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3 years ago

Which operations are the most efficient to use to solve the problem and what is the solution?

1 answer:

6 0

To find the width of Lance's banner, use division (area divided by length). Then use multiplication to times the width by 3 so you can find Evelyn's banner width. (I assume you know how to multiply and divide, but just comment if you'd like me to show you :) )

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Rosalie is organizing a circus performance to raise money for a charity. She is trying to decide how much to charge for tickets.

Ray Of Light [21]

Answer:

she should charge $7.395 per ticket in order to make the most money.

Step-by-step explanation:

From the given information:

If Rosalie charges $5 then 1175 people will attend the circus performance

If Rosalie charges $7 then 935 people will attend the circus performance

Let x be the cost and y to be the number of people that will attend the performance . Then, we will have two points which are;

(5, 1175) and (7, 935)

The slope(m) of this points = $\dfrac{\Delta y}{\Delta x}$

= $\dfrac{y_2-y_1}{x_2-x_1}$

$=\dfrac{935-1175}{7-5}$

Slope (m) = $\dfrac{-240}{2}$

Slope (m) = -120

However; we can now have the linear equation:

$y-y_1 = m(x-x_1)$

$y-1175= -120(x-5)$

$y-1175= -120x+600$

$y= -120x+600+1175$

$y= -120x+1775$

The linear function is : y = -120x + 1775

Now; the total amount of money she can now earn is:

f(x) = xy

f(x) = x(-120x + 1775)

f(x) = -120x² + 1775x

The above expression is a quadratic equation; Using the quadratic formula; we have:

$=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

where; a = -120 ; b = +1775 and c = 0

$=\dfrac{-(1775) \pm \sqrt{(1775)^2-4(-120)(0)}}{2(-120)}$

$=\dfrac{-(1775) + \sqrt{(1775)^2-4(-120)(0)}}{2(-120)} \ \ \ \ \ OR \ \ \ \ \dfrac{-(1775) -\sqrt{(1775)^2-4(-120)(0)}}{2(-120)}$

$=\dfrac{-(1775) + \sqrt{(1775)^2}}{(-240)} \ \ \ \ \ OR \ \ \ \ \dfrac{-(1775) -\sqrt{(1775)^2}}{(-240)}$

$=\dfrac{-(1775) + (1775)}{(-240)} \ \ \ \ \ OR \ \ \ \ \dfrac{-(1775) - (1775)}{(-240)}$

$=\dfrac{0}{(-240)} \ \ \ \ \ OR \ \ \ \ \dfrac{-3550}{(-240)}$

= 0 OR 14.79

Since; we are considering the value greater than zero

x = 14.79

maximum value of x = 14.79/2 = 7.395

Thus ; she should charge $7.395 per ticket in order to make the most money.

6 0

3 years ago

Please help!!!! Select the correct location on the graph.

Ludmilka [50]

Answer:

Its the one on the left

Step-by-step explanation:

7 0

3 years ago

Find a_1 and d for an arithmetic sequence with these terms. a_4=9and a_7=18

jonny [76]

$\bf \begin{array}{llll}
term&value\\
-----&-----\\
a_4&9\\
a_5&9+d\\
a_6&(9+d)+d\\
&9+2d\\
a_7&(9+2d)+d\\
&9+3d=\underline{18}
\end{array}\\\\
-------------------------------\\\\
9+3d=18\implies 3d=9\implies d=\cfrac{9}{3}\implies \boxed{d=3}$

$\bf n^{th}\textit{ term of an arithmetic sequence}\\\\
a_n=a_1+(n-1)d\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
n=7\\
d=3
\end{cases}
\\\\\\
a_7=a_1+(7-1)d\implies 18=a_1+(7-1)3\implies 18=a_1+18
\\\\\\
18-18=a_1\implies \boxed{0=a_1}$

3 0

4 years ago

A coin is tossed 5 times. Find the probability that exactly 1 is a tail. Find the probability that at most 2 are tails.

Natalija [7]

Answer:

Step-by-step explanation:

<h2>First question</h2>

The only possibilities where there is exactly 1 tail are:

(t,h,h,h,h)
(h,t,h,h,h)
(h,h,t,h,h)
(h,h,h,t,h)
(h,h,h,h,t)

those are 5 favorable outcomes.

where <em>h represent heads</em> and <em>t represent tails. </em>There are $2^5 32$ total number of outcomes after tossing the coin 5 times. Because every time you toss the coin, you have 2 possibilities, and as you do it 5 times, those are $2^5$ options. We can conclude from this that

The probability that exactly 1 is a tail is $5/32$ .

<h2>Second question</h2>

We already know the total number of outcomes; 32. Now we need to find the number of favorable outcomes. In order to do that, we can divide our search in three cases: <em>1.-there are no tails, 2.-exactly 1 is a tail, 3.- exactly 2 are tails.</em>

The first case is 1 when every coin is a head. The second case we already solved it, and there are 5. The third case is the interesting one, we can count the outcomes as we did in the previous questions, but that's only because there are not too many outcomes. Instead we are going to use combinations:

We need to have <u>2</u> tails, the other coins are going to be heads. We made <u>5</u> tosses, then the possible combinations are $C_{5,2} = \frac{5!}{3!2!} = \frac{120}{6*2} = 10$

Finally, we conclude that there are 1 + 5 + 10 favorable outcomes, and this implies that

The probability that at most 2 are tails is $\frac{16}{32} = \frac{1}{2}$ .

4 0

4 years ago

A trinomial that contains the variable k the coefficient of the second degree term is 1, the coefficient of the first degree ter

lapo4ka [179]

$k^{2}-7k-17$

4 0

4 years ago