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DaniilM [7]
2 years ago
15

Vincent started the year with $231 in his account at the health spa. Each time he uses the spa, $6 is taken out of the account.

How many more times can he use the spa if he has already had $48 taken out of his account this year?
Mathematics
1 answer:
SpyIntel [72]2 years ago
6 0

Answer:

Vincent would have 30.5 more visits, but since you can't have half a visit technically speaking, round up to 31 if it is allowed.

Step-by-step explanation:

$231 - $48 = $183

$183 divided by $6 = 30.5

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In a random sample of 380 cars driven at low altitudes, 42 of them exceeded a standard of 10 grams of particulate pollution per
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Answer:

The test statistic for testing if the proportion of high-altitude vehicles exceeding the standard is greater than the proportion of low-altitude vehicles exceeding the standard is 3.234.

Step-by-step explanation:

We are given that in a random sample of 380 cars driven at low altitudes, 42 of them exceeded a standard of 10 grams of particulate pollution per gallon of fuel consumed.

In an independent random sample of 90 cars driven at high altitudes, 24 of them exceeded the standard.

Let p_1 = <u><em>population proportion of cars driven at high altitudes who exceeded a standard of 10 grams</em></u>.

p_2 = <u><em>population proportion of cars driven at low altitudes who exceeded a standard of 10 grams</em></u>.

So, Null Hypothesis, H_0 : p_1\leq p_2      {means that the proportion of high-altitude vehicles exceeding the standard is smaller than or equal to the proportion of low-altitude vehicles exceeding the standard}

Alternate Hypothesis, H_A : p_1>p_2      {means that the proportion of high-altitude vehicles exceeding the standard is greater than the proportion of low-altitude vehicles exceeding the standard}

The test statistics that will be used here is <u>Two-sample z-test statistics</u> for proportions;

                             T.S.  =  \frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }  ~  N(0,1)

where, \hat p_1 = sample proportion of cars driven at high altitudes who exceeded a standard of 10 grams = \frac{24}{90} = 0.27

\hat p_2 = sample proportion of cars driven at low altitudes who exceeded a standard of 10 grams = \frac{42}{380} = 0.11

n_1 = sample of cars driven at high altitudes = 90

n_2 = sample of cars driven at low altitudes = 380

So, the test statistics =  \frac{(0.27-0.11)-(0)}{\sqrt{\frac{0.27(1-0.27)}{90}+\frac{0.11(1-0.11)}{380} } }      

                                   =  3.234

The value of z-test statistics is 3.234.

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