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andreyandreev [35.5K]
3 years ago
6

Tyler wears a jacket to school every 3rd day and a scarf to school every 8th day. What will be the first day Tyler will wear his

jackets and scarf to school on the same day.
Mathematics
1 answer:
disa [49]3 years ago
6 0

Answer:

24th day will  be the first day on which Tyler will wear his jackets and scarf to school

Step-by-step explanation:

We can find the first day on which the Tyler will wear his jackets and scarf to school by taking the LCM of(3,8)

<u>Finding the LCM of(3,8)</u>

List all prime factors for each number.

Prime Factorisation of 3 shows:

3 is prime  =>  3^1

Prime Factorisation of 8 is:

2 x 2 x 2  =>  2^3

For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.

The new super-set list is

2, 2, 2, 3

Multiply these factors together to find the LCM.

LCM = 2 x 2 x 2 x 3 = 24

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A is correct

Step-by-step explanation:

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The answer would be 309! I hope this helps!

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There were 6 purple socks and 4 oraange socks without looking and then another without looking (or replacing the first). What is
shusha [124]

Answer:

The probability that he picked 2 purpled socks is <u>0.33</u>.

Step-by-step explanation:

Given:

Number of purple socks, n(P) = 6

Number of orange socks, n(O) = 4

Two socks are picked without replacement.

Now, total number of socks, n(T)=N(P)+n(O)=6+4=10

Probability of picking the first cap as purple cap is given as:

P(P1)=\frac{n(P)}{n(T)}\\\\P(P1)=\frac{6}{10}=\frac{3}{5}

Since there is no replacement, the number of socks decreases by 1. Also, if the first sock picked is purple, then number of purple socks is also decreased by 1.

Therefore, probability of picking the second cap as purple cap is given as:

P(P2)=\frac{n(P)-1}{n(T)-1}\\\\P(P2)=\frac{5}{9}

Now, probability that both the picked caps are purple is given by their probability product. This gives,

P(P1\ and\ P2)=P(P1)\times P(P2)\\\\P(P1\ and\ P2)=\frac{3}{5}\times\frac{5}{9}\\\\P(P1\ and\ P2)=\frac{3}{9}=\frac{1}{3}=0.33

Therefore, the probability that he picked 2 purpled socks is 0.33

4 0
3 years ago
The population of Farmersville can be modeled by the inequality y &gt; 2,000x + 2,100, where x is the number of years since 2010
Kaylis [27]

Answer:

The population can be greater than 6,100 but less than 6,300, and thus, we cannot determine if the population of Farmersville will be greater than 6,300 in 2012.

Step-by-step explanation:

Inequality for the population of Farmsville:

The inequality for the population of Farmsville in x years after 2010 is given by:

y(x) > 2000x + 2100

Will the population of Farmersville be greater than 6,300 in 2012?

2012 is 2012-2010 = 2 years after 2010, so we have to find y(2).

y(2) > 2000(2) + 2100

y(2) > 6100

The population can be greater than 6,100 but less than 6,300, and thus, we cannot determine if the population of Farmersville will be greater than 6,300 in 2012.

3 0
3 years ago
The number of people , d, in thousands applying for medical benefits per week in a particular city can be modules by the equatio
Mekhanik [1.2K]

Answer:

<h2>4773 peoples.</h2>

Step-by-step explanation:

Given the number of people d, in thousands applying for medical benefits per week in a particular city c modeled by the equation d(t)=2.5 sin(0.76t+0.3)+3.8 where t is the time in years, the maximum number of people tat will apply will occur at d(t)/dt = 0

Differentiating the function given with respect to t, we will have;

d(t)=2.5 sin(0.76t+0.3)+3.8

First we need to know that differential of any constant is zero.

Using\ chain\ rule\\\frac{d(t)}{dt} = 2.5cos(0.76t+0.3) * 0.76 + 0\\ \\\frac{d(t)}{dt} = 1.9cos(0.76t+0.3)

If \frac{d(t)}{dt} =0 then;

1.9cos(0.76t+0.3) = 0\\\\cos(0.76t+0.3)  = 0\\\\0.76t+0.3  = cos^{-1} 0\\\\0.76+3t = 90\\\\3t = 90-0.76\\3t = 89.24\\\\t = 89.24/3\\\\t = 29.75years

To know the maximum number of people in thousands that apply for benefits per year in the city, we wil substitute the value of t = 29.75 into the modeled equation

d(t)=2.5 sin(0.76t+0.3)+3.8\\d(29.75) = 2.5 sin(0.76(29.75)+0.3)+3.8\\d(29.75) = 2.5 sin(22.61+0.3)+3.8\\\\d(29.75) = 2.5 sin(22.91)+3.8\\\\d(29.75) = 0.9732+3.8\\d(29.75) = 4.7732\\\\

Since d is in thousands, the maximum number of people in thousands will be 4.7732*1000 = 4773.2 which is approximately 4773 peoples.

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3 years ago
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