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Triss [41]
3 years ago
13

How to do this problem 2 + 1/8 - 1 1/3

Mathematics
1 answer:
Svetllana [295]3 years ago
6 0
17/8 -11/3
51/24-88/24
-37/24 simplified is -1 13/24
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Events A and B are disjointed.
Mama L [17]

<u>7/11</u> is the final answer.

use the formula for disjointed events: P(A or B) = P(A) + P(B)

=================================

P(A or B) = P(A) + P(B)

P(A or B) = 4/11 + 3/11

P(A or B) = 7/11

8 0
2 years ago
A recent survey by the American Automobile Association showed that a family of two adults and two children on vacation in the Un
masha68 [24]

Answer: We are given:

\mu=247,\sigma=60

We need to find the z scores for the following vacation expense amounts:

$197, $277, $310

We know that z score formula is:

z=\frac{x-\mu}{\sigma}

When x = 197, the z score is:

z=\frac{197-247}{60}

        =\frac{-50}{60}

        =-0.83

When x = 277, the z score is:

z=\frac{277-247}{60}

        =\frac{30}{60}

        =0.5

When x = 310, the z score is:

z=\frac{310-247}{60}

        =\frac{63}{60}

        =1.05

Therefore, the z scores for the vacation expense amounts $197 per day, $277 per day, and $310 per day are -0.83, 0.5 and 1.05 respectively


8 0
3 years ago
Read 2 more answers
HELP ME ASAP PLSSSSSSSS
MrRa [10]

Answer:

This is just like congreunt and similarity

8 0
3 years ago
A ray is a defined term because it
vodomira [7]
'is described using an undefined term'

because for defined terms,
<span>"definitions are formed using known words or terms to describe a new word."</span>



4 0
3 years ago
The following formula for the sum of the cubes of the first n integers is proved in Appendix E. Use it to evaluate the limit in
Marina86 [1]

Answer:

\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2})

And when we apply the limit we got that:

\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2}) =1

Step-by-step explanation:

Assuming this complete problem: "The following formula for the sum of the cubes of the first n integers is proved in Appendix E. Use it to evaluate the limit . 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2"

We have the following formula in order to find the sum of cubes:

\lim_{n\to\infty} \sum_{n=1}^{\infty} i^3

We can express this formula like this:

\lim_{n\to\infty} \sum_{n=1}^{\infty}i^3 =\lim_{n\to\infty} [\frac{n(n+1)}{2}]^2

And using this property we need to proof that: 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2

\lim_{n\to\infty} [\frac{n(n+1)}{2}]^2

If we operate and we take out the 1/4 as a factor we got this:

\lim_{n\to\infty} \frac{n^2(n+1)^2}{n^4}

We can cancel n^2 and we got

\lim_{n\to\infty} \frac{(n+1)^2}{n^2}

We can reorder the terms like this:

\lim_{n\to\infty} (\frac{n+1}{n})^2

We can do some algebra and we got:

\lim_{n\to\infty} (1+\frac{1}{n})^2

We can solve the square and we got:

\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2})

And when we apply the limit we got that:

\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2}) =1

3 0
3 years ago
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