Can someone please help me for question 9, 10, 11, 12, and 13?

When constructing parallel lines with a compass and straight edge how should you start construction

Andrews [41]

Draw two lines that never intersects and those are parallel line. A compass and straightedge helps you make the lines straight.

8 0

3 years ago

For a soccer game Mr Morant prepared 32 hot dogs he sol1/2 to the fans

Umnica [9.8K]

That’s a lot of hot dogs. Where’s the rest of the problem?

8 0

3 years ago

The probability that a lab specimen contains high levels of contamination is 0.14. A group of 4 independent samples are checked.

Shtirlitz [24]

Answer:

a) 0.5470 = 54.70% probability that none contain high levels of contamination.

b) 0.3562 = 35.62% probability that exactly one contains high levels of contamination.

c) 0.4530 = 45.30% probability that at least one contains high levels of contamination.

Step-by-step explanation:

For each sample, there are only two possible outcomes. Either they contain high levels of contamination, or they do not. The samples are independent. So we use the binomial probability distribution to solve this question.

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.

The probability that a lab specimen contains high levels of contamination is 0.14.

This means that $p = 0.14$

A group of 4 independent samples are checked.

This means that $n = 4$

(a) What is the probability that none contain high levels of contamination?

This is P(X = 0)

$P(X = 0) = C_{4,0}.(0.14)^{0}.(0.86)^{4} = 0.5470$

0.5470 = 54.70% probability that none contain high levels of contamination.

(b) What is the probability that exactly one contains high levels of contamination?

This is P(X = 1)

$P(X = 1) = C_{4,1}.(0.14)^{1}.(0.86)^{3} = 0.3562$

0.3562 = 35.62% probability that exactly one contains high levels of contamination.

(c) What is the probability that at least one contains high levels of contamination?

Either none of the samples contain high levels of contamination, or at least one does. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

We want to find $P(X \geq 1)$ . So

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.5470 = 0.4530$

0.4530 = 45.30% probability that at least one contains high levels of contamination.

8 0

3 years ago

Sanjay sold 2/5 of his mangoes. What percentage of his mangoes did he remain with?

Zinaida [17]

Answer:

60%

Step-by-step explanation:

3/5 left x 100 = 60%

4 0

3 years ago

Read 2 more answers

Identify the phrase that cannot be described by the same absolute value as the other three. Explain your reasoning.

r-ruslan [8.4K]

The answer would probably be 18 degrees below normal because the absolute values of the rest are 8 not 18 which is the absolute value of 18 degrees below normal.

6 0

4 years ago