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Alona [7]
3 years ago
7

Hector and Francis a spent total of $47.50 at the mall. If Hector spent 50% more than Francis, how much did Francis spend at the

mall?
Mathematics
2 answers:
Zigmanuir [339]3 years ago
5 0
Francis spent $19 and Hector spent $28.50
Vitek1552 [10]3 years ago
4 0

Answer:

$19.

Step-by-step explanation:

Let H be the amount spent by Hector and F be the amount spent by Francis.

We have been given that Hector and Francis a spent total of $47.50 at the mall. We can represent this information in an equation as:

H+F=47.50...(1)

We are also told that Hector spent 50% more than Francis. We can represent this information in an equation as:

H=F+(\frac{50}{100}*F)...(2)

H=F+0.50F...(2)

H=1.50F...(2)    

We will use substitution method to solve system of linear equations.

Upon substituting equation (2) in equation (1) we will get,

1.50F+F=47.50

2.50F=47.50

Let us divide both sides of our equation by 2.50.

\frac{2.50F}{2.50}=\frac{47.50}{2.50}

F=19

Therefore, Francis spent $19 at the mall.

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

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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:

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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.

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Thus, there are no outliers in the data set.

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