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3 years ago

Need help with a math question PLEASE HELP

Mathematics

1 answer:

navik [9.2K]3 years ago

4 0

Answer:

(-1, -3)

Step-by-step explanation:

We suppose your notation means you want to reflect given point P across the horizontal line y=1.

The x-coordinate will remain the same.

The new y-coordinate will be such that y=1 is the midpoint between the original and its reflection:

(5 + y)/2 = 1

5 + y = 2 . . . . multiply by 2

y = 2 -5 = -3 . . . subtract 5

The reflected point is (-1, -3).

___

The same sort of math applies whenever you have a midpoint and want to find the other end point. Double the midpoint value and subtract the end point you have in order to find the other end point.

You might be interested in

Find the circumference of a circle where r=1.5

il63 [147K]

Answer:

there are three different types of answers to this and all of them have to do with the perspective of pi

1) if you don't want to see the decimals and you want to leave pi alone your answer would be this

3pi

2) if you want to solve for pi it would be

9.43 respectively

3) if you want pi to just be narrowed down to 3 digits 3.14 your answer would be

9.42

Step-by-step explanation:

formula

C=2*pi*R

plug in

2*pi*1.5

5 0

3 years ago

Read 2 more answers

Help fast plsssssssssss

bearhunter [10]

-9-(-12)=-9+12=3
-9-12=-9+(-12)=-21

4 0

3 years ago

Read 2 more answers

Give the solution set for the inequality 7x < 7(x - 2) in interval notation.

eduard

ANSWER:

The solution set for the inequality 7x < 7(x - 2) is null set $\varnothing$

SOLUTION:

Given, inequality expression is 7x < 7 × (x – 2)

We have to give the solution set for above inequality expression in the interval notation form.

Now, let us solve the inequality expression for x.

Then, 7x < 7 × (x – 2)

7x < 7 × x – 2 × 7

7x < 7x – 14

7x – (7x – 14) < 0

7x – 7x + 14 < 0

0 + 14 < 0

14 < 0

Which is false, so there exists no solution for x which can satisfy the given equation.

So, the interval solution for given inequality will be null set

Hence, the solution set is $\varnothing$

3 0

3 years ago

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.

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.

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

The fifth term of a geometric series is 405 and the sixth term is 1215. Find the sum of the first nine terms. Don't know how to

Margarita [4]

Since the sixth term is 1215 and the 5th term is 405, we know each term is 3 times more than the last.
Let 'n' represent the first term, then the sum of the first 9 terms can be written as:
n+3n+9n+27n+81n+243n+729n+2187n+6561n
which sinplifies to 9841n
since each term is 3 times the last, the first term must be 405 divided by 3, 4 times in total, which is 5
Now subsitute 'n' for 5, and we get 9841(5)
which is 49205 which is the answer.

Hope this helps

5 0

3 years ago