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vovikov84 [41]
3 years ago
11

A flat rectangular piece of aluminum has a perimeter of 62 inches. The length is 15

Mathematics
1 answer:
Salsk061 [2.6K]3 years ago
6 0

Answer:

So, the required width of rectangular piece of aluminium is 8 inches

Step-by-step explanation:

We are given:

Perimeter of rectangular piece of aluminium = 62 inches

Let width of rectangular piece of aluminium = w

and length of rectangular piece of aluminium  = w+15

We need to find width i.e value of x

The formula for finding perimeter of rectangle is: Perimeter=2(Length+Width)\\

Now, Putting values in formula for finding Width w:

Perimeter=2(Length+Width)\\\\62=2(w+15+w)\\62=2(2w+15)\\62=4w+30\\62-30=4w\\4w=32\\w=\frac{32}{4}\\w=8

After solving we get the width of rectangular piece :w = 8

So, the required width of rectangular piece of aluminium is 8 inches

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Almost all medical schools in the United States require students to take the Medical College Admission Test (MCAT). To estimate
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Answer:

Probability of having student's score between 505 and 515 is 0.36

Given that z-scores are rounded to two decimals using Standard Normal Distribution Table

Step-by-step explanation:

As we know from normal distribution: z(x) = (x - Mu)/SD

where x = targeted value; Mu = Mean of Normal Distribution; SD = Standard Deviation of Normal Distribution

Therefore using given data: Mu (Mean) = 510, SD = 10.4 we have z(x) by using z(x) = (x - Mu)/SD as under:

In our case, we have x = 505 & 515

Approach 1 using Standard Normal Distribution Table:

z for x=505: z(505) = (505-510)/10.4 gives us z(505) = -0.48

z for x=515: z(515) = (515-510)/10.4 gives us z(515) = 0.48

Afterwards using Normal Distribution Tables and rounding the values to two decimals we find the probabilities as under:

P(505) using z(505) = 0.32

Similarly we have:

P(515) using z(515) = 0.68

Now we may find the probability of student's score between 505 and 515 using:

P(505 < x < 515) = P(515)-P(505) = 0.68 - 0.32 = 0.36

PS: The standard normal distribution table is being attached for reference.

Approach 2 using Excel or Google Sheets:

P(x) = norm.dist(x,Mean,SD,Commutative)

P(505) = norm.dist(505,510,10.4,1)

P(515) = norm.dist(515,510,10.4,1)

Probability of student's score between 505 and 515= P(515) - P(505) = 0.36

Download pdf
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Step-by-step explanation:

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