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2 years ago

Here is a diagram that represents (3.1) ⋅ (1.4).

Mathematics

2 answers:

zimovet [89]2 years ago

8 0

Answer:

1) 3.1 2) 1.24 C) 4.34

Step-by-step explanation:

Area = l(w)

Area of rectangle A:

3.1 * 1 = 3.1

Area of rectangle B:

0.4 * 3.1 = 1.24

Both rectangles put together:

First, let's add 1 to 0.4... 1.4

1.4 * 3.1 = 4.34

I really, really, really hope I helped you. Good Luck :D

Firdavs [7]2 years ago

4 0

Step-by-step explanation:

1. solution,

length of rectangle A(l)=3.1units

breadth of rectangle A(b)=1units

now,

area of rectangle A=l×b

=3.1units×1units

=3.1 sq units

2.solution,

length of rectangle B(l)=3.1units

breadth of rectangle B (b)=0.4units

now,

area of rectangleB=l×b

=3.1units×0.4units

=1.24 sq units

3.solution,

length=3.1units

breadth=1.4units

area =lxb

=3.1units x 1.4units

=4.34units

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Robert takes out a loan for $7200 at a 4.3% rate for 2 years. What is the loan future value?

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Answer:

$7,526.94 future value with interest.

Step-by-step explanation:

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2 years ago

Return to the credit card scenario of Exercise 12 (Section 2.2), and let C be the event that the selected student has an America

Nadya [2.5K]

Answer:

A. P = 0.73

B. P(A∩B∩C') = 0.22

C. P(B/A) = 0.5

P(A/B) = 0.75

D. P(A∩B/C) = 0.4

E. P(A∪B/C) = 0.85

Step-by-step explanation:

Let's call A the event that a student has a Visa card, B the event that a student has a MasterCard and C the event that a student has a American Express card. Additionally, let's call A' the event that a student hasn't a Visa card, B' the event that a student hasn't a MasterCard and C the event that a student hasn't a American Express card.

Then, with the given probabilities we can find the following probabilities:

P(A∩B∩C') = P(A∩B) - P(A∩B∩C) = 0.3 - 0.08 = 0.22

Where P(A∩B∩C') is the probability that a student has a Visa card and a Master Card but doesn't have a American Express, P(A∩B) is the probability that a student has a has a Visa card and a MasterCard and P(A∩B∩C) is the probability that a student has a Visa card, a MasterCard and a American Express card. At the same way, we can find:

P(A∩C∩B') = P(A∩C) - P(A∩B∩C) = 0.15 - 0.08 = 0.07

P(B∩C∩A') = P(B∩C) - P(A∩B∩C) = 0.1 - 0.08 = 0.02

P(A∩B'∩C') = P(A) - P(A∩B∩C') - P(A∩C∩B') - P(A∩B∩C)

= 0.6 - 0.22 - 0.07 - 0.08 = 0.23

P(B∩A'∩C') = P(B) - P(A∩B∩C') - P(B∩C∩A') - P(A∩B∩C)

= 0.4 - 0.22 - 0.02 - 0.08 = 0.08

P(C∩A'∩A') = P(C) - P(A∩C∩B') - P(B∩C∩A') - P(A∩B∩C)

= 0.2 - 0.07 - 0.02 - 0.08 = 0.03

A. the probability that the selected student has at least one of the three types of cards is calculated as:

P = P(A∩B∩C) + P(A∩B∩C') + P(A∩C∩B') + P(B∩C∩A') + P(A∩B'∩C') +

P(B∩A'∩C') + P(C∩A'∩A')

P = 0.08 + 0.22 + 0.07 + 0.02 + 0.23 + 0.08 + 0.03 = 0.73

B. The probability that the selected student has both a Visa card and a MasterCard but not an American Express card can be written as P(A∩B∩C') and it is equal to 0.22

C. P(B/A) is the probability that a student has a MasterCard given that he has a Visa Card. it is calculated as:

P(B/A) = P(A∩B)/P(A)

So, replacing values, we get:

P(B/A) = 0.3/0.6 = 0.5

At the same way, P(A/B) is the probability that a student has a Visa Card given that he has a MasterCard. it is calculated as:

P(A/B) = P(A∩B)/P(B) = 0.3/0.4 = 0.75

D. If a selected student has an American Express card, the probability that she or he also has both a Visa card and a MasterCard is written as P(A∩B/C), so it is calculated as:

P(A∩B/C) = P(A∩B∩C)/P(C) = 0.08/0.2 = 0.4

E. If a the selected student has an American Express card, the probability that she or he has at least one of the other two types of cards is written as P(A∪B/C) and it is calculated as:

P(A∪B/C) = P(A∪B∩C)/P(C)

Where P(A∪B∩C) = P(A∩B∩C)+P(B∩C∩A')+P(A∩C∩B')

So, P(A∪B∩C) = 0.08 + 0.07 + 0.02 = 0.17

Finally, P(A∪B/C) is:

P(A∪B/C) = 0.17/0.2 =0.85

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3 years ago

Suppose f (x) = x^2. Find the graph of f(x+2)

NISA [10]

$\bf ~~~~~~~~~~~~\textit{function transformations}
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\\\\
~~~~y={{ A}}({{ B}}x+{{ C}})+{{ D}}
\\\\
f(x)={{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\\\
f(x)={{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\\\
f(x)={{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
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--------------------$

$\bf \bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}\\
~~~~~~\textit{reflection over the x-axis}
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\bullet \textit{ flips it sideways if }{{ B}}\textit{ is negative}\\
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2 years ago

13 out of 60 people surveyed were left handed calculate the angle you would use in a pie chart

PtichkaEL [24]

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13 out of 60 = 13 / 60

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3 years ago

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Rzqust [24]

Answer:

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Step-by-step explanation:

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2 years ago