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wolverine [178]
3 years ago
8

Complete the table of values of y = 14 x + 1.

Mathematics
1 answer:
Nesterboy [21]3 years ago
5 0

Answer:

-4,0

0,1

8,3

======

x: -4, 0, 8

f(x) = 1/4(x) + 1

Plug in x values into f(x)...

f(-4) = 0

f(0) = 1

f(8) = 3

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What is the interquartile range of the numbers 9, 28, 16, 2, 33, 6, 10
Mumz [18]

Answer:

IQR = 22

Step-by-step explanation:

IQR Formula:

IQR = Q3 - Q1

In order you find Q3 and Q1, please follow these steps:

1. First, you need to order the list of numbers from least to greatest:

2, 6, 9, 10, 16, 28, 33

2. Then, you need to find the median, or the middle number.

2, 6, 9, 10, 16, 28, 33

3. In order to find IQR, you must find the first and third quartiles.

2, 6, 9, 10, 16, 28, 33

Q1 = 6

Q3 = 28

This is because 6 basically means that all the numbers leading to 6 would account for 25% of the data while all the numbers leading to 28 would account for 75% of the data, hence why these are called quartiles.

Now since you have Q1 and Q3, you follow the formula.

28 - 6 = 22

IQR = 22

6 0
3 years ago
Write the standard form of the equation of each line given the slope and y- intercept
SOVA2 [1]
Standard form is y=mx+b 

y=9x+4


7 0
3 years ago
Round the following numbers to the nearest hundreds place.
Eddi Din [679]
62,900
21,000
126,300
856,700
7 0
3 years ago
Show that the following functions are probability density functions for some value of k and determine k. Then, determine the mea
lord [1]

Answer:

a) 17.5

b) 15.6

c) 13.3

d) 21.51

Step-by-step explanation:

The given function is equal to:

f(x)=kx^2

where

\int\limits^y_0 {kx^{2} } \, =1

where y=23

Clearing k=0.00025

a) Ex=\int\limits^y_0 {xf(x)} \, dx =\int\limits^y_0 {x*0.00025x^{2} } \, dx =17.5

b)Vx=Ex^{2} -(Ex)^{2} =\int\limits^y_0 {x^{2}f(x) } \, dx-17.5^{2}  =\int\limits^y_0 {x^{2} *0.00025x^{2} } \, dx -17.5^{2} =321.82-306.25=15.6

c) The function is equal to:

f(x)=k(1+2x)

\int\limits^y_0 {k(1+2x)} \, =1

where y=20

k=0.0024

Ex=\int\limits^y_0 {xf(x)} \, dx =\int\limits^y_0 {x*0.0024(1+2x)} \, dx =13.3

d) Vx=Ex^{2} -(Ex)^{2} =\int\limits^y_0 {x^{2} f(x)} \, dx -13.3^{2}=\int\limits^y_0 {x^{2} *0.0024(1+2x)} \, dx-13.3^{2}   =198.4-176.89=21.51

8 0
3 years ago
Plz help! Is this right? Question is below!
GalinKa [24]

Answer:

infinitely many solutions

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
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