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Delicious77 [7]
3 years ago
10

Yanika is teaching a swim class. She needs to divide the class into teams for a relay race. There is 18 students.

Mathematics
1 answer:
coldgirl [10]3 years ago
8 0
A possible sizes two teams of 9 or 9 teams of 2, 6 teams of 3 or 3 teams of 6

B yes 3 lanes could divide her 18 students by putting 6 people in each lane 
18/3= 6
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Elen planned to read a book of 40 pages in two days. On the first day, she read 4 hours reading 7 pages per hour. On the second
amm1812

Ok so 7 times 4 is 28 so she read 28 the first day. For the second day since 40-28= 12, 12 divided by 2 is 6. So she read 6 pages per hour.

7 0
3 years ago
Read 2 more answers
50 / (-8) / (-5) = ?
ser-zykov [4K]

<em>☽------------❀-------------☾</em>

<em>Hi there!</em>

<em>~</em>

<em></em>50 \div (-8) \div (-5) = 1.25<em></em>

<em>❀Hope this helped you!❀</em>

<em>☽------------❀-------------☾</em>

<em></em>

8 0
4 years ago
A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 111.4-cm and a standard dev
Kipish [7]

Answer:

P(111.2-cm < ¯ x < 111.4-cm) = 0.4726

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

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

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 111.4, \sigma = 0.5, n = 23, s = \frac{0.5}{\sqrt{23}} = 0.1043

Find the probability that the average length of a randomly selected bundle of steel rods is between 111.2-cm and 111.4-cm.

This is the pvalue of Z when X = 111.4 subtracted by the pvalue of Z when X = 111.2. So

X = 111.4

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

By the Central Limit Theorem

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

Z = \frac{111.4 - 111.4}{0.1043}

Z = 0

Z = 0 has a pvalue of 0.5.

X = 111.2

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

Z = \frac{111.2 - 111.4}{0.1043}

Z = -1.92

Z = -1.92 has a pvalue of 0.0274.

0.5 - 0.0274 = 0.4726

P(111.2-cm < ¯ x < 111.4-cm) = 0.4726

5 0
3 years ago
At a sale, coats were sold for $2 each. This price was 5% of a coat's original price. How much did a coat originally cost?
erica [24]

Answer:

$40

Step-by-step explanation:

This isn't the proper way to do this but

5×20=100 so you multiply the $2 by 20

4 0
3 years ago
The length of a rectangle is 5 units more than the width. The area of the rectangle is 36 units. What is the width, in units, of
PIT_PIT [208]

Answer:

4

Step-by-step explanation:

Let y be the width of the rectangle

The length of the rectangle is 5 unit more than the width. This is written as:

Length = y + 5

Area = 36

Recall:

Area of rectangle = length x width

36 = (y + 5) x y

36 = y^2 + 5y

Rearrange the expression

y^2 + 5y — 36 = 0

To solve this problem by factorization, multiply the first term (i.e y^2) and last term ( i.e — 36) together. This gives — 36y^2

Next, find two factors of —36y^2 such that when we add them together it will result to the second term (5y). These factors are —4y and 9y. Now we substitute —4y and 9y in place of 5y in the equation. This is illustrated below:

y^2 + 5y — 36 = 0

y^2 —4y + 9y — 36 = 0

We factorize as follows:

y(y — 4) + 9(y —4) = 0

(y + 9) (y — 4) =0

y + 9 = 0 or y — 4 = 0

y = —9 or y = 4

Since the measurement can not be negative, therefore y (i.e the width) is 4

4 0
3 years ago
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