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Anit [1.1K]
3 years ago
7

Jonah read 5 1/2 chapters in his book in 90 minutes how long did it take him to read one chapter

Mathematics
1 answer:
Scrat [10]3 years ago
7 0

Answer:

around 16 minutes. you partition an hour and a half (all out) by what number of sections he read (5.5

Step-by-step explanation:

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djverab [1.8K]
The answer is 6, 4 divided by 2/3 =6
if you need to show your work just multiply by the reciprocal :)
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A poll asks shoppers at the mall how many cups of water they typically drink in a day. The results of the poll are listed below.
castortr0y [4]
First you find the median, the middle term (if there are an even number of terms it is the average of the middle two terms.)  Since you have 21 terms, odd, you just use the middle term, in this case 8.  The first quartile works the same way, but just on the left half between the start and the median...so the first quartile is 6.
5 0
3 years ago
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A rectangular field is 75 yards wide and 105 yards long
nika2105 [10]

Answer:

width = 72 yards

length = 108 yards

Step-by-step explanation:

Given:

  • Width = 75 yards
  • Length = 105 yards

<u>Area of the field</u> with the given values:

\begin{aligned}\textsf{Area of a rectangle}&=\sf width \times length\\& = \sf 75 \times 105\\& = \sf 7875\:\: yd^2\end{aligned}

To maintain the <u>same perimeter</u>, but <u>change the area</u>, either:

  • decrease the width and increase the length by the same amount, or
  • increase the width and decrease the length by the same amount.

In geometry, length pertains to the <u>longest side</u> of the rectangle while width is the <u>shorter side</u>.  Therefore, we should choose:

  • decrease the <u>width</u> and increase the <u>length</u> by the <u>same amount</u>.

<u>Define the variables</u>:

  • Let x = the amount by which to decrease/increase the width and length.

Therefore:

\implies \sf width \times length < 7875\:\:yd^2

\implies (75-x)(105+x) < 7875

Solve the inequality:

\begin{aligned}(75-x)(105+x) & < 7875\\7875-30x-x^2 & < 7875\\-x^2-30x & < 0\\-x(x+30) & < 0\\x(x+30) & > 0\\\implies x & > 0 \:\: \textsf{ or }\:\:x < - 30\end{aligned}

Therefore, as distance is positive only and the maximum width is 75 yd (since we are subtracting from the original width):

\begin{cases}\textsf{width} = 75 - x\\\textsf{length} = 105 + x\end{cases}

\textsf{where } 0 < x < 75

Therefore, to find the width and length of another rectangular field that has the same perimeter but a smaller area than the first field, simply substitute a value of x from the restricted interval into the found expressions for width and length:

<u>Example 1</u>:

  • Let x = 3

⇒ Width = 75 - 3 = 72 yd

⇒ Length = 105 + 3 = 108 yd

⇒ Perimeter = 2(72 + 108) = 360 yd

⇒ Area = 72 × 108 = 7776 yd²

<u>Example 2</u>:

  • Let x = 74

⇒ Width = 75 - 74 = 1 yd

⇒ Length = 105 + 74 = 179 yd

⇒ Perimeter = 2(1 + 179) = 360 yd

⇒ Area = 1 × 179 = 179 yd²

4 0
1 year ago
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Please help me find the answer fast no jokes.
saw5 [17]

Answer:

I think the volume would be 490π in^3

Step-by-step explanation:

V = π7^2*10

π*49*10 = 490π

7 0
3 years ago
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The telephone company purchased 192 yards of wire for $2,304 in January. They
Aleks [24]

Answer:

Step-by-step explanation:

400yd($2304/192yd)=$4800

6 0
3 years ago
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