Which of the following lines best represents the fraction 9/10 on a number line?

Please help, will give brainliest to first

Lelechka [254]

Answer:

AN = 16 units

Step-by-step explanation:

So we can use the fact that $\frac{AN}{FN} =\frac{OR}{FR}$

Since this is an isosceles triangle, segments FO and FR are both equal to 7

Manipulating the equation above to solve for AN we have $AN = \frac{(FN)(OR)}{FR}$

Substituting in the known values we have $AN=\frac{(10)(11.2)}{7} =16$

8 0

3 years ago

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Which geometric term is modeled by the stop sign

lakkis [162]

A stop sign has 8 sides and all angles are the same, so it is a regular octagon.

7 0

3 years ago

The Wall Street Journal reported that Walmart Stores Inc. is planning to lay off 2300 employees at its Sam's Club warehouse unit

Advocard [28]

Answer:

(a) The mean is 52 and the median is 55.

(b) The first quartile is 44 and the third quartile is 60.

(c) The value of range is 33 and the inter-quartile range is 16.

(d) The variance is 100.143 and the standard deviation is 10.01.

(e) There are no outliers in the data set.

(f) Yes

Step-by-step explanation:

The data provided is:

S = {55, 56, 44, 43, 44, 56, 60, 62, 57, 45, 36, 38, 50, 69, 65}

(a)

Compute the mean of the data as follows:

$\bar x=\frac{1}{n}\sum x\\=\frac{1}{15}[55+ 56+ 44+ 43+ 44+ 56+ 60+ 62+ 57+ 45 +36 +38 +50 +69+ 65]\\=\frac{780}{15}\\=52$

Thus, the mean is 52.

The median for odd set of values is the computed using the formula:

$Median=(\frac{n+1}{2})^{th}\ obs.$

Arrange the data set in ascending order as follows:

36, 38, 43, 44, 44, 45, 50, 55, 56, 56, 57, 60, 62, 65, 69

There are 15 values in the set.

Compute the median value as follows:

$Median=(\frac{15+1}{2})^{th}\ obs.=(\frac{16}{2})^{th}\ obs.=8^{th}\ observation$

The 8th observation is, 55.

Thus, the median is 55.

(b)

The first quartile is the middle value of the upper-half of the data set.

The upper-half of the data set is:

36, 38, 43, 44, 44, 45, 50

The middle value of the data set is 44.

Thus, the first quartile is 44.

The third quartile is the middle value of the lower-half of the data set.

The upper-half of the data set is:

56, 56, 57, 60, 62, 65, 69

The middle value of the data set is 60.

Thus, the third quartile is 60.

(c)

The range of a data set is the difference between the maximum and minimum value.

Maximum = 69

Minimum = 36

Compute the value of Range as follows:

$Range =Maximum-Minimum\\=69-36\\=33$

Thus, the value of range is 33.

The inter-quartile range is the difference between the first and third quartile value.

Compute the value of IQR as follows:

$IQR=Q_{3}-Q_{1}\\=60-44\\=16$

Thus, the inter-quartile range is 16.

(d)

Compute the variance of the data set as follows:

$s^{2}=\frac{1}{n-1}\sum (x_{i}-\bar x)^{2}\\=\frac{1}{15-1}[(55-52)^{2}+(56-52)^{2}+...+(65-52)^{2}]\\=100.143$

Thus, the variance is 100.143.

Compute the value of standard deviation as follows:

$s=\sqrt{s^{2}}=\sqrt{100.143}=10.01$

Thus, the standard deviation is 10.01.

(e)

An outlier is a data value that is different from the remaining values.

An outlier is a value that lies below 1.5 IQR of the first quartile or above 1.5 IQR of the third quartile.

Compute the value of Q₁ - 1.5 IQR as follows:

$Q_{1}-1.5QR=44-1.5\times 16=20$

Compute the value of Q₃ + 1.5 IQR as follows:

$Q_{3}+1.5QR=60-1.5\times 16=80$

The minimum value is 36 and the maximum is 69.

None of the values is less than 20 or more than 80.

Thus, there are no outliers in the data set.

(f)

Yes, the data provided indicates that the Walmart is meeting its goal for reducing the number of hourly employees

6 0

3 years ago

With food prices becoming a great issue in the world; wheat yields are even more important. Some of the highest yielding dry lan

Rzqust [24]

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,

$P(z$

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

Hence, the yield of 61.6 bushels per acre or more would save the seed.

7 0

3 years ago

How do you solve this with distributive property 5(2x + 10)

DaniilM [7]

5 times 2x + 5 times 10
10x+50

8 0

3 years ago

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