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3 years ago

What is the interquartile range for the data set??

Mathematics

1 answer:

kramer3 years ago

8 0

Answer:

Step-by-step explanation:

First, order the numbers in order. Then find the "middle numbers" and omit them after you have divided the number set into two equals parts. Then find the median of those number sets and then subtract.

You should get 71-6 at the end which is 65.

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Because of her work schedule, Molly Kay plans to take classes at 8:00, 10:00, and 11:00. Five courses that fit her program are o

pychu [463]

Answer: There are 90 different schedules are possible for Molly Kay.

Step-by-step explanation:

Since we have given that

Three time scheduled at 8:00, 10:00, 11:00.

Number of courses offer at 8:00 = 5

Number of courses offer at 10:00 = 9

Number of courses offer at 11:00 = 2

So, the number of different schedules are possible for Molly Kay is given by

$^{5}C_1\times ^9C_1\times ^2C_1\\\\=5\times 9\times 2\\\\=45\times 2\\\\=90$

Hence, there are 90 different schedules are possible for Molly Kay.

7 0

3 years ago

W(x)=9x+8,evaluate w(5)

natima [27]

Use the substitution method

w(x)=9x+8

w(5)=9(5)+8

Do the parenthesis first then add because of using PEMDAS

P= Parentheses

E= Exponents

M= Multiplication

D= Division

A= Addition

S= Subtraction

45+8

=53

Answer: w(5)=53

7 0

3 years ago

Read 2 more answers

Evaluate UV if possible.

Mnenie [13.5K]

Answer: B

Step-by-step explanation:

Multiply U and V:

$\begin{pmatrix}-9&1&0\\ \:4&0&2\\ \:0&1&5\end{pmatrix}\begin{pmatrix}2\\ \:-2\\ \:1\end{pmatrix}$

$\mathrm{Multiply \ the \ rows \ of \ the \ first \ matrix \ by \ the \ columns \ of \ the \ second \ matrix:}$

$\begin{aligned}&\left(\begin{array}{ccc}-9 & 1 & 0\end{array}\right)\left(\begin{array}{c}2 \\-2 \\1\end{array}\right)=(-9) \cdot 2+1 \cdot(-2)+0 \cdot 1 \\&\left(\begin{array}{ccc}4 & 0 & 2\end{array}\right)\left(\begin{array}{c}2 \\-2 \\1\end{array}\right)=4 \cdot 2+0 \cdot(-2)+2 \cdot 1 \\&\left(\begin{array}{lll}0 & 1 & 5\end{array}\right)\left(\begin{array}{c}2 \\-2 \\1\end{array}\right)=0 \cdot 2+1 \cdot(-2)+5 \cdot 1\end{aligned}$

$=\begin{pmatrix}\left(-9\right)\cdot \:2+1\cdot \left(-2\right)+0\cdot \:1\\ 4\cdot \:2+0\cdot \left(-2\right)+2\cdot \:1\\ 0\cdot \:2+1\cdot \left(-2\right)+5\cdot \:1\end{pmatrix}$