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Reil [10]
3 years ago
14

What is the first quartile of the data displayed in this box-and-whisker plot? A)49 B)41 C)37 D)35

Mathematics
2 answers:
Mariulka [41]3 years ago
7 0

Answer:

The correct option is C. The first quartile of the data displayed in this box-and-whisker plot is 37.

Step-by-step explanation:

In a box plot,

Point 1: Left end point of the line represents the minimum value of the data set.

Point 2: Left side of the box represent first quartile (Q₁) of the data set.

Point 3: Line inside the box represents Median of the data set.

Point 4: Right side of the box represent third quartile (Q₃) of the data set.

Point 5: Right end point of the line represents the maximum value of the data set.

From the given box-and-whisker plot it is clear that left side of the box is at 37, So, the first quartile of the data displayed in this box-and-whisker plot is 37.

Q_1=37

Therefore correct option is C.

fiasKO [112]3 years ago
5 0
C. 37
It is the lower quartile marked on the box-and-whisker plot and is the median of all data below the median
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iogann1982 [59]

Area of a square is side length^2

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4 0
3 years ago
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Records for the past several years show that 46% of customers at a local game shop use a customer loyalty card. Due to a recent
Ratling [72]

Answer:

1. Null hypotheses: p=0.49

Alternative hypotheses: p>0.49

2. p=0.5704 or 57.04%

Step-by-step explanation:

1.

We have to writer the hypotheses for testing the proportion has increased from 49%. So, the population proportion that is to be tested is 0.49.

As we already know that the null hypotheses always contain equality so,

Null hypotheses: p=0.49

and we have to test whether the proportion has increased from 0.49, so,

Alternative hypotheses: p>0.49

Thus, the hypotheses for testing the proportion has increased from 49% are

Null hypotheses: p=0.49

Alternative hypotheses: p>0.49

2.

We have to find the proportion of sampled customers using loyalty card.

The given sample indicate that 77 out of 135 customers are using loyalty card. So,

p=x/n

where x is the number of favorable outcome and n is total number of outcome.

Here, x=77 and n=135. So,

p=77/135=0.5704 or 57.04%

6 0
3 years ago
Help me so confused?
Eva8 [605]

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7 0
3 years ago
Algebra help?<br><br> 343 = (x/6) ^2/3
Vladimir79 [104]

I can only assume that you meant, "Solve for x:"

Apply the exponent 3/2 to both sides of this equation.  The result will be

     3/2

343 = x/6.    

Multiplying both sides by 6 isolates x:

          3/2

6*343           = x                         Since 7^3 = 343,  the expression for x

                                                  can be rewritten as

           3/2

6*(7^3)          =   x                      which can be further simplified, as follows:


x =  6^(3/2)*7^(9/2), or:

x = 6^(3/2)*7^(8/2)*√7, or

x = 6^(3/2)*7^4*√7

3 0
3 years ago
A basic cellular phone plan costs $4 per month for 70 calling minutes. Additional time costs $0.10 per minute. The formula C= 4+
Harrizon [31]

Answer:

For a monthly cost of at least $7 and at most $8, you can have between 100 and 110 calling minutes.

Step-by-step explanation:

The problem states that the monthly cost of a celular plan is modeled by the following function:

C(x) = 4 + 0.10(x-70)

In which C(x) is the monthly cost and x is the number of calling minutes.

How many calling minutes are needed for a monthly cost of at least $7?

This can be solved by the following inequality:

C(x) \geq 7

4 + 0.10(x - 70) \geq 7

4 + 0.10x - 7 \geq 7

0.10x \geq 10

x \geq \frac{10}{0.1}

x \geq 100

For a monthly cost of at least $7, you need to have at least 100 calling minutes.

How many calling minutes are needed for a monthly cost of at most 8:

C(x) \leq 8

4 + 0.10(x - 70) \leq 8

4 + 0.10x - 7 \leq 8

0.10x \leq 11

x \leq \frac{11}{0.1}

x \leq 110

For a monthly cost of at most $8, you need to have at most 110 calling minutes.

For a monthly cost of at least $7 and at most $8, you can have between 100 and 110 calling minutes.

8 0
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