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Scrat [10]
3 years ago
12

Anyone wanna have a convo? If you keep the conversation up I'll give brainliest :)

Mathematics
2 answers:
Dmitry [639]3 years ago
8 0

Answer:

Step-by-step explanation:

uhm sure

Mnenie [13.5K]3 years ago
7 0

Answer:

I will have a convo with you no prob's

Step-by-step explanation:

You might be interested in
I dont know how to find x​
Verdich [7]
Simplify both sides:

x
—- = -9
4

now isolate the variable (x):

x= -9 • 4
x= -36
4 0
3 years ago
What is the 93rd term of the arithmetic sequence -6, 13, 23 and how do I find it?​
e-lub [12.9K]

Answer:

Step-by-step explanation:

the formula for an arithmetic sequence is

a, a+d,a+3d,a+3d etc, where d is the common difference

we have the terms -6, 13,23

first term is -6

-6+19=13

however, 13+10=23

this is not an arithmetic sequence

7 0
3 years ago
3x + 12 = -3 What is the value of X
Blizzard [7]

Answer:

x = -5

Step-by-step explanation:

i took algebra a couple years ago :)

4 0
3 years ago
Read 2 more answers
We roll a fair die repeatedly until we see the number four appear and then we stop. The outcome of the experiment is the number
muminat

Answer:

0

Step-by-step explanation:

given that we roll a fair die repeatedly until we see the number four appear and then we stop.

the number 4 can appear either in I throw, or II throw or .... indefinitely

So X = the no of throws can be from 1 to infinity

This is a discrete distribution countable.

Sample space= {1,2,.....}

b) Prob ( 4 never appears) = Prob (any other number appears in all throws)

= \frac{5}{6} *\frac{5}{6} *\frac{5}{6} *\frac{5}{6} *...\\=(\frac{5}{6} )^n

where n is the number of throws

As n tends to infinity, this becomes 0 because 5/6 is less than 1.

Hence this probability is approximately 0

Or definitely 4 will appear atleast once.

8 0
3 years ago
Assume the hold time of callers to a cable company is normally distributed with a mean of 5.5 minutes and a standard deviation o
Ierofanga [76]

Answer:

The percent of callers are 37.21 who are on hold.

Step-by-step explanation:

Given:

A normally distributed data.

Mean of the data, \mu = 5.5 mins

Standard deviation, \sigma = 0.4 mins

We have to find the callers percentage who are on hold between 5.4 and 5.8 mins.

Lets find z-score on each raw score.

⇒ z_1=\frac{x_1-\mu}{\sigma}   ...raw score,x_1 = 5.4

⇒ Plugging the values.

⇒ z_1=\frac{5.4-5.5}{0.4}

⇒ z_1=-0.25  

For raw score 5.5 the z score is.

⇒ z_2=\frac{5.8-5.5}{0.4}  

⇒ z_2=0.75

Now we have to look upon the values from Z score table and arrange them in probability terms then convert it into percentages.

We have to work with P(5.4<z<5.8).

⇒ P(5.4

⇒ P(-0.25

⇒ P(z

⇒ z(1.5)=0.7734 and z(-0.25)=0.4013.<em>..from z -score table.</em>

⇒ 0.7734-0.4013

⇒ 0.3721

To find the percentage we have to multiply with 100.

⇒ 0.3721\times 100

⇒ 37.21 %

The percent of callers who are on hold between 5.4 minutes to 5.8 minutes is 37.21

4 0
3 years ago
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