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3 years ago

10

A survey was conducted at a local restaurant about the type of seating preferred by males and females. The results of the survey

are shown i
the table.
Indoor Outdoor Total
Males
23
65
88
Females
47
13
60
Total
70
78
148
Approximately what percentage of those surveyed preferred indoor seating?
A. 47%
B. 53%
C. 70%
D. 78%

1 answer:

Lubov Fominskaja [6]3 years ago

6 0

Answer:

c: 70

Step-by-step explanation:

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Each taco will cost you $1.50.

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1 year ago

The given matrix is the augmented matrix for a linear system. Use technology to perform the row operations needed to transform t

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Answer:

$x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\$

Step-by-step explanation:

As the given Augmented matrix is

$\left[\begin{array}{ccccc}9&-2&0&-4&:8\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]$

Step 1 :

$r_{1}$ ↔ $r_{1} - r_{2}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]$

Step 2 :

$r_{3}$ ↔ $r_{3} - 8r_{1}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\0&124&-54&77&:-82\end{array}\right]$

Step 3 :

$r_{2}$ ↔ $\frac{r_{2}}{7}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&124&-54&77&:-82\end{array}\right]$

Step 4 :

$r_{1}$ ↔ $r_{1} + 14r_{2}$ , $r_{3}$ ↔ $r_{3} - 124r_{2}$

$\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&- \frac{254}{7} &\frac{663}{7} &:-\frac{1690}{7} \end{array}\right]$

Step 5 :

$r_{3}$ ↔ $\frac{r_{3}. 7}{254}$

$\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&1&-\frac{663}{254} &:-\frac{1690}{254} \end{array}\right]$

Step 6 :

$r_{1}$ ↔ $r_{1} - 4r_{3}$ , $r_{2}$ ↔ $r_{2} + \frac{1}{7} r_{3}$

$\left[\begin{array}{ccccc}1&0&0&-\frac{71}{127} &:\frac{176}{127} \\0&1&0&-\frac{131}{254} &:\frac{284}{127} \\0&0&1&-\frac{663}{254} &:\frac{845}{127} \end{array}\right]$

∴ we get

$x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\$

6 0

3 years ago

Consider this right triangle.<br> 21<br> 29<br> 20<br> Enter the ratio equivalent to s

AleksAgata [21]

Answer:

Part 1) $sin(B)=\frac{21}{29}$

Part 2) $csc(A)=\frac{29}{20}$

Part 3) $cot(A)=\frac{21}{20}$

Step-by-step explanation:

<u><em>The complete question is</em></u>

Consider this right triangle. 21 29 20 Write the ratio equivalent to: Sin B - CscA- Cot B

The picture of the question in the attached figure

Part 1) Write the ratio equivalent to: Sin B

we know that

In the right triangle ABC

$sin(B)=\frac{AC}{AB}$ ----> by SOH (opposite side divided by the hypotenuse)

substitute the values

$sin(B)=\frac{21}{29}$

Part 2) Write the ratio equivalent to: Csc A

we know that

In the right triangle ABC

$csc(A)=\frac{1}{sin(A)}$

$sin(A)=\frac{BC}{AB}$ -----> by SOH (opposite side divided by the hypotenuse)

substitute the values

$sin(A)=\frac{20}{29}$

therefore

$csc(A)=\frac{29}{20}$

Part 3) Write the ratio equivalent to: Cot A

we know that

In the right triangle ABC

$cot(A)=\frac{1}{tan(A)}$

$tan(A)=\frac{BC}{AC}$ -----> by TOA (opposite side divided by the adjacent side)

substitute the values

$tan(A)=\frac{20}{21}$

therefore

$cot(A)=\frac{21}{20}$

4 0

3 years ago

Factor the polynomial by grouping : x^3+5x^2-6x-30

andrew-mc [135]

The idea is to pair up the terms, factor each sub-group, and then pull out the overall GCF to fully factor.

x^3+5x^2-6x-30

(x^3+5x^2)+(-6x-30)

x^2(x+5)+(-6x-30)

x^2(x+5)-6(x+5)

(x^2-6)(x+5)

-------------

Answer: (x^2-6)(x+5)

4 0

3 years ago

Please help me this is. big

kramer

Answer:

C. = 69.5

Step-by-step explanation:

The median or the average between 80 and 59 is 69.5

Hope this helps! :)

8 0

3 years ago