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3 years ago

a total of 60 raffle tickets were sold at a school fair Bethany has a 20% chance of wining how many tickets did she buy

Mathematics

1 answer:

Studentka2010 [4]3 years ago

7 0

Answer:

Bethany brought 3 tickets

Step-by-step explanation:

Because 20% of 60 is 3

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La probabilidad de que un dia cualquiera, carlos almuerce pollo frito es de 0,4. La probabilidad de que almuerce hambuergues es

MAXImum [283]

Answer:

0.6

Step-by-step explanation:

La fórmula para calcular la probabilidad de dos eventos se da como:

P (A⋃B) = P (A) + P (B) - P (A ⋂ B)

De la pregunta, sabemos que:

Probabilidad de que Carlos haya comido pollo frito: P (C) = 0.4.

Probabilidad de que Carlos coma una hamburguesa para el almuerzo: P (H) = 0.3.

Probabilidad de que Carlos almorce pollo frito y hamburguesa el mismo día: P (C ⋂ H) = 0.1.

Probabilidad de que Carlos coma pollo frito o hamburguesa para el almuerzo:

P (C ⋃ H) =?

Por lo tanto, aplicando nuestra fórmula:

P (C ⋃ H) = P (C) + P (H) - P (C ⋂ H)

P (C ⋃ H) = 0.4 + 0.3 - 0.1

P (C ⋃ H) = 0.6

6 0

3 years ago

Below is the graph of f(x)=2In(x). how would you describe the graph of g(x)=4In(x)

Andreyy89

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\

\begin{array}{rllll} 
% left side templates
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\end{array}$

$\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{ period of }\frac{2\pi }{{{ B}}}
\end{array}$

now... with that template in mind, let's see yours

$\bf \begin{array}{llll}
g(x)=&4ln(x)\implies &2(2)ln(x)\\
&\uparrow &\uparrow \\
&A&A
\end{array}$

A is twice as large as in f(x), thus the graph shrinks twice as much in g(x)

7 0

4 years ago

Read 2 more answers

Allowing 20 % discount on the marked price of an article and levying 15 % VAT. a buyer has to pay Rs 9.200 for the article. Find

nata0808 [166]

Answer:

here I'm confused that whether the selling price is Rs.9,200 or Rs.9.200

any ways you can take the help of below procedure

Step-by-step explanation:

here,

let marked price be X

discount(d)=20%

VAT=15%

SP with VAT= 9200

now,

SP without VAT= SP - VAT of SP

= 9200-15/100 ×9200

= 9200- 1380

=Rs. 1380

again,

SP= MP - D of MP

or,7820= x- 20/100 × x

or, 7820× 100 = 100x- 20x

or,782000=80x

or, 782000/80=X

or, X= 9775.

hence MP is Rs 9775

8 0

3 years ago

In triangle JKL, tan(bº) =

VikaD [51]

This is the working outs

8 0

3 years ago

A classroom bulletin board is 7 feet by 4 feet.If there is a picture of a student every 6 inches along the edge including 1 in e

vovikov84 [41]

When you illustrate the problem, it would look like the diagram shown in the picture. There are pictures, each with a length of 6 inches, that are placed all around the perimeter. To solve the number of pictures, the solution is as follows:

Size of picture = 6 inches * 1 ft/12 inches = 0.5 ft/picture
Pictures along the length = 7 ft * 1 picture/0.5 ft = 14 pictures
Pictures along the width = 4 ft * 1 picture/0.5 ft = 8 pictures
Since perimeter is 2L + 2W, the total number of pictures is:
Total pictures = 2(14) + 2(8) = 44 pictures

7 0

3 years ago