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Musya8 [376]
2 years ago
11

3v+4(v-2)= -29 solve for v​

Mathematics
1 answer:
Lesechka [4]2 years ago
4 0

Answer:

v = -3

Step-by-step explanation:

1) Expand.

3v + 4v - 8 = -29

2) Simplify 3v + 4v - 8 to 7v - 8.

7v - 8 = -29

3) Add 8 to both sides.

7v = -29 + 8

4) Simplify -29 + 8 to -21.

7v = -21

5) Divide both sides by 7.

v = -21/7

6) Simplify 21/7 to 3.

v = -3

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A tablet PC contains 3217 music files. The distribution of file size is highly skewed with many small files. Suppose the true me
m_a_m_a [10]

Answer:

Let X the random variable who represents the file sizeof music. We know the following info:

\mu =2.3,\sigma =3.25

We select a sample of n=50 nails. That represent the sample size.  

Since the sample size is large enough n >30, we can use the central limit theorem. From this theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And we can approximate the distribution of the sample mean as a normal distribution and no matter if the distribution for X is right skewed or no.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Let X the random variable who represents the file sizeof music. We know the following info:

\mu =2.3,\sigma =3.25

We select a sample of n=50 nails. That represent the sample size.  

Since the sample size is large enough n >30, we can use the central limit theorem. From this theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And we can approximate the distribution of the sample mean as a normal distribution and no matter if the distribution for X is right skewed or no.

8 0
3 years ago
How tall is a tree which is 15 feet shorter than a pole that is 3 times as tall as the tree?
Citrus2011 [14]
<h2>Height of tree is 7.5 feet</h2>

Step-by-step explanation:

Let t be the height of pole.

Given that the tree is 5 feet shorter than a pole.

Height of pole = t + 15

Also given that the pole is 3 times as tall as the tree.

Height of pole = 3t

So we have

              t + 15 = 3t

                 2t = 15

                   t = 7.5 feet

Height of tree = 7.5 feet

7 0
3 years ago
Jerome found the lengths of each side of triangle QRS as shown, but did not simplify his answers. Simplify the lengths of each s
emmainna [20.7K]

Answer:

Triangle QRS is an isosceles triangle because QR = RS.

Step-by-step explanation:

option D for Edgu hope this helps :)

9 0
3 years ago
Read 2 more answers
Can.anyone.help.....
alina1380 [7]

Answer:

0.8

Step-by-step explanation:

4/5 which is 0.8, or cross multiply to get the answer.

3 0
3 years ago
In microbiology, colony-forming units (CFUs) are used to measure the number of microorganisms present in a sample. To determine
zimovet [89]

Answer:

(a)  0.2650

(b)  0.0111

(c)  0.0105

(d)  0.0006

Step-by-step explanation:

Given that:

In microbiology, colony-forming units (CFUs) are used to measure the number of microorganisms present in a sample.

Suppose that the number of CFUs that appear after incubation follows a Poisson distribution with mean μ = 15.       &;

If the area of the agar plate is 75cm²;

what is the probability  of observing fewer than 4 CFUs in a 25 cm² area of the plate.

We can determine the mean number of CFUs that appear on a 25cm² area of the plate as follows;

75cm²/25cm² = 3

Since;

mean  μ = 15  

mean number of CFUs that appear on a 25cm² = 15/3 = 5 CFUs

Thus ; the probability of observing fewer than 4 CFUs in a 25 cm² area of the plate is estimated as:

= P(X < 4)

Using the EXCEL FUNCTION ( = poisson.dist(3, 5, TRUE) )

we have ;

P(X < 4) = 0.2650

b) If you were to count the total number of CFUs in 5 plates, what is the probability you would observe more than 95 CFUs?

Given that the total number of CFUs = 5 plates; then the mean number of CFUs in 5 plates =  15×5 = 75 CFUs

The probability is therefore = P( X > 95 )

= 1 - P(X ≤ 95)

= 1 - poisson.dist(95,75,TRUE) ( by using the excel function)

= 0.0111

c) Repeat the probability calculation in part (b) but now use the normal approximation.

Let assume that the mean and the variance of the poisson distribution are equal

Then;

X \sim N (\mu = 75 , \sigma^2 = 75)

We are to repeated the probability calculation in part (b) from above;

So:

P( X > 95 )

use the normal approximation

From standard normal variable table:

P(Z > 2.3094)

Using normal table

P(Z > 2.3094) = 0.0105

(d)  Find the difference between this value and your answer in part (b).

So;

the difference between the value in part c and part b is;

=  0.0111 - 0.0105

= 6*10^{-4}

= 0.0006 to four decimal places

5 0
3 years ago
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