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svetlana [45]
3 years ago
12

Cost for bowling at Blue Oaks bowling alley are $2.50 for shoes and $4.25 for each game bowled. Austin bowled 5 games. what was

his total cost to bowl?
SOLVE FOR X AND PLEASE SHOW WORK!!!!
Mathematics
1 answer:
yanalaym [24]3 years ago
5 0
2.5 + 4.25x

2.5 + 4.25(5) = 2.5 + 21.25
= 23.75

Hope this help

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Which of the following is equal to the fraction below?
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The managers of 21 supermarkets counted the number of cars in their parking lots on the same day. The results are shown in the l
mixer [17]

The IQR is 42.5

Step-by-step explanation:

Interquartile range is the difference of third and first quartile.

First of all we have to find the median for that purpose the data has to be arranged in ascending order. The data is already in ascending order.

As the number of values are odd

n=21

The median will be: (\frac{n+1}{2}) th\ term

Putting n=21

(\frac{21+1}{2})th\ term\\=(\frac{22}{2})th\ term\\= 11th\ term

The 11th term is 133

So median = 133

Now the data is divided into two halves

One is: 98, 100, 101, 102, 108, 109, 111,118, 129, 132

2nd is: 135, 135, 145, 146, 146, 156, 170 176, 180, 180

Q1 will be the median of first half and Q3 will be the median of 2nd half.

As now the halves contain even number of values, the medians will be the average of middle two values

<u>For First Half:</u>

98, 100, 101, 102, <u>108, 109</u>, 111,118, 129, 132

Q_1 = \frac{108+109}{2}\\Q_1 = \frac{217}{2}\\Q_1 = 108.5

<u>For Second Half:</u>

135, 135, 145, 146, 146, 156, 170 176, 180, 180

Q_2 = \frac{146+156}{2}\\Q_2 = \frac{302}{2}\\Q_2 = 151

Now

<u>Interquartile Range:</u>

IQR = Q_3-Q_1\\= 151-108.5\\=42.5

Hence,

The IQR is 42.5

Keywords: Median, IQR

Learn more about median at:

  • brainly.com/question/10940255
  • brainly.com/question/10941043

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Maria would like to leave a tip (D) of at least 15% of her total dinner bill (3). Which inequality could Maria use to calculate
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Step-by-step explanation:

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3 years ago
(A)using geometry vocabulary, describe a sequence of transformations that maps figure P (-1,2)(-1,4) (-4,2) (-4,4) onto figure Q
andrey2020 [161]

Before we proceed on determining the transformation happening on this problem, it's better to see first the location of the figure by drawing it in a cartesian coordinate plane. We have

If we observe the figures and the coordinates of the plot, we can see that there is a difference of 1 on the x coordinates of P and y coordinates of Q. Therefore, the first transformation that we consider here is the movement of figure P by 1 unit to the left. We have

\begin{gathered} P_1=(-1-1,2_{})=(-2,2) \\ P_2=(-1-1,4)=(-2,4) \\ P_3=(-4-1,2)=(-5,2) \\ P_4=(-4-1,4)=(-5,4) \end{gathered}

This transformation changes the location of figure P into

The next transformation will be the rotation of the red dotted figure on the figure above by 90 degrees counterclockwise. With this transformation, the coordinates will transform as

P_{ccw,90}=(-y,x)

Hence, for the rotation, we have the new coordinates.

\begin{gathered} P_1^{\prime}=(-2,-2) \\ P_2^{\prime}=(-4,-2) \\ P_3^{\prime}=(-2,-5) \\ P_4^{\prime}=(-4,-5) \end{gathered}

The transformed image, which is represented as NMPO, will now be at

For the last transformation, we will be reflecting the figure NMPO over the <em>y</em> axis. This changes the coordinates as

P_{\text{rotation,y}-\text{axis}}=(-x,y)

We now have the new coordinates:

\begin{gathered} P^{\doubleprime}_1=(2,-2)=Q_1_{}_{} \\ P_2^{\doubleprime}=(4,-2)=Q_3 \\ P_3^{\doubleprime}=(2,-5)=Q_2 \\ P_4^{\doubleprime}_{}=(4,-5)=Q_4_{} \end{gathered}

As you can see, they have the same coordinates as figure Q.

The mapping rules for the sequence described above are as follows:

First transformation (moving one unit to the left (x-1,y))

\begin{gathered} P_1(-1,2)\rightarrow P_1(-1-1,2)\rightarrow P_1(-2,2) \\ P_2(-1,4)\rightarrow P_1(-1-1,4)\rightarrow P_2(-2,4) \\ P_3(-4,2)\rightarrow P_1(-4-1,2)\rightarrow P_3(-5,2) \\ P_4(-4,4)\rightarrow P_1(-4-1,4)\rightarrow P_4(-5,4) \end{gathered}

Second transformation (rotation counter clockwise (-y,x))

\begin{gathered} P_1(-2,2)\rightarrow P^{\prime}_1(-2,-2)_{} \\ P_2(-2,4)\rightarrow P^{\prime}_2(-4,-2) \\ P_3(-5,2)\rightarrow P^{\prime}_3(-2,-5)_{} \\ P_4(-5,4)\rightarrow P^{\prime}_4(-4,-5)_{} \end{gathered}

Third Transformation (reflection over y-axis (-x,y))

\begin{gathered} P^{\prime}_1(-2,-2)\rightarrow P^{\doubleprime}_1(-(-2),-2)\rightarrow P^{\doubleprime}_1=(2,-2)=Q_1 \\ P^{\prime}_2(-4,-2)\rightarrow P^{\doubleprime}_1(-(-4),-2)\rightarrow P^{\doubleprime}_1=(4,-2)=Q_3 \\ P^{\prime}_3(-2,-5)\rightarrow P^{\doubleprime}_1(-(-2),-5)\rightarrow P^{\doubleprime}_1=(2,-5)=Q_2 \\ P^{\prime}_4(-4,-5)\rightarrow P^{\doubleprime}_1(-(-4),-5)\rightarrow P^{\doubleprime}_1=(4,-5)=Q_4 \end{gathered}

7 0
1 year ago
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