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3 years ago

Question 24 of 58

Mathematics

1 answer:

Misha Larkins [42]3 years ago

3 0

Answer: Mary

Mary is more trustworthy because she interviews more students, therefore has a more accurate range of answers.

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Eastern Meats pays its delivery workers regular time for up to 36 hours per week and

Cloud [144]

Answer:

42-36=6

42+6=48

545.28 / 48=11.36

11.36

8 0

3 years ago

Tan x (degrees) = 0.3<br> what is the value of x?

jenyasd209 [6]

X= shift tan 0.3
x=16.6992≈17

3 0

3 years ago

The sum of two numbers is 64 and the difference is 6. What are the numbers?

Ne4ueva [31]

Answer:

29 and 35

Step-by-step explanation:

x+y=64

x-y=6

Solve by substitution.

x=6+y

6+y+y=64

6+2y=64

y=29

Then plug in 29 for y.

x+29=64 ......35

7 0

2 years ago

Select the correct answer.

Marysya12 [62]

Your answer will be A because the absolute value of -7 is 7 that is smaller than 8 8 smaller than 9 and -10 absolute value is 10 so A displayed in the right number sequence

3 0

3 years ago

The Copy Shop has made 20 copies of a document for you. Since the defective rate is 0.1, you think there may be some defective c

Pepsi [2]

Answer:

Binomial

There is a 34.87% probability that you will encounter neither of the defective copies among the 10 you examine.

Step-by-step explanation:

For each copy of the document, there are only two possible outcomes. Either it is defective, or it is not. This means that we can solve this problem using the binomial probability distribution.

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 combinatios 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

Of the 20 copies, 2 are defective, so $p = \frac{2}{20} = 0.1$ .

What is the probability that you will encounter neither of the defective copies among the 10 you examine?

This is P(X = 0) when $n = 10$ .

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

$P(X = 0) = C_{10,0}.(0.1)^{0}.(0.9)^{10} = 0.3487$

There is a 34.87% probability that you will encounter neither of the defective copies among the 10 you examine.

8 0

3 years ago