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3 years ago

The two-way table shows the ages of the players on different sOCcer teams.

Mathematics

1 answer:

Tomtit [17]3 years ago

6 0

**Answer**
The true statement is: The probability that a randomly selected player on Team C is 10 years old is 5/16

please mark me brainliest, happy holidays :)

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How would the number be written in standard form? 30,000+500+70+6

Lisa [10]

30,576 because 30 fills the thousands and 5,7,6 fill the rest

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4 years ago

Read 2 more answers

a solar lease customer built up an excess of 6,500 kilowatts hour (kwh) during the summer using his solar panels. when he turned

Jlenok [28]

Question:

A solar lease customer built up an excess of 6,500 kilowatts hour (kwh) during the summer using his solar panels. when he turned his electric heat on, the excess be used up at 50 kilowatts hours per day .

(a) If E represents the excess left and d represent the number of days. Write an equation for E in terms of d

(b) How much of excess will be left after one month (1 month = 30 days)

Answer:

a. $E = 6500 - 50d$

b. $E = 5000$

Step-by-step explanation:

Given

Excess = 6500kwh

Rate = 50kwh/day

Solving (a): E in terms of d

The Excess left (E) in d days is calculated using:

$E = Excess - Rate * days$

The expression uses minus because there's a reduction in the excess kwh on a daily basis.

Substitute values for Excess, Rate and days

$E = 6500 - 50 * d$

$E = 6500 - 50d$

Solving (b); The value of E when d = 30.

Substitute 30 for d in $E = 6500 - 50d$

$E = 6500 - 50 * 30$

$E = 6500 - 1500$

$E = 5000$

<em>Hence, there are 5000kwh left after 30 days</em>

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3 years ago

Convert 12.4 hours to hours and minutes.

Katen [24]

Answer:

12.4=12 hours and 24 minutes

Step-by-step explanation:

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3 years ago

Read 2 more answers

Suppose the number of children in a household has a binomial distribution with parameters n=12n=12 and p=50p=50%. Find the proba

nadya68 [22]

Answer:

a) 20.95% probability of a household having 2 or 5 children.

b) 7.29% probability of a household having 3 or fewer children.

c) 19.37% probability of a household having 8 or more children.

d) 19.37% probability of a household having fewer than 5 children.

e) 92.71% probability of a household having more than 3 children.

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem, we have that:

$n = 12, p = 0.5$

(a) 2 or 5 children

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 5) = C_{12,5}.(0.5)^{5}.(0.5)^{7} = 0.1934$

$p = P(X = 2) + P(X = 5) = 0.0161 + 0.1934 = 0.2095$

20.95% probability of a household having 2 or 5 children.

(b) 3 or fewer children

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{12,0}.(0.5)^{0}.(0.5)^{12} = 0.0002$

$P(X = 1) = C_{12,1}.(0.5)^{1}.(0.5)^{11} = 0.0029$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.0537$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0002 + 0.0029 + 0.0161 + 0.0537 = 0.0729$

7.29% probability of a household having 3 or fewer children.

(c) 8 or more children

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 8) = C_{12,8}.(0.5)^{8}.(0.5)^{4} = 0.1208$

$P(X = 9) = C_{12,9}.(0.5)^{9}.(0.5)^{3} = 0.0537$

$P(X = 10) = C_{12,10}.(0.5)^{10}.(0.5)^{2} = 0.0161$

$P(X = 11) = C_{12,11}.(0.5)^{11}.(0.5)^{1} = 0.0029$

$P(X = 12) = C_{12,12}.(0.5)^{12}.(0.5)^{0} = 0.0002$

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.1208 + 0.0537 + 0.0161 + 0.0029 + 0.0002 = 0.1937$

19.37% probability of a household having 8 or more children.

(d) fewer than 5 children

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{12,0}.(0.5)^{0}.(0.5)^{12} = 0.0002$

$P(X = 1) = C_{12,1}.(0.5)^{1}.(0.5)^{11} = 0.0029$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.0537$

$P(X = 4) = C_{12,4}.(0.5)^{4}.(0.5)^{8} = 0.1208$

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0002 + 0.0029 + 0.0161 + 0.0537 + 0.1208 = 0.1937$

19.37% probability of a household having fewer than 5 children.

(e) more than 3 children

Either a household has 3 or fewer children, or it has more than 3. The sum of these probabilities is 100%.

From b)

7.29% probability of a household having 3 or fewer children.

p + 7.29 = 100

p = 92.71

92.71% probability of a household having more than 3 children.

5 0

3 years ago

An isosceles triangle has side lengths of 15.6 units and 20.33 units. What is the greatest possible value of the perimeter of th

ArbitrLikvidat [17]

Answer:

56.26

Step-by-step explanation:

An isosceles triangle has two sides that are the same length, and one side that is the same length. We have two sizes, so two find the greatest possible perimeter we must assume that the greater size is the value for the two sides, and the smaller size the value for the single side. In other words:

$x = 15.6 + 2(20.33)$

$x = 56.26$

6 0

2 years ago