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zmey [24]
3 years ago
13

Solve P = 2w + 2h for w.

Mathematics
1 answer:
VMariaS [17]3 years ago
3 0
I'd say its the second one
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O is the midpoint of Reason: Defintiin of a midpoint
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What is the value of X I don’t understand this
mash [69]

Answer:

X=4

Step-by-step explanation:

28+8=36

36/9=4

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4 0
3 years ago
A particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams. The heavies
RoseWind [281]

Answer:

The heaviest 5% of fruits weigh more than 747.81 grams.

Step-by-step explanation:

We are given that a particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams.

Let X = <u><em>weights of the fruits</em></u>

The z-score probability distribution for the normal distribution is given by;

                              Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean weight = 733 grams

            \sigma = standard deviation = 9 grams

Now, we have to find that heaviest 5% of fruits weigh more than how many grams, that means;

                    P(X > x) = 0.05       {where x is the required weight}

                    P( \frac{X-\mu}{\sigma} > \frac{x-733}{9} ) = 0.05

                    P(Z > \frac{x-733}{9} ) = 0.05

In the z table the critical value of z that represents the top 5% of the area is given as 1.645, that means;

                               \frac{x-733}{9}=1.645

                              {x-733}}=1.645\times 9

                              x = 733 + 14.81

                              x = 747.81 grams

Hence, the heaviest 5% of fruits weigh more than 747.81 grams.

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3 years ago
Solve this equation with variables on both sides- 3x=5x+18 show all steps
Dima020 [189]
3x=5x+18
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0=2x+18
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-18=2x
-18÷2=2x÷2
-9=x
Hope this helps!
Vote me brainliest!
8 0
3 years ago
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Answer:

Use dimensional analysis to answer each of the following questions. To find the distance light travels in a year for part a, convert sec to yr by using unit ratios with min, hr, and days in their denominators. Thus, a light year is approximately 5,874,589,152,000 mi

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