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Vlada [557]
2 years ago
15

Mia, jade, and Amelia sell car wash tickets. Jade sells 8 more tickets then Mia. Amelia sells twice as many tickets as Mia. The

friends sell a total of 296 tickets. How many tickets does each friend sell?
Mathematics
1 answer:
Ede4ka [16]2 years ago
8 0

Answer:

Mia sold 72 tickets, Jade sold 80 tickets, and Amelia sold 144 tickets.

Step-by-step explanation:

You can solve this question by coming up with equations for each friend and then substituting them into the equation M + J + A = 296

Jade sells 8 more tickets than Mia, so Jade's equation for how many tickets she sold is the amount of tickets that Mia sold + 8:

J = M + 8

Amelia sells twice as many tickets as Mia, so her equation is the amount of tickets Mia sold * 2:

A = 2 * M

Now we can substitute these equations into the overall one.

M + (M + 8) + (2M) = 296

Now we can combine like terms and solve for M.

2M + 8 + 2M = 296

4M + 8 = 296

4M = 288

M = 72

We now have the amount of tickets Mia sold, so we can insert this number into the equations for Amelia and Jade.

J = M + 8

J = 72 + 8

J = 80

A = M * 2

A = 72 * 2

A = 144

So Mia sold 72 tickets, Jade sold 80 tickets, and Amelia sold 144 tickets.

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Answer:

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The measures of spread start with 55 to 72 for 'restaurant a' where the upper and lower quartiles  are 70-60 showing a distribution 10-17 into the whisker plot.

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For the outer quartile the comparison to restaurant a is twice as small meaning restaurant a is greater by over double and shown as   53 <a < 60 which when showing both outer quartiles we can use the amounts being 7 and 70<a<72 = 2

and show closed dots on equality line number line separately showing a<-7  and a>2 for each outer exterior quartile. For restaurant b this shows opposite similar equalities 65<b< 68 = b>-3 and   82<b<89 = b>7

So the differences are smaller for b for left side by 4 and larger for b by 5 up on the right side.

We compare both to restaurant c and find c =   60<c<61 = c>-1 and 70<c<71 = c>1 so the differences are that the outer quartile for c through the other quartiles = -2> c < -6 . This shows how smaller c is compared to a and b outer quartiles). We can also prove that while the outer quartiles are much more smaller for c than a and b we cna prove that c actually has an inner quartile more similar to c and closer distribution of b as the median is more closer for a and c where b has a greater output for median as restaurant b has the higher fat content and greater distribution within the inner quartiles over all.

Summary findings Restaurant c outer quartile is moderately skewed as they show -1 -0.5  on the left side and 1-0.5 on the right side. Restaurant a has a closer median inner quartile to c and closer distribution of inner distribution of c. It's output outer quartile distribiution is a distribution that is higher than b but a smaller inner quartile compared to b, when this happens then the distribution spread shows less range and fewer products to account for.

So i think restaurant b has the higher fat content to its menu.

Where restaurant a must be the healthiest as it holds the lowest range and larger gap is such range for healthier food.ie) when compared to the others. 

Meanings of what we are asked.

In a box and whisker plot: the ends of the box are the upper and lower quartiles, so the box spans the interquartile range. the median is marked by a vertical line inside the box. the whiskers are the two lines outside the box that extend to the highest and lowest observations.

Skewness refers to distortion or asymmetry in a symmetrical bell curve, or normal distribution, in a set of data. If the curve is shifted to the left or to the right, it is said to be skewed. Skewness can be quantified as a representation of the extent to which a given distribution varies from a normal distribution.

As a general rule of thumb:

If skewness is less than -1 or greater than 1, the distribution is highly skewed.

If skewness is between -1 and -0.5 or between 0.5 and 1, the distribution is moderately skewed.

If skewness is between -0.5 and 0.5, the distribution is approximately symmetric.

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