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romanna [79]
3 years ago
7

Select the correct answer from each drop-down menu. The first quartile of the data set represented by the box plot is . The medi

an of the data set is

Mathematics
1 answer:
atroni [7]3 years ago
6 0

Answer:

The first quartile of the data set represented by the box plot is 12

The median of the data set is 16

Step-by-step explanation:

The computation is shown below:

Quartile means the third digit form the starting

While on the other hand the median represent the mid value of the data set that is shown in the inner side of the box plot as on the vertical line

So, the median of the data set is 16

The same is to be considered

Kindly find the attachment below:

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Can someone help me plz
Karo-lina-s [1.5K]

Answer:

y = 2x - 10

Step-by-step explanation:

The slope is 4 + 4 / 7 - 3 = 8 / 4 = 2

y = 2x + b

-4 = 6 + b

b = -10

Thank, 5 star, braniliest if helpful

3 0
2 years ago
Thirty-two jelly beans are shared by 8 students. How many jelly beans will<br> each student get?
Katyanochek1 [597]

Answer:

4

Step-by-step explanation:

because 32 divided by 8 = 4

8 0
3 years ago
12 girls in a class of 15 girls have short hair and the rest have long hair . What percent of the girls has long hair ?​
kati45 [8]

Step-by-step explanation:

long hair

15 - 12 = 3

long hair percentage

= 3 ÷ 15 × 100

= 20%

4 0
3 years ago
Suppose X, Y, and Z are random variables with the joint density function f(x, y, z) = Ce−(0.5x + 0.2y + 0.1z) if x ≥ 0, y ≥ 0, z
dexar [7]

Answer:

The value of the constant C is 0.01 .

Step-by-step explanation:

Given:

Suppose X, Y, and Z are random variables with the joint density function,

f(x,y,z) = \left \{ {{Ce^{-(0.5x + 0.2y + 0.1z)}; x,y,z\geq0  } \atop {0}; Otherwise} \right.

The value of constant C can be obtained as:

\int_x( {\int_y( {\int_z {f(x,y,z)} \, dz }) \, dy }) \, dx = 1

\int\limits^\infty_0 ({\int\limits^\infty_0 ({\int\limits^\infty_0 {Ce^{-(0.5x + 0.2y + 0.1z)} } \, dz }) \, dy } )\, dx = 1

C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0 {e^{-0.2y }(\int\limits^\infty_0 {e^{-0.1z} } \, dz  }) \, dy  }) \, dx = 1

C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0{e^{-0.2y}([\frac{-e^{-0.1z} }{0.1} ]\limits^\infty__0 }) \, dy  }) \, dx = 1

C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0 {e^{-0.2y}([\frac{-e^{-0.1(\infty)} }{0.1}+\frac{e^{-0.1(0)} }{0.1} ])  } \, dy  }) \, dx = 1

C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0 {e^{-0.2y}[0+\frac{1}{0.1}]  } \, dy  }) \, dx =1

10C\int\limits^\infty_0 {e^{-0.5x}([\frac{-e^{-0.2y} }{0.2}]^\infty__0  }) \, dx = 1

10C\int\limits^\infty_0 {e^{-0.5x}([\frac{-e^{-0.2(\infty)} }{0.2}+\frac{e^{-0.2(0)} }{0.2}]   } \, dx = 1

10C\int\limits^\infty_0 {e^{-0.5x}[0+\frac{1}{0.2}]  } \, dx = 1

50C([\frac{-e^{-0.5x} }{0.5}]^\infty__0}) = 1

50C[\frac{-e^{-0.5(\infty)} }{0.5} + \frac{-0.5(0)}{0.5}] =1

50C[0+\frac{1}{0.5} ] =1

100C = 1 ⇒ C = \frac{1}{100}

C = 0.01

3 0
2 years ago
How do I factorise out the negative?
SCORPION-xisa [38]
I think the anser is c that what i think
3 0
3 years ago
Read 2 more answers
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