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2 years ago

This Venn diagram shows sports played by 10 students

Mathematics

2 answers:

exis [7]2 years ago

7 0

Answer:

B. 1/10 or 0.10

Step-by-step explanation:

The question asks what's the probability that a student picked randomly will be playing both basketball and soccer.

The answer is right in the diagram.

We have only one student who plays both basketball and soccer: Ella

Since we have 10 students in the selected group, the probably you'll pick Ella is:

1 / 10 = 0.10 = 10%

So, the answer is B.

Kipish [7]2 years ago

5 0

Answer:

Just did it A

Step-by-step explanation:

belive it

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A school principal has to choose randomly among the six best students in each grade to be the school captain every month. For th

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Since there are 6 students out of which one needs to be selected, the principal chose two die on which there are six numbers each numbered from 1 , 2, 3, 4, 5, 6.

Since there are two dice, the total possible outcome is 36.

Hence, the probability of getting one number each is 1/36

Hence, the principal used a fair method because each result is an equally likely possible outcome.

Option B is correct.

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Un equipo de fútbol se conforma de cuatro líneas de jugadores: portero (9%), defensas (37%), medios (27%) y delanteros (27%), do

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Answer:

According to the proble, the total number of goals are 1+11+15+23 = 50: 1 by the goalkeeper, 11 by defense, 15 by midfielders and 23 by strikers.

So, each probability can be found by using standard probabilities

$P=\frac{events}{total outcomes}$

<h3>(a) Defense</h3>

$P=\frac{11}{50} \approx 0.22$

Therefore, the defense has a probability of 22% of score that goal.

<h3>(b) Midfielders</h3>

$P=\frac{15}{50} \approx 0.30$

Therefore, midfielders have a probability of 30% of scoring that goal.

<h3>(c) Strikers.</h3>

$P=\frac{23}{50} \approx 0.46$

Therefore, strikers have a probability of 46% of scoring that goal.

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2 years ago

The Acme Candy Company claims that 60% of the jawbreakers it produces weigh more than 0.4 ounces. Suppose that 800 jawbreakers

vichka [17]

Answer:

Yes, it would be statistically significant

Step-by-step explanation:

The information given are;

The percentage of jawbreakers it produces that weigh more than 0.4 ounces = 60%

Number of jawbreakers in the sample, n = 800

The mean proportion of jawbreakers that weigh more than 0.4 = 60% = 0.6 = $\mu _ {\hat p}$ =p

The formula for the standard deviation of a proportion is $\sigma _{\hat p} =\sqrt{\dfrac{p(1-p)}{n} }$

Solving for the standard deviation gives;

$\sigma _{\hat p} =\sqrt{\dfrac{0.6 \cdot (1-0.6)}{800} } = 0.0173$

Given that the mean proportion is 0.6, the expected value of jawbreakers that weigh more than 0.4 in the sample of 800 = 800*0.6 = 480

For statistical significance the difference from the mean = 2× $\sigma _{\hat p}$ = 2*0.0173 = 0.0346 the equivalent number of Jaw breakers = 800*0.0346 = 27.7