All categories

2 years ago

Can i get some help on this question?

Mathematics

2 answers:

Yakvenalex [24]2 years ago

7 0

Divide both sides by 8.5 to get b on its own

68/8.5 = 8
So b = 8

Anton [14]2 years ago

4 0

Answer:

Bye I hope this helps you pass!

You might be interested in

What does the y-intercept represent?

lara31 [8.8K]

Where a line goes across the y axis
Or in the formula y = mx + c, the y axis is represented by c

8 0

2 years ago

Read 2 more answers

Please someone help quickly thank you very much

Tema [17]

Answer:

I think the answer is 81.

Step-by-step explanation:

Vertical angles are congruent.

x + 3 = 84

- 3 = -3

--------------

x = 81

3 0

3 years ago

Two integers from 1 through 40 are chosen by a random number generator. write your answer as a fraction in simplest form a/b. p(

jasenka [17]

Probability is 1/40.

5 0

3 years ago

The mean of (x + 5), (x + 3), (2x + 1) and (x - 6) is 12. What is the<br> value of x?

navik [9.2K]

Answer:

x=9

Step-by-step explanation:

answers too short so cigdtusrysyksrytf7hiuf6s4td6figiio

7 0

2 years ago

Suppose that a batch of 100 items contains 6 that are defective and 94 that are not defective. If X is the number of defective i

Tcecarenko [31]

Answer:

a) P(X = 0) = 0.5223

b) P(X > 2) = 0.0125

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The order that the items are selected is not important, so the combinations formula is used to solve this problem.

Combinations formula:

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

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

In this problem, we have that

100 items

6 defective

94 not defective.

a) P{X=0}

None defective.

Desired outcomes

Combinations of 10 from a set of 94. So

$D = C_{94,10} = \frac{94!}{10!84!} = 9041256800000$

Total outcomes:

Combinations of 10 from a set of 100. So

$T = C_{100,10} = \frac{100!}{10!90!} = 17310309000000$

P(X = 0)

$p = \frac{D}{T} = \frac{9041256800000}{17310309000000} = 0.5223$

(b) P(X>2}.

Either two or less are defective, or more than two are defective. The sum of the probabilities of these events is decimal 1. So

$P(X \leq 2) + P(X >2) = 1$

$P(X > 2) = 1 - P(X \leq 2)$

In which

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

P(X = 0)

$P(X = 0) = \frac{D}{T} = \frac{9041256800000}{17310309000000} = 0.5223$

P(X = 1)

Desired outcomes

Combinations of 9 from a set of 94(non defecive) and 1 from a set of 6(defective). So

$D = C_{84,9}*C_{6,1} = \frac{94!}{9!85!}*\frac{6!}{1!5!} = 6382063700000$

Total outcomes:

Combinations of 10 from a set of 100. So

$T = C_{100,10} = \frac{100!}{10!90!} = 17310309000000$

P(X = 1)

$P(X = 1) = \frac{D}{T} = \frac{6382063700000}{17310309000000} = 0.3687$

P(X = 2)

Desired outcomes

Combinations of 8 from a set of 94(non defecive) and 2 from a set of 6(defective). So

$D = C_{94,8}*C_{6,2} = \frac{94!}{8!86!}*\frac{6!}{2!4!} = 1669726000000$

Total outcomes:

Combinations of 10 from a set of 100. So

$T = C_{100,10} = \frac{100!}{10!90!} = 17310309000000$

P(X = 2)

$P(X = 2) = \frac{D}{T} = \frac{1669726000000}{17310309000000} = 0.0965$

Finally

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.5223 + 0.3687 + 0.0965 = 0.9875$

$P(X > 2) = 1 - P(X \leq 2) = 1 - 0.9875 = 0.0125$

4 0

2 years ago