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Vlada [557]
3 years ago
13

The trifecta at most racetracks consists of selecting the​ first-, second-, and​ third-place finishers in a particular race in t

heir proper order. If there are nine entries in the trifecta​ race, how many tickets must you purchase to guarantee a​ win?
Mathematics
1 answer:
ziro4ka [17]3 years ago
3 0

The total tickets to be purchased to guarantee the win =  504 tickets

Step-by-step explanation:

Step 1 :

Number of entries in the trifecta race = 9

The win is to select the first finisher, second finisher and third finisher in their proper order.

We need to find the number of tickets to be purchased to guarantee the win

Step 2 :

Number of ways to select the first finisher = 9

Number of ways to select the second finisher = 8 [the first is selected and fixed. So the number of available finishes is reduced by 1]

Number of ways to select the third finisher = 7

Hence the total tickets to be purchased to guarantee the win = 9 × 8 × 7 = 504

Step 3 :

Answer :

The total tickets to be purchased to guarantee the win =  504 tickets

             

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7 0
3 years ago
2. For a particular model of a car, a dealer offers 3 sizes of engines, 2 types of stereos, 18 body
Gnesinka [82]

The given details follow the product rule. The different model of the car available are 756.

Given that:

Engines = 3

Stereos = 2

Body=18

Color = 7

The different possibility is calculated using the following product rule:

Total =Engines \times Stereos \times Body \times Color

So, we have:

Total = 3\times 2\times 18 \times 7

Total = 756

Hence, there are 756 different models available.

Read more about product rules at:

brainly.com/question/3944602

3 0
2 years ago
Please help to answer this question I’m really confused
igor_vitrenko [27]

Answer:

It would be B!

Step-by-step explanation:

D has an obtuse angle, so it can be eliminated.

C has angles that are not vertical, but rather adjacent. Same goes for A!

Hope this helps!

3 0
2 years ago
Please help, due tomorrow. will give brainlist
Tatiana [17]

Step-by-step explanation:

By Remainder Theorem, f(5) = 91.

=> (5)³ - 4(5) + k = 91

=> 125 - 20 + k = 91

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8 0
3 years ago
Consider a random sample of ten children selected from a population of infants receiving antacids that contain aluminum, in orde
saveliy_v [14]

Answer:

a. Null hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is equal to the mean plasma aluminum level of the population of infants not receiving antacids.

Complementary alternative hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is different from the mean plasma aluminum level of the population of infants not receiving antacids.

b. (32.1, 42.3)

c. p-value < .00001

d. The null hypothesis is rejected at the α=0.05 significance level

e. Reformulated null hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is equal to the mean plasma aluminum level of the population of infants not receiving antacids.

Reformulated complementary alternative hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is higher than the mean plasma aluminum level of the population of infants not receiving antacids.

p-value equals < .00001. The null hypothesis is rejected at the α=0.05 significance level. This suggests that being given antacids<em> </em>greatly increases the plasma aluminum levels of children.

Step-by-step explanation:

a. Null hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is equal to the mean plasma aluminum level of the population of infants not receiving antacids.

Complementary alternative hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is different from the mean plasma aluminum level of the population of infants not receiving antacids. This may imply that being given antacids significantly changes the plasma aluminum level of infants.

b. Since the population standard deviation σ is unknown, we must use the t distribution to find 95% confidence limits for μ. For a t distribution with 10-1=9 degrees of freedom, 95% of the observations lie between -2.262 and 2.262. Therefore, replacing σ with s, a 95% confidence interval for the population mean μ is:

(X bar - 2.262\frac{s}{\sqrt{10} } , X bar + 2.262\frac{s}{\sqrt{10} })

Substituting in the values of X bar and s, the interval becomes:

(37.2 - 2.262\frac{7.13}{\sqrt{10} } , 37.2 + 2.262\frac{7.13}{\sqrt{10} })

or (32.1, 42.3)

c. To calculate p-value of the sample , we need to calculate the t-statistics which equals:

t=\frac{(Xbar-u)}{\frac{s}{\sqrt{10} } } = \frac{(37.2-4.13)}{\frac{7.13}{\sqrt{10} } } = 14.67.

Given two-sided test and degrees of freedom = 9, the p-value equals < .00001, which is less than 0.05.

d. The mean plasma aluminum level for the population of infants not receiving antacids is 4.13 ug/l - not a plausible value of mean plasma aluminum level for the population of infants receiving antacids. The 95% confidence interval for the population mean of infants receiving antacids is (32.1, 42.3) and does not cover the value 4.13. Therefore, the null hypothesis is rejected at the α=0.05 significance level. This suggests that being given antacids <em>greatly changes</em> the plasma aluminum levels of children.

e. Reformulated null hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is equal to the mean plasma aluminum level of the population of infants not receiving antacids.

Reformulated complementary alternative hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is higher than the mean plasma aluminum level of the population of infants not receiving antacids.

Given one-sided test and degree of freedom = 9, the p-value equals < .00001, which is less than 0.05. This result is similar to result in part (c). the null hypothesis is rejected at the α=0.05 significance level. This suggests that being given antacids<em> greatly increases</em> the plasma aluminum levels of children.

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3 years ago
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