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alexandr402 [8]
2 years ago
13

7-3y=y-6(2+1/2y) how do I solve for y

Mathematics
1 answer:
IrinaK [193]2 years ago
4 0
  -3y + 7 = y - 6(¹/₂y + 2)
  -3y + 7 = y - 6(¹/₂y) - 6(2)
  -3y + 7 = y - 3y - 12
  -3y + 7 = -2y - 12
+ 3y        + 3y
           7 = y - 12
      + 12      + 12
          19 = y
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A socket set is an integral part of a technician's tool box because ____. A. the extension allows easier access to nuts/bolts in
Ann [662]

Answer: C

Explanation:

. there are multiple sockets that can be interchanged easily to accommodate many different size nuts and bolts

8 0
3 years ago
Question: Researchers in Pakistan wanted to better understand the effects of anthracycline (a chemotherapeutic drug) on the hear
Sedbober [7]

Using the normal distribution, it is found that:

1. His z-score was of Z = -1.88.

2. There is a 0.0301 = 3.01% probability that a randomly selected person has a smaller E/A ratio than the patient in question 1.

3. Z-score of z = 1.85, there is a 0.0322 = 3.22% probability that a randomly selected patient has a higher E/A ratio.

4. Due to the higher absolute value of the z-score, the first patient had a more extraordinary result.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

  • It measures how many standard deviations the measure is from the mean.
  • After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

  • The mean is of \mu = 1.35.
  • The standard deviation is of \sigma = 0.33.

Item 1:

Considering his ratio, we have that X = 0.73, hence:

Z = \frac{X - \mu}{\sigma}

Z = \frac{0.73 - 1.35}{0.33}

Z = -1.88

His z-score was of Z = -1.88.

Item 2:

The probability is the <u>p-value of Z = -1.88</u>, hence, there is a 0.0301 = 3.01% probability that a randomly selected person has a smaller E/A ratio than the patient in question 1.

Item 3:

Score of X = 1.96, hence:

Z = \frac{X - \mu}{\sigma}

Z = \frac{1.96 - 1.35}{0.33}

Z = 1.85

The probability is 1 subtracted by the p-value of Z = 1.85, hence, 1 - 0.9678 = 0.0322 = 3.22% probability that a randomly selected patient has a higher E/A ratio.

Item 4:

Due to the higher absolute value of the z-score, the first patient had a more extraordinary result.

More can be learned about the normal distribution at brainly.com/question/24663213

7 0
2 years ago
Sarah has 27
Setler79 [48]
Sarah can make 2, however she will have 3 feet leftover
5 0
2 years ago
Read 2 more answers
A local sorority sold hot dogs and bratwursts at the spring fling picnics. The first day they sold 8 dozen hot dogs and 13 dozen
kicyunya [14]

Answer:

1 hot dog costs $0.75

1 bratwurst costs $1.35

Step-by-step explanation:

Let x and y be the price per dozen of hot dogs and bratwursts respectively.

The first day they sold 8 dozen hot dogs and 13 dozen bratwursts for ​$282.60

8x + 13y = 282.60

The second day they sold 10 dozen hot dogs and 15 dozen bratwursts for a total of ​$333.00

10x + 15y = 333

and we have the linear system

<em>8x + 13y = 282.60 </em>

<em>10x + 15y = 333 </em>

which can be written in matrix form as

\bf \left(\begin{array}{cc}8&13\\10&15\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}282.60\\333\end{array}\right)

The solution would be given by

\bf \left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{cc}8&13\\10&15\end{array}\right)^{-1}\left(\begin{array}{c}282.60\\333\end{array}\right)

We have  

\bf \left(\begin{array}{cc}8&13\\10&15\end{array}\right)^{-1}=\left(\begin{array}{cc}-3/2&13/10\\1&-4/5\end{array}\right)

hence

\bf \left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{cc}-3/2&13/10\\1&-4/5\end{array}\right)\left(\begin{array}{c}282.60\\333\end{array}\right)=\left(\begin{array}{c}9\\ 16.2\end{array}\right)

Now,

if a dozen hot dogs cost $9, 1 hot dog costs 9/12 = $0.75

if a dozen bratwursts cost $16.2, 1 bratwurst costs 16.2/12 = $1.35

4 0
3 years ago
A certain rectangle is 5 times as long as it is wide. Suppose the length and width are both tripled. The perimeter of the second
astraxan [27]
It is tripled as well. You are tripling every distance and thus tripling the sum of the distances as well. 
3 0
3 years ago
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