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NeTakaya
3 years ago
5

Fraction word problems Your strip of paper represents one lap on a track. Three students ran a relay and took turns running equa

l parts of the track. The race was three- fourths of a lap long. How much of the track did each ran ?
Mathematics
1 answer:
artcher [175]3 years ago
5 0

Answer:

A quarter (\dfrac{1}{4}) of one lap of the track.

Step-by-step explanation:

Three students ran ran a relay and took turns running equal parts of the track.

The race was three-fourths of a lap long.

Let the length of one lap of the track=x

The length of the race=\frac{3}{4}x

Since each of the students ran equal part,

Length run by each student

=\dfrac{3}{4}x \div 3\\=\dfrac{3}{4}x X \dfrac{1}{3}\\=\dfrac{1}{4}x

Therefore, each student ran a quarter (\dfrac{1}{4}) of one lap of the track.

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Below an equation is modeled . What value of x makes this equation true?
Strike441 [17]
5x + 6 = 1
Subtract 6 from both sides.
5x = -5
Divide from both sides.
x = -1 (ANSWER)

hope this helps!!
5 0
3 years ago
Plz answer my question it is urgent..!!
oee [108]

Answer:

In the step-by-step explanation!

Step-by-step explanation:

Not sure if it is too late but here:

1.)

\frac{1}{\alpha}+\frac{1}{\beta} \\\frac{\beta}{\alpha \beta } +\frac{\alpha}{\alpha \beta } \\\frac{\alpha +\beta }{\alpha \beta }

2.)

\frac{1 }{\alpha } *\frac{1}{\beta} = \frac{1*1}{\alpha*\beta} =\frac{1}{\alpha \beta}

3.)

\frac{1}{\alpha}-\frac{1}{\beta}\\ \frac{\beta}{\alpha \beta } -\frac{\alpha }{\alpha \beta } \\\frac{\beta -\alpha }{\alpha \beta }

Hopes this help! Please give me Brainliest!

6 0
3 years ago
What is the measure of \angle x∠xangle, x?<br> Angles are not necessarily drawn to scale.
krok68 [10]

Answer:

x=135°

Step-by-step explanation:

WZ is 180 degrees. Simply subtract 48 from 180.

180-45=135.

Hope this helps!

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8 0
2 years ago
Question: Researchers in Pakistan wanted to better understand the effects of anthracycline (a chemotherapeutic drug) on the hear
Sedbober [7]

Using the normal distribution, it is found that:

1. His z-score was of Z = -1.88.

2. There is a 0.0301 = 3.01% probability that a randomly selected person has a smaller E/A ratio than the patient in question 1.

3. Z-score of z = 1.85, there is a 0.0322 = 3.22% probability that a randomly selected patient has a higher E/A ratio.

4. Due to the higher absolute value of the z-score, the first patient had a more extraordinary result.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

  • It measures how many standard deviations the measure is from the mean.
  • After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

  • The mean is of \mu = 1.35.
  • The standard deviation is of \sigma = 0.33.

Item 1:

Considering his ratio, we have that X = 0.73, hence:

Z = \frac{X - \mu}{\sigma}

Z = \frac{0.73 - 1.35}{0.33}

Z = -1.88

His z-score was of Z = -1.88.

Item 2:

The probability is the <u>p-value of Z = -1.88</u>, hence, there is a 0.0301 = 3.01% probability that a randomly selected person has a smaller E/A ratio than the patient in question 1.

Item 3:

Score of X = 1.96, hence:

Z = \frac{X - \mu}{\sigma}

Z = \frac{1.96 - 1.35}{0.33}

Z = 1.85

The probability is 1 subtracted by the p-value of Z = 1.85, hence, 1 - 0.9678 = 0.0322 = 3.22% probability that a randomly selected patient has a higher E/A ratio.

Item 4:

Due to the higher absolute value of the z-score, the first patient had a more extraordinary result.

More can be learned about the normal distribution at brainly.com/question/24663213

7 0
2 years ago
John is working part-time in an office. He works 30 hours per week and earns $600 each week. He currently has $2,700 in a saving
Dmitry [639]
The answer is 8 weeks.
7 0
3 years ago
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