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alex41 [277]
2 years ago
14

Together, Katya and Mimi have 480 pennies in their piggy banks. After Katya lost ½ of her pennies and Mimi lost 2/3 of her penni

es, they both had an equal number of pennies left. How many pennies did they lose altogether
Mathematics
2 answers:
PilotLPTM [1.2K]2 years ago
8 0

288 pennies

<h3>Further explanation</h3>

<u>Given:</u>

  • Together, Katya and Mimi have 480 pennies in their piggy banks.
  • After Katya lost ½ of her pennies and Mimi lost ²/₃ of her pennies, they both had an equal number of pennies left.

<u>Question:</u>

How many pennies did they lose altogether?

<u>The Process:</u>

Let k = Katya's pennies and m = Mimi's pennies.

<u>Step-1:</u> Finding out the amount of money in the beginning

At first, Katya and Mimi have 480 pennies in their piggy banks.

Equation-1: \boxed{ \ k + m = 480 \ }

After Katya lost ½ of her pennies and Mimi lost ²/₃ of her pennies, they both had an equal number of pennies left. That is, the number of pennies remaining from Katya is \boxed{k - \frac{1}{2}k = \frac{1}{2}k}, while Mimi is \boxed{m - \frac{2}{3}m = \frac{1}{3}m}.

Equation-2: \boxed{ \ \frac{1}{2}k = \frac{1}{3}m \ }

From Equation-2, we get \boxed{ \ k = \frac{2}{3}m \ } and substitute into Equation-1.

\boxed{ \ \frac{2}{3}m + m = 480 \ }

\boxed{ \ \frac{2}{3}m + \frac{3}{3}m = 480 \ }

\boxed{ \ \frac{5}{3}m = 480 \ }

\boxed{ \ m = 480 \times \frac{3}{5} \ } \rightarrow \boxed{\boxed{ \ m = 288 \ }}

Substitute the value of m into \boxed{ \ k = \frac{2}{3}m \ }.

\boxed{ \ k = \frac{2}{3} \times 288 \ } \rightarrow \boxed{\boxed{ \ k = 192 \ }}

Therefore, we got Katya's money of 192 pennies and Mimi's money of 288 pennies at first.

<u>Step-2:</u> Finding out how many pennies did they lose altogether

Katya lost ½ of her pennies and Mimi lost ²/₃ of her pennies.

\boxed{ \ = \Big(\frac{1}{2} \times 192 \Big) + \Big(\frac{2}{3} \times 288 \Big) \ }

\boxed{ \ = 96 + 192 \ }

\boxed{ \ = 288 \ }

Thus, they did lose 288 pennies altogether.

- - - - - - - - - -

Notes

From \boxed{ \ k = \frac{2}{3}m \ }, we can find out their initial money ratio is Katya's: Mimi's = 2: 3.

<h3>Learn more</h3>
  1. What fraction of the entire garden is planted in flowers brainly.com/question/8482599
  2. Mr. Frye distributed his money equally among his 4 children for their weekly allowance. brainly.com/question/13174288
  3. How much money did Jack keep?  brainly.com/question/8538355

Keywords: together, Katya and Mimi, 480 pennies, piggy banks, lost, an equal number, left, lose

vesna_86 [32]2 years ago
6 0

Answer:

288 pennies

Step-by-step explanation:

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<h3>Given</h3>
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<h3>To find</h3>
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<h3>Solution</h3>

<u>Equations as per given are:</u>

  • s + l = 1300
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<u>Subtract equation 1 from equation 3:</u>

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<u>Considering this in the equation 2:</u>

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<u>Substituting s in the equation 1:</u>

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<u>Answer:</u> Love seat costs $500

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2 years ago
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Answer:

Option E) 61.6

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 100 bushels per acre

Standard Deviation, σ = 30 bushels per acre

We assume that the distribution of yield is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

P(X>x) = 0.90

We have to find the value of x such that the probability is 0.90

P(X > x)  

P( X > x) = P( z > \displaystyle\frac{x - 100}{30})=0.90  

= 1 -P( z \leq \displaystyle\frac{x - 100}{30})=0.90  

=P( z \leq \displaystyle\frac{x - 100}{30})=0.10  

Calculation the value from standard normal table, we have,  

P(z

\displaystyle\frac{x - 100}{30} = -1.282\\x = 61.55 \approx 61.6  

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1 year ago
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2 years ago
The average annual amount American households spend for daily transportation is $6312 (Money, August 2001). Assume that the amou
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Answer:

(a) The standard deviation of the amount spent is $3229.18.

(b) The probability that a household spends between $4000 and $6000 is 0.2283.

(c) The range of spending for 3% of households with the highest daily transportation cost is $12382.86 or more.

Step-by-step explanation:

We are given that the average annual amount American households spend on daily transportation is $6312 (Money, August 2001). Assume that the amount spent is normally distributed.

(a) It is stated that 5% of American households spend less than $1000 for daily transportation.

Let X = <u><em>the amount spent on daily transportation</em></u>

The z-score probability distribution for the normal distribution is given by;

                          Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = average annual amount American households spend on daily transportation = $6,312

           \sigma = standard deviation

Now, 5% of American households spend less than $1000 on daily transportation means that;

                      P(X < $1,000) = 0.05

                      P( \frac{X-\mu}{\sigma} < \frac{\$1000-\$6312}{\sigma} ) = 0.05

                      P(Z < \frac{\$1000-\$6312}{\sigma} ) = 0.05

In the z-table, the critical value of z which represents the area of below 5% is given as -1.645, this means;

                           \frac{\$1000-\$6312}{\sigma}=-1.645                

                            \sigma=\frac{-\$5312}{-1.645}  = 3229.18

So, the standard deviation of the amount spent is $3229.18.

(b) The probability that a household spends between $4000 and $6000 is given by = P($4000 < X < $6000)

      P($4000 < X < $6000) = P(X < $6000) - P(X \leq $4000)

 P(X < $6000) = P( \frac{X-\mu}{\sigma} < \frac{\$6000-\$6312}{\$3229.18} ) = P(Z < -0.09) = 1 - P(Z \leq 0.09)

                                                            = 1 - 0.5359 = 0.4641

 P(X \leq $4000) = P( \frac{X-\mu}{\sigma} \leq \frac{\$4000-\$6312}{\$3229.18} ) = P(Z \leq -0.72) = 1 - P(Z < 0.72)

                                                            = 1 - 0.7642 = 0.2358  

Therefore, P($4000 < X < $6000) = 0.4641 - 0.2358 = 0.2283.

(c) The range of spending for 3% of households with the highest daily transportation cost is given by;

                    P(X > x) = 0.03   {where x is the required range}

                    P( \frac{X-\mu}{\sigma} > \frac{x-\$6312}{3229.18} ) = 0.03

                    P(Z > \frac{x-\$6312}{3229.18} ) = 0.03

In the z-table, the critical value of z which represents the area of top 3% is given as 1.88, this means;

                           \frac{x-\$6312}{3229.18}=1.88                

                         {x-\$6312}=1.88\times 3229.18  

                          x = $6312 + 6070.86 = $12382.86

So, the range of spending for 3% of households with the highest daily transportation cost is $12382.86 or more.

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