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3 years ago

6

A store advertises a sale as "Get 30% off your highest priced item when you buy 2 or more items." Lee buys items with prices of

$30 and $20. Find the percent of discount on the total purchase

1 answer:

agasfer [191]3 years ago

5 0

Answer:

60

Step-by-step explanation:

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Mark and don are planning to sell each of their marble collections at a garage sale. if don has 4 more than 5 times the number o

natita [175]

Mark and Don have to sell 11 and 59 marbles respectively.

<em><u>Explanation</u></em>

Lets assume, the number of marbles Mark has is $x$

As, Don has 4 more than 5 times the number of marbles mark has, so the number of marbles Don has $= 5x+ 4$

Question says that the total number of marbles is 70. So, the equation will be...

$x+(5x+4)= 70\\ \\ 6x+4= 70 \\ \\ 6x = 70-4\\ \\ 6x=66 \\ \\ x= \frac{66}{6}= 11$

<em>So, the number of marbles Mark has = 11 </em>

<em>and the number of marbles Don has = $(5*11)+4 = 55+4= 59$ </em>

6 0

3 years ago

Heights of men have a bell-shaped distribution, with a mean of 176 cm and a standard deviation of 7 cm. Using the Empirical Rule

Vaselesa [24]

Answer:

a) 68% of the men fall between 169 cm and 183 cm of height.

b) 95% of the men will fall between 162 cm and 190 cm.

c) It is unusual for a man to be more than 197 cm tall.

Step-by-step explanation:

The 68-95-99.5 empirical rule can be used to solve this problem.

This values correspond to the percentage of data that falls within in a band around the mean with two, four and six standard deviations of width.

<em>a) What is the approximate percentage of men between 169 and 183 cm? </em>

To calculate this in an empirical way, we compare the values of this interval with the mean and the standard deviation and can be seen that this interval is one-standard deviation around the mean:

$\mu-\sigma=176-7=169\\\mu+\sigma=176+7=183$

Empirically, for bell-shaped distributions and approximately normal, it can be said that 68% of the men fall between 169 cm and 183 cm of height.

<em>b) Between which 2 heights would 95% of men fall?</em>

This corresponds to ±2 standard deviations off the mean.

$\mu-2\sigma=176-2*7=162\\\\\mu+2\sigma=176+2*7=190$

95% of the men will fall between 162 cm and 190 cm.

<em>c) Is it unusual for a man to be more than 197 cm tall?</em>

The number of standard deviations of distance from the mean is

$n=(197-176)/7=3$

The percentage that lies outside 3 sigmas is 0.5%, so only 0.25% is expected to be 197 cm.

It can be said that is unusual for a man to be more than 197 cm tall.

3 0

3 years ago

If x2 = 4, what are the possible numbers for x?

vichka [17]

Answer:

2

Step-by-step explanation:

x=2 so 2×2=4

(mark me as brainlest)

4 0

2 years ago

Find the value of x and the measure of each angle in the parallelogram

Morgarella [4.7K]

Answer:

$\angle D= 162\\\\\angle E = 18\\\\\angle F = 162\\\\\angle G = 18$

Step-by-step explanation:

By properties of parallelograms we know that opposite are the same so we have

$12x=4(x+3)\\\\4(3x)=4(x+3)\\\\3x=x+3\\\\2x=3\\\\x=\frac{3}{2}$

so for the angles we have

$\angle G=\angle E=12(\frac{3}{2})=18\\\\\angle D=\angle F = 180-\angle G=180-18=162$

5 0

3 years ago

For the function given state the period f(t) =6sin(3t-pi/6)-1

lapo4ka [179]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{ A}}sin({{ B}}x+{{ C}})+{{ D}}
\\\\
f(x)=&{{ A}}cos({{ B}}x+{{ C}})+{{ D}}\\\\
f(x)=&{{ A}}tan({{ B}}x+{{ C}})+{{ D}}

\end{array}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{ A}}|\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\end{array}$

$\bf \begin{array}{llll}


\bullet \textit{vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{function period or frequency}\\
\qquad \frac{2\pi }{{{ B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\
\qquad \frac{\pi }{{{ B}}}\ for\ tan(\theta),\ cot(\theta)
\end{array}$

now, with that template in mind, let's take a peek at yours

$\bf \begin{array}{lllcclll}
f(t)=&6sin(&3t&-\frac{\pi }{6})&-1\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}\\\\
-----------------------------\\\\
period\qquad \cfrac{2\pi }{B}\iff\cfrac{2\pi }{3}$

8 0

3 years ago