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Maru [420]
3 years ago
9

What is the area of rectangle PQRS?

Mathematics
2 answers:
NISA [10]3 years ago
5 0
D. 42.75 square inches
Katen [24]3 years ago
4 0

Hi,

The area of rectangle PQRS is:

D. 42.75 square inches

a=wl

9.5*4.5=42.75

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Mr. Dermot saves an equal amount of money each month for 6 months. He is saving to buy a motorcycle that costs $2,400. Draw a ba
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Consider this equation: –2x – 4 + 5x = 8 Generate a plan to solve for the variable. Describe the steps that will be used.
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We are given the function <span>–2x – 4 + 5x = 8  and is asked in the problem to solve for the variable x in the function. In this case, we can first group the like terms and put them in their corresponding sides:

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Then, do the necessary operations.

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The weight of items produced by a machine is normally distributed with a mean of 8 ounces and a standard deviation of 2 ounces.
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<span>First we calculate z using the formula:

z = (x - μ)/σ</span>

Where:

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μ = mean, 8

σ = standard dev, 2

Substituting known values:<span>
z = (10 - 8)/2
z = 2/2
z = 1

Using the tables of the normal distribution to find the p-value with z = 1

p = 0.8413

Since we want "greater than 10”, we need to subtract the probability from 1 therefore

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3 years ago
The width of a rectangle is only 15% of its length. If the perimeter of the rectangle is 46, what is the length
sveta [45]

Answer:

20 units

Step-by-step explanation:

Let the length be x. According to the question,

  • Length = x
  • Width = 15% of the length

➝ Width = 15% of the length

➝ Width = 15/100x

➝ Width = 3/20x

We have the perimeter of the rectangle that is 46 units.

\longrightarrow \sf {Perimeter_{(Rec.)} = 2(L + W) } \\

\longrightarrow \sf {46= 2\Bigg \lgroup x + \dfrac{3}{20}x \Bigg \rgroup } \\

\longrightarrow \sf {46= 2\Bigg \lgroup x + \dfrac{3}{20}x \Bigg \rgroup } \\

\longrightarrow \sf {46= 2\Bigg \lgroup \dfrac{20x + 3x}{20} \Bigg \rgroup } \\

\longrightarrow \sf {46= 2\Bigg \lgroup \dfrac{23x}{20} \Bigg \rgroup } \\

\longrightarrow \sf {\dfrac{46}{2}=  \dfrac{23x}{20}} \\

\longrightarrow \sf {23=  \dfrac{23x}{20}} \\

\longrightarrow \sf {23 \times 20 = 23x} \\

\longrightarrow \sf {460= 23x} \\

\longrightarrow \sf {\cancel{\dfrac{460}{23}} = x} \\

\longrightarrow \underline{\boxed{ \bf {20\; units = x}}} \\

<u>Therefore</u><u>,</u><u> </u><u>length</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>rectang</u><u>le</u><u> </u><u>is</u><u> </u><u>2</u><u>0</u><u> </u><u>units</u><u>.</u>

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4 years ago
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