\frac{20}{\sqrt{100}} = 4

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 183 - 4 = 179 pounds.

The upper end of the interval is the sample mean added to M. So it is 183 + 4 = 187 pounds.

The 95% confidence interval for the mean summer weight (including carry-on luggage) of Frontier Airlines passengers is between 179 and 187 pounds. This means that we are 95% sure that the mean summer weight of all Frontier Airlines passengers is between these two values.

Question b:

Critical value is the same(same sample size and confidence level).

The margin of error is:

M = T\frac{s}{\sqrt{n}} = 1.9842\frac{23}{\sqrt{100}} = 4.6

The lower end of the interval is the sample mean subtracted by M. So it is 190 - 4.6 = 185.4 pounds.

The upper end of the interval is the sample mean added to M. So it is 190 + 4.6 = 194.6 pounds.

The 95% confidence interval for the mean winter weight (including carry-on luggage) of Frontier Airlines passengers is between 185.4 pounds and 194.6 pounds. This means that we are 95% sure that the mean winter weight of all Frontier Airlines passengers is between these two values.

c. The new FAA recommendations are 190 pounds for summer and 195 pounds for winter. Comment on these recommendations in light of the confidence interval estimates from Parts (a) and (b).

They are respected, as the upper bound of both intervals is below the new FAA recommendations.

7 0
3 years ago
If I have a garden that's ten yards and I want to plant a plant that requires 2 yards how many plants can I plant
Damm [24]

Pretty sure it's 5. 5X2=10. Sooo

6 0
3 years ago
HELP ME PLLLLZ dis is PEMDAS
hram777 [196]

Answer:

Give Branliest PLz

Answer is 15

Step-by-step explanation:

Half of 40 is 20 cause 40/2=20

Five less than 20 is 20-5 which= 15

So ur answer is 15

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