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2 years ago

So alot of the text got cut off so im going to re wright the problem

Mathematics

1 answer:

gregori [183]2 years ago

4 0

Answer:

A. x+y<10

B. x+y≤10

Step-by-step explanation:

A is correct because they can be less than.

B is correct because there is less than or equal too of numbers of volunteers.

C, D, and E cannot be correct however, because that means that they can be over at any number. Therefore, these three would be incorrect.

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478 × 965 in distributive property

kakasveta [241]

478(900) + 478(95)

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3 years ago

Read 2 more answers

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horrorfan [7]

do you sell genius brain

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2 years ago

Describe the solutions of 2-5>n in words

inna [77]

$2-5\ \textgreater \ n \ \textgreater \ Switch\;sides \ \textgreater \ n\ \textless \ 2-5 \ \textgreater \ Simplify \ \textgreater \ 2-5 \ \textgreater \ -3$

$n\ \textless \ -3$

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5 0

3 years ago

If you are a dog lover, having your dog with you may reduce your stress level. Does having a friend with you reduce stress? To e

Pepsi [2]

The outlier of a dataset is a data element that is relatively far from the remaining data elements

99 is an outlier of pet group
See attachment for the parallel box plots

(a) Prove that 99 is an outlier for Pet

We have:

Pets: 58 64 65 68 69 69 69 70 70 72 76 79 85 86 99

$n = 15$

The quartiles positions are:

$Q_1 = \frac{n + 1}{4}$

$Q_1 = \frac{15 + 1}{4}$

$Q_1 = \frac{16}{4}$

$Q_1 = 4th$

$Q_3 = Q_1 \times 3$

$Q_3 = 4th \times 3$

$Q_3 = 12th$

So, we have:

$Q_1 = 4th$

$Q_3 = 12th$

From the pet group:

The data elements at the 4th and 12th positions are 68 and 79

So, we have:

$Q_1 = 68$

$Q_3= 79$

The lower and upper limits of the outlier are:

$L = Q_1 - 1.5 \times (Q_3 - Q_1)$

$U = Q_3 + 1.5 \times (Q_3 - Q_1)$

So, we have:

$L = 68 - 1.5 \times (79 - 68)$

$L = 51.5$

$U = 79+ 1.5 \times (79 - 68)$

$U = 95.5$

This means that data below 51.5 or above 95.5 are outliers.

Hence, 99 is an outlier because 99 is greater than 95.5

(b) The parallel box plot

The three groups are:

Pets: 58 64 65 68 69 69 69 70 70 72 76 79 85 86 99

Erlento: 88 80 80 81 92 87 88 81 82 80 87 92 87 80 82

Alone: 62 70 73 75 77 80 84 84 84 87 87 87 90 91 99

See attachment for the parallel box plots

Read more about box plots and outliers at:

brainly.com/question/14940764

5 0

3 years ago

Given that the commuter used public transportation, find the probability that the commuter had a commute of 60 6060 or more minu

AlladinOne [14]

Complete Question

The two-way table below gives the thousands of commuters in Massachusetts in 2015 by transportation method and one-way length of commute.

$\left|\begin{array}{c|c|c|c|c|c|c}$Transport Mode/Minutes&$Less than 15&15-29&30-44& 45-59&$60 or more &$Total\\---&---&---&---&---&---&---\\$Private vehicle & 636&908&590&257&256&2647\\$Public Transportation&9&54&96&62&108&329 \\$Other&115&70&23&7&7&222\\---&---&---&---&---&---&---\\$Total&760&1032&709&326&371 &3198 \end{array}\right|$

Answer:

$\dfrac{108}{329}$

Step-by-step explanation:

From the table above

Total Number of commuters who used public transportation=329

Number of commuters who spent 60 or more minutes on public transportation=108

Therefore:

Given that a commuter used public transportation, the probability that the commuter had a commute of 60 or more minutes:

P ( 60 + minutes ∣ public transportation ) = $\dfrac{108}{329}$

8 0

3 years ago