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kakasveta [241]
3 years ago
7

A card is dealt from a 52-card deck. Find the probability that it it not a 9

Mathematics
1 answer:
Lunna [17]3 years ago
7 0
The probability that it is a nine is calculated by 4/52 (because there are four nines)
this equals 1/13, that means there is a 12/13ths chance you will not draw a nine
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The library has allotted $480 for spring landscaping and has decided
Mazyrski [523]

Answer:

a. x = roses

   y= lavender

b. x + y=36

c. 25x + 5y = 480

d. 19 rose bushes and 1 lavender bush

5 0
3 years ago
Factor the expression below.
Leni [432]

Answer: B

Step-by-step explanation:

When you factor an expression, you want to take out a factor of the numbers and variables.

Since this expression is a bit hard to factor as itself, we can FOIL the answers to see the correct expressions.

(4x-5)(4x-5)

16x²-20x-20x+25

16x²-40x+25

8 0
3 years ago
There are n machines in a factory, each of them has defective rate of 0.01. Some maintainers are hired to help machines working.
frosja888 [35]

Answer:

a) 1 - ∑²⁰ˣ_k=0 ((e^-λ) × λ^k) / k!

b) 1 - ∑^(80x/d)_k=0 ((e^-λ) × λ^k) / k!

c) ∑²⁰ˣ_k=0 (3^k)/k! = 0.99e³

Step-by-step explanation:  

Given that;

if n ⇒ ∞

p ⇒ 0

⇒ np = Constant = λ,  we can apply poisson approximation

⇒ Here 'p' is small ( p=0.01)

⇒ if (n=large) we can approximate it as prior distribution

⇒ let the number of defective items be d

so p(d) = ((e^-λ) × λ) / d!

NOW

a)

Let there be x number of repairs, So they will repair 20x machines on time. So if the number of defective machine is greater than 20x they can not repair it on time.

λ[n0.01]

p[ d > 20x ] = 1 - [ d ≤ 20x ]

= 1 - ∑²⁰ˣ_k=0 ((e^-λ) × λ^k) / k!

b)

Similarly in this case if number of machines d > 80x/3;

Then it can not be repaired in time

p[ d > 80x/3 ]

1 - ∑^(80x/d)_k=0 ((e^-λ) × λ^k) / k!

c)

n = 300, lets do it for first case i.e;

p [ d > 20x } ≤ 0.01

1 - ∑²⁰ˣ_k=0 ((e^-λ) × λ^k) / k! = 0.01

⇒ ∑²⁰ˣ_k=0 ((e^-λ) × λ^k) / k! = 0.99

⇒ ∑²⁰ˣ_k=0 (λ^k)/k! = 0.99e^λ

∑²⁰ˣ_k=0 (3^k)/k! = 0.99e³

8 0
3 years ago
The triangles are similar.
erik [133]
5×4=20 , 12×4=48
53÷4 = 13 then
x=13
6 0
3 years ago
Read 2 more answers
Which figure has the same area as the parallelogram shown below? A. A triangle with a base of 4 mm and a height of 20 mm B. A tr
Arisa [49]
<h2>Answer:  A trapezoid with bases of 6 mm and 14 mm and a height of 8 mm </h2>

The parallelogram in the figure has an area of 80mm^{2}, according to the following formula, which works for all rectangles and parallelograms:

A_{parallelogram}=(b)(h)   (1)

Where b is the base and h is the height

The<u> area of a triangle</u> is given by the following formula:

A_{triangle}=\frac{1}{2}(b)(h)   (2)

So, for option A:

A_{triangle}=\frac{1}{2}(4mm)(20mm)=40mm^{2} \neq 80mm^{2}    

Now, the <u>area of a trapezoid </u>is:

A_{trapezoid}=\frac{1}{2}(b_{1}+ b_{2})(h)   (3)

For option B:

A_{trapezoid}=\frac{1}{2}(15mm+25mm)(2 mm)=40mm^{2} \neq 80mm^{2}    

For option C:

A_{trapezoid}=\frac{1}{2}(6mm+14mm)(8 mm)=80mm^{2}>>>>This is the correct option!

For option D:

A_{rectangle}=(30mm)(8mm)=240mm^{2} \neq 80mm^{2}    

<h2>Therefore the correct option is C</h2>
5 0
3 years ago
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