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Pavel [41]
3 years ago
8

This season, Lisa's lacrosse team has won $\frac 23$ of their home games (games played at Lisa's school), but just $\frac 25$ of

their away games (games played at other schools). In total, Lisa's team has won $26$ games out of $49$ games they have played. How many home games has Lisa's team played? Explain how you solved the problem.
Mathematics
1 answer:
8090 [49]3 years ago
3 0
The team has played 24 home games.

Let h be the number of home games and a be the number of away games.  The total number of games is 49; this gives us
h+a=49

They won 2/3 of the home games and 2/5 of the away games; there were 26 games won.  This gives us
2/3h + 2/5a = 26

In the first equation we isolate h by subtracting a from both sides:
h+a-a=49-a
h=49-a

Substitute this into the second equation:
2/3(49-a)+2/5a = 26

Using the distributive property, we have
2/3*49 - 2/3*a + 2/5a = 26
98/3 - 2/3a + 2/5a = 26

Finding a common denominator to combine like terms, we have
98/3 - 10/15a + 6/15a = 26
98/3 - 4/15a = 26

We want to convert the whole number to thirds as well; 26 = 26*3/3 = 78/3:
98/3 - 4/15a = 78/3

Subtracting 98/3 from both sides:
98/3 - 4/15a - 98/3 = 78/3 - 98/3
-4/15a = -20/3

Divide both sides by -4/15:
a = -20/3 ÷ -4/15
a = -20/3 × - 15/4 = 300/12 = 25

There were 25 away games.

This means there were 49-25 = 24 home games.
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The valid conclusions for the manager based on the considered test is given by: Option

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One sample z-test is performed if the sample size is large enough (n  > 30) and we want to know if the sample comes from the specific population.

For this case, we're specified that:

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Forming hypotheses:

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where \mu_0 is the hypothesized population mean of the money his customer spends in his store.

The z-test statistic we get is:

z = \dfrac{\overline{x} - \mu_0}{\sigma/\sqrt{n}} = \dfrac{160 - 150}{30.20/\sqrt{40}} \approx 2.094

The test is single tailed, (right tailed).

The critical value of z at level of significance 0.025 is 1.96

Since we've got 2.904 > 1.96, so we reject the null hypothesis.

(as for right tailed test, we reject null hypothesis if the test statistic is > critical value).

Thus, we accept the alternate hypothesis that customer spends more in his store than the national average.

Learn more about one-sample z-test here:

brainly.com/question/21477856

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