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marysya [2.9K]
3 years ago
6

Henry knows that the area of a rectangle is 30 square inches. The perimeter is 22 inches. If the length is 1 inch longer than th

e width, what are the length and width of Henry's rectangle? Explain how you know.
Mathematics
1 answer:
dsp733 years ago
5 0
The length is 6 inches and the width is 5 inches
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Answer:

x=13

Step-by-step explanation:

Switch sides:

2x−9=17

Add 9 to both sides:

2x−9+9=17+9

Simplify

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2 years ago
The reaction time (in seconds) to a certain stimulus is a continuous random variable with pdf f(x) 5 5 3 2 ? 1 x2 1
muminat

The probability that the reaction time for this density function is at most 2.5 seconds is equal to 0.9.

<h3>What is a density function?</h3>

A density function can be defined as a type of function which is used to represent the density of a continuous random variable that lies within a specific range.

<h3>How to calculate the probability that reaction time is at most 2.5 seconds?</h3>

P(X ≤ 2.5) = Fx(2.5)

Fx(2.5) = 3/2 - 3/2(2.5)

Fx(2.5) = 3/2 - 3/5

Fx(2.5) = 0.9.

Read more on density function here: brainly.com/question/14448717

#SPJ4

Complete Question:

The reaction time (in seconds) to a certain stimulus is a continuous random variable with pdf:

f(x) = \left \{ {{\frac{3}{2x^2} \;1 \leq x\leq 3} \atop {0}\;otherwise} \right

What is the probability that reaction time is at most 2.5 seconds?

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2 years ago
Hiro bought a small carton of milk at lunch. If the approximate dimensions of the milk carton are shown what is the minimum amou
faltersainse [42]

Answer:

C.\ 104\ in^2

Step-by-step explanation:

At first, the question looks like an optimization problem, but since all the dimensions of the carton are given, we only have to compute the total area of the given figure.

Let's calculate the front (and back) areas, which are rectangles

A_1=(6.5)(3)=19.5\ in^2

Now with the lateral rectangles which happen to have the very same dimensions

A_2=19.5\ in^2

Next, we compute the front and back triangles of base 3 in and height 1.5 in

A_3=\frac{1}{2}(3)(1.5)=2.25\ in^2

Now, the lateral inclined rectangles of base 3 in and height 2 in

A_4=(3)(2)=6\ in^2

Finally, the base rectangle who happens to be a square of side 3 in

A_5=(3)(3)= 9\ in^2

This last area, unlike all others, is not doubled because its counterpart is inside the carton and is not part of the lateral area

Our total area of cardboard is

A_t=2(19.5)+2(19.5)+2(2.25)+2(6)+9=103.5\ in^2

The closest option to this answer is

C.\ 104\ in^2

5 0
3 years ago
Read 2 more answers
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