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11111nata11111 [884]
1 year ago
7

Is 7 rational or irrational

Mathematics
2 answers:
faust18 [17]1 year ago
8 0

Answer: rational

Step-by-step explanation:

Roman55 [17]1 year ago
3 0

Answer:

It's rational

Step-by-step explanation:

It's an integer and all integers are rational

You might be interested in
Write an inequality to compare the integers -5 and -6
Dovator [93]
-5 > -6

I think this is what it would be because -5 is greater than (>) -6

Hope this is right and it helps :)
5 0
1 year ago
Rewrite in simplest radical form x 5/6 x 1/6
Natasha_Volkova [10]

Answer:

x^{\frac{5}{6}}/x^{\frac{1}{6}} = \sqrt[3]{x^2}

Step-by-step explanation:

Given

x^{\frac{5}{6}}/x^{\frac{1}{6}}

Required

Rewrite in simplest radical form

Using laws of indices:

a^m/a^n = a^{m-n}

This implies that

x^{\frac{5}{6}}/x^{\frac{1}{6}} = x^{\frac{5}{6} - \frac{1}{6}}

Solve Exponents

x^{\frac{5}{6}}/x^{\frac{1}{6}} = x^{\frac{5 - 1}{6} }

x^{\frac{5}{6}}/x^{\frac{1}{6}} = x^{\frac{4}{6} }

Simplify exponent to lowest fraction

x^{\frac{5}{6}}/x^{\frac{1}{6}} = x^{\frac{2}{3} }

Using laws of indices:

a^{\frac{m}{n}} = \sqrt[n]{a^m}

This implies that

x^{\frac{5}{6}}/x^{\frac{1}{6}} = \sqrt[3]{x^2}

This is as far as the expression can be simplified

3 0
1 year ago
Suppose that a manager is interested in estimating the average amount of money customers spend in her store. After sampling 36 t
musickatia [10]

Answer:

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

The 90% confidence interval for this case would be (38.01, 44.29) and is given.

The best interpretation for this case would be: We are 90% confident that the true average is between $ 38.01 and $ 44.29 .

And the best option would be:

The store manager is 90% confident that the average amount spent by all customers is between S38.01 and $44.29

Step-by-step explanation:

Assuming this complete question: Which statement gives a valid interpretation of the interval?

The store manager is 90% confident that the average amount spent by the 36 sampled customers is between S38.01 and $44.29.

There is a 90% chance that the mean amount spent by all customers is between S38.01 and $44.29.

There is a 90% chance that a randomly selected customer will spend between S38.01 and $44.29.

The store manager is 90% confident that the average amount spent by all customers is between S38.01 and $44.29

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

The 90% confidence interval for this case would be (38.01, 44.29) and is given.

The best interpretation for this case would be: We are 90% confident that the true average is between $ 38.01 and $ 44.29 .

And the best option would be:

The store manager is 90% confident that the average amount spent by all customers is between S38.01 and $44.29

8 0
2 years ago
Weiming, Farhan and Sam had some comic books.
mixas84 [53]

Answer:

Farhan had 9 comic books

Sam had 51 comic books and lived in his Mother's basement

Step-by-step explanation:

W + F = 36

W = 3F

3F + F = 36

F = 9

W = 27

S + F = 60

S + 9 = 60

S = 51

5 0
1 year ago
Suppose 50 percent of the customers at Pizza Palooza order a square pizza, 70 percent order a soft drink, and 35 percent order b
Firdavs [7]

Answer:

Ordering a soft drink is independent of ordering a square pizza.

Step-by-step explanation:

20% more customers order a soft drink than pizza, therefore they cannot be intertwined.

Given: P(A)=0.5 & P(B)=.7

P(A∩B) =  P(A) × P(B)

=  0.5 × .7

=  0.35

P(A∪B) =  P(A) + P(B) - P(A∩B)

=  0.5 + .7 - 0.35

=  0.85

P(AΔB) =  P(A) + P(B) - 2P(A∩B)

=  0.5 + .7 - 2×0.35

=  0.5

P(A') =  1 - P(A)

=  1 - 0.5

=  0.5

P(B') =  1 - P(B)

=  1 - .7

=  0.3

P((A∪B)') =  1 - P(A∪B)

=  1 - 0.85

=  0.15

7 0
1 year ago
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