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3 years ago

Bruce walks 34 mile in 15 hour. What is his speed in miles per hour?

Mathematics

1 answer:

erastova [34]3 years ago

3 0

Answer:

S=D/T

34/15=2.26 mph

Step-by-step explanation:

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Helppppp PLZZZZZZZZZZZZZZZZZZZZZZ i need help

Olin [163]

Answer/Step-by-step explanation:

A) The temperature in Chicago could be -10*F because -10 is to the right of -13 on the number line.

Higher Temperatures -->

o----------->I

I---I---I---I---I---I---I---I---I---I---I---I---I---I---I---I---I---I---I---I---I---I---I---I---I

-13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

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3 years ago

Read 2 more answers

You want to compute a 90% confidence interval

Vesnalui [34]

Answer:

you do it your self just saying you good at it believe in yourself

7 0

3 years ago

What is the value of 4/5 minus 2/3?<br><img src="https://tex.z-dn.net/?f=4%2F5%20%20-%20%202%2F3" id="TexFormula1" title="4/5 -

Dafna1 [17]

2/15

4/5-2/3

convert so they have the same denominator

12/15-10/15

solve

2/15

4 0

2 years ago

Can someone help me figure this out

Setler79 [48]

Answer:

<h2>See the explanation.</h2>

Step-by-step explanation:

It is told that, 330 students that are surveyed, are 9th graders.

Hence, 0.55 represents 330 students, that is 1 represents $\frac{330}{0.55}$ = 600 students.

The number of 9th grade students whose favorite elective is music is represented by 0.18.

1 ≡ 600

0.18 ≡ $600\times0.18$ = 108.

The percentage of students whose favorite elective is home economics and are also in 10th grade is $\frac{0.06}{1} \times100$ = 6%.

5 0

3 years ago

Solve for XX. Assume XX is a 2×22×2 matrix and II denotes the 2×22×2 identity matrix. Do not use decimal numbers in your answer.

sveticcg [70]

The question is incomplete. The complete question is as follows:

Solve for X. Assume X is a 2x2 matrix and I denotes the 2x2 identity matrix. Do not use decimal numbers in your answer. If there are fractions, leave them unevaluated.

$\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ · X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ =<em>I</em>.

First, we have to identify the matrix <em>I. </em>As it was said, the matrix is the identiy matrix, which means

<em>I</em> = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

So, $\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ · X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

Isolating the X, we have

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ - $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

Resolving:

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}2-1&8-0\\-6-0&-9-1\end{array}\right]$

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Now, we have a problem similar to A.X=B. To solve it and because we don't divide matrices, we do X=A⁻¹·B. In this case,

X= $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ ⁻¹· $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Now, a matrix with index -1 is called Inverse Matrix and is calculated as: A . A⁻¹ = I.

So,

$\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ · $\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]$ = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

9a - 3b = 1

7a - 6b = 0

9c - 3d = 0

7c - 6d = 1

Resolving these equations, we have a= $\frac{2}{11}$ ; b= $\frac{7}{33}$ ; c= $\frac{-1}{11}$ and d= $\frac{-3}{11}$ . Substituting:

X= $\left[\begin{array}{ccc}\frac{2}{11} &\frac{-1}{11} \\\frac{7}{33}&\frac{-3}{11} \end{array}\right]$ · $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Multiplying the matrices, we have

X= $\left[\begin{array}{ccc}\frac{8}{11} &\frac{26}{11} \\\frac{39}{11}&\frac{198}{11} \end{array}\right]$

6 0

3 years ago