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saveliy_v [14]
3 years ago
11

Need answer thx in advanace!

Mathematics
1 answer:
Andrews [41]3 years ago
4 0

Answer:

13

Step-by-step explanation:

The term is in an arithmetic progression.

That means there is 6 between each term and the next one.

Givens

a = -30

d = 6

last term an = 42

Formula

L = a + (n - 1)*d

Solution

42 = -30 + (n -1 )*6               Add 30 to both sides

42+30 = (n - 1) * 6                Combine the terms on the left.

72 = (n - 1)*6                        Divide both sides by 6

72/6 = n - 1

12     = n - 1                          Add 1 to both sides

12 + 1 = n

n = 13

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AlexFokin [52]

C

100 divided by 2 (boy and girl) is 50, then 4 goes into 100 25 times so a chance of being a girl in a family of 4 is 25% or .25

5 0
2 years ago
A probability experiment is conducted in which the sample space of the experiment is upper s equals startset 4 comma 5 comma 6 c
olchik [2.2K]
The correct question statement is:

A probability experiment is conducted in which the sample space of the experiment is S = {4,5,6,7,8,9,10,11,12,13,14,15}. Let event E={7,8,9,10,11,12,13,14,15}. Assume each outcome is equally likely. List the outcomes in E^{c}. Find P(E^{c}).

Solution:

Part 1:

E^{c} means compliment of the set E. A compliment of a set can be obtained by finding the difference of the set from the universal set. The universal set is the set which contains all the possible outcomes of the events which is S in this case.

So, compliment of E will be equal to S - E. S - E will result in all those elements of S which are not present in E. So, we can write:

E^{c}=S-E \\  \\ 
E^{c}=(4,5,6,7,8,9,10,11,12,13,14,15)-(7,8,9,10,11,12,13,14,15) \\  \\ 
E^{c}=(4,5,6)

Thus the set compliment of E will contain the elements {4,5,6}.So

E^{c} = {4,5,6}

Part 2)

P(E^{c}) means probability that if we select any number from the Sample Space S, it will belong the set E compliment.

P(E^{c}) = (Number of Elements in E^{c})/Number of elements in S

Number of elements in set S = n(S) = 12
Number of elements in set E^{c} = n(E^{c})=3

So, 

P(E^{c})= \frac{n(E^{c}) }{n(S)} \\  \\ 
P(E^{c})= \frac{3}{12} \\  \\ 
P(E^{c})= \frac{1}{4}
7 0
3 years ago
The nicotine content in a single cigarette of a particular brand has a distribution with mean 0.9 mg and standard deviation 0.1
Ad libitum [116K]

Answer:

The answer to the question is;

The probability that the resulting sample mean of nicotine content will be less than 0.89 is 0.1587 or 15.87 %.

Step-by-step explanation:

The mean of the distribution = 0.9 mg

The standard deviation of the sample = 0.1 mg

The size of the sample = 100

The mean of he sample = 0.89

The z score for sample mean is given by

Z =\frac{X-\mu}{\sigma/ \sqrt{n} } where

X = Mean of the sample

μ = Mean of the population

σ = Standard deviation of the population

Therefore Z = \frac{0.89-0.90}{0.1/\sqrt{100} } = -1

From the standard probabilities table we have the probability for  a z value of -1.0 = 0.1587

Therefore the probability that the resulting sample mean will be less than 0.89 = 0.1587 That is the probability that the mean is will be less than 0.89 is 15.87 % probability.

             

7 0
4 years ago
Malia works at a movie theater and is paid hourly. One month, she earned $388.80. The total amount she earns can be found using
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Malia : y = 8.1x.....this tells me she makes 8.10 per hr worked

Bill : if he earned 144.90 for working 18 hrs.....144.90/18 = 8.05 per hr worked.

so Malia makes 5 cents more per hr worked.

Is that the answer ur looking for ?
4 0
4 years ago
Read 2 more answers
Distribute. -3(-2x - 1)
ANTONII [103]

I believe the answer should be 6x+3

6 0
3 years ago
Read 2 more answers
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