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3 years ago

the first quartile of a data set is 51 and the third quartule is 67. which of these values in the data set is an outlie

Mathematics

1 answer:

Verizon [17]3 years ago

7 0

If one of the quartiles is an outlier, then there is at least another one. not sure tell me if it helps

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Jennifer has a taco stand.she found that her daily costs can be modeled by C(x)=x^2-40x+610. if she wants to keep her daily cost

Svetllana [295]

Plug in 300 for C(x), to get:
300= $x^{2} -40x=610$
Then subtract to get: 0= $0= x^{2} -40x+310$
Then use the quadratic formula and plug everything in. Use a calculator to simplify.

You'll get two answers: 29.486 and -0.513.
Since you can't sell <em>"negative" </em>tacos, use the positive solution
Around 29-30 tacos.

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3 years ago

❗❗❗THERE IS A PICTURE PLEASE ANSWER❗❗❗ SHOW HOW YOU GOT THE ANSWER

Feliz [49]

Answer:

1/2 is poured out and 1 1/2 is left

Step-by-step explanation:

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3 years ago

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Hadley’s softball team has a phone tree in case a game is canceled. The coach calls 3 players. Then each of those players calls

Setler [38]

1 2 3
1 2 3 1 2 3 1 2 3
123 123 123 123 123 123 123 123 123
3 3 3 3 3 3 3 3 3
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3 years ago

Determine Upper P (Upper F or Upper G )using the general addition rule. Select the correct choice below and fill in any answer b

Lilit [14]

Answer:

S= {8 , 9 , 10 , 11 , 12 , 13 , 14 , 15 , 16 , 17 , 18 , 19 }

event F= {10 , 11 , 12 , 13 , 14}

event G= {14 , 15 , 16 , 17}

a) List of outcomes in F or G

answer;

(F or G) = { 10 , 11, 12 , 13 , 14 , 15 , 16 , 17}

b) P ( F or G) = by counting the number of outcomes in F or G

Ans ( F or G) = No. of outcomes in ( F or G) / Total no. of outcomes

= 8/12 = 2/3

c) P ( F or G) using general addition rule

Ans P (F or G) = P(F) = P(G) - P( F and G)

= 5/12 + 4/12 - 1/12

= (5+4-1)/12 = 8/12 =2/3

Step-by-step explanation:

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3 years ago

An elementary school is offering 3 language classes: one in Spanish, one in French,and one in German. The classes are open to an

Alona [7]

Answer:

A. 0.5

B. 0.32

C. 0.75

Step-by-step explanation:

There are

28 students in the Spanish class,
26 in the French class,
16 in the German class,
12 students that are in both Spanish and French,
4 that are in both Spanish and German,
6 that are in both French and German,
2 students taking all 3 classes.

So,

2 students taking all 3 classes,
6 - 2 = 4 students are in French and German, bu are not in Spanish,
4 - 2 = 2 students are in Spanish and German, but are not in French,
12 - 2 = 10 students are in Spanish and French but are not in German,
16 - 2 - 4 - 2 = 8 students are only in German,
26 - 2 - 4 - 10 = 10 students are only in French,
28 - 2 - 2 - 10 = 14 students are only in Spanish.

In total, there are

2 + 4 + 2 + 10 + 8 + 10 +14 = 50 students.

The classes are open to any of the 100 students in the school, so

100 - 50 = 50 students are not in any of the languages classes.

A. If a student is chosen randomly, the probability that he or she is not in any of the language classes is

$\dfrac{50}{100} =0.5$

B. If a student is chosen randomly, the probability that he or she is taking exactly one language class is

$\dfrac{8+10+14}{100}=0.32$

C. If 2 students are chosen randomly, the probability that both are not taking any language classes is

$0.5\cdot 0.5=0.25$

So, the probability that at least 1 is taking a language class is

$1-0.25=0.75$

3 0

3 years ago