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andreev551 [17]
3 years ago
11

a) Read section 1.5 in the Yakir textbook. If you were a teacher and had 30 students in your class and wanted to know the class

average on the first quiz, would you use a parameter or a statistic
Mathematics
1 answer:
OlgaM077 [116]3 years ago
7 0

Answer:

Parameter

Step-by-step explanation:

Required

Parameter of Statistic

From the question, we understand that the teacher is to calculate the class average.

To calculate the class average, the teacher will use the mean function/formula, which is calculated as:

Mean = \frac{\sum x}{n}

Generally, mean is an example of a parameter.

<em>So, we can conclude that the teacher will use parameer</em>

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A fair coin is flipped 10 times. What is the probability that exactly 4 of the 10 flips come up heads
kifflom [539]

105/2^9

Step-by-step explanation:

The probability of getting a head in a single toss

p=12

The probability of not getting a head in a single toss

q=1−12=12

Now, using Binomial theorem of probability,

The probability of getting exactly r=4 heads in total n=10 tosses

=10C4(1/2)4(1/2)10−4

=10×9×8×7/4! 1/2^4 1/2^6

=2^44⋅9⋅35/24(2^10)

=105/2^9

5 0
2 years ago
Read 2 more answers
PLEASE HELP!!! WILL AWARD BRAINLIEST!!! you work a part-time job earning 6.80/hr with tips that average 2.30/hr you work 20 hr/w
solmaris [256]
143.7  I believe since you take 6.80*20=136+46=2.3*20

136+46=182

182*7.65%=168.08

168.08*8.95%=153.04

153.04*6.1%=143.7
4 0
3 years ago
Read 2 more answers
These triangles are similar.<br> True<br> Or<br> False
iren2701 [21]

Answer:

<u><em>True</em></u> because they both are triangles, they both have tree side but one is just smaller than the other.

3 0
3 years ago
The speed with which utility companies can resolve problems is very important. GTC, the Georgetown Telephone Company, reports it
brilliants [131]

Answer:

(a) 11.25 and 1.68  

(b) 0.1651

(c) 0.3903

(d) 0.6865

Step-by-step explanation:

We are given that GTC, the Georgetown Telephone Company, reports it can resolve customer problems the same day they are reported in 75% of the cases and suppose the 15 cases reported today are representative of all complaints.

This situation can be represented through Binomial distribution as;

P(X=r)= \binom{n}{r}p^{r}(1-p)^{n-r} ; x = 0,1,2,3,....

where,  n = number of trials (samples) taken = 15

             r = number of success

             p = probability of success which in our question is % of cases in

                  which customer problems are resolved on the same day, i.e.;75%

So, here X ~ Binom(n=15,p=0.75)

(a) Expected number of problems to be resolved today = E(X)

            E(X) = \mu = n * p = 15 * 0.75 = 11.25

    Standard deviation = \sigma = \sqrt{n*p*(1-p)} = \sqrt{15*0.75*(1-0.75)} = 1.68

(b) Probability that 10 of the problems can be resolved today = P(X = 10)

     P(X = 10) = \binom{15}{10}0.75^{10}(1-0.75)^{15-10}

                    = 3003*0.75^{10} *0.25^{5} = 0.1651

(c) Probability that 10 or 11 of the problems can be resolved today is given by = P(X = 10) + P(X = 11)

    = \binom{15}{10}0.75^{10}(1-0.75)^{15-10}+\binom{15}{11}0.75^{11}(1-0.75)^{15-11}

    = 3003*0.75^{10} *0.25^{5} + 1365*0.75^{11} *0.25^{4} = 0.3903

(d) Probability that more than 10 of the problems can be resolved today is

    given by = P(X > 10)

P(X > 10) = P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)  

= \binom{15}{11}0.75^{11}(1-0.75)^{15-11}+\binom{15}{12}0.75^{12}(1-0.75)^{15-12} + \binom{15}{13}0.75^{13}(1-0.75)^{15-13}+\binom{15}{14}0.75^{14}(1-0.75)^{15-14} + \binom{15}{15}0.75^{15}(1-0.75)^{15-15}

= 1365*0.75^{11} *0.25^{4} + 455*0.75^{12} *0.25^{3}+105*0.75^{13} *0.25^{2} + 15*0.75^{14} *0.25^{1}+1*0.75^{15} *0.25^{0}

= 0.6865

3 0
3 years ago
Endent Practice
Vikki [24]

Answer:

Part A: The unit that would be best to describes the situations is to the mile

Part B: 4 significant figures

Part C: 1,450 miles

Part D: He is not correct

Step-by-step explanation:

Part A: By unit conversion, we have;

1 mile = 1,760 yards

Given that the distance from Ottawa, Canada to Jackson Mississippi is 2,552,000 yards it would best simplified into miles

Part B: The number of significant digits in the reporters estimate are the non-zero digits before the zeros in the number figure which are 2, 5, 5, and 2 or 4 significant figures

Part C: 2,552,000 yards = 2,552,000 yards/(1,760 yards/mile) = 1,450 miles

Part D: Given that the distance from Jackson to Tucson = 7,159,680 feet, we have;

1 mile = 5,280 feet

Therefore;

7,159,680 feet = 7,159,680 feet/(5,280 feet/mile) = 1,356 miles

Therefore, the claim that the distance from Jackson to Tucson which is 1,356 miles is further than the distance from Ottawa, Canada to Jackson Mississippi which is 1,450 miles is not correct because;

The distance from Jackson to Tucson 1,356 miles \ngtr 1,450 miles which is the distance from Jackson to Ottawa

6 0
3 years ago
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