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Firdavs [7]
3 years ago
13

A store offers a 30% discount on a coat that originally sold for $120. The coat still does not sell, so the store offers another

30% o the sale price. What is the difference between this price and the price of the coat had it been marked down 60% all at once?A. $5.80 B.$7.40C. C. $10.8 D. no difference; the resulting prices are the same
Mathematics
1 answer:
Romashka [77]3 years ago
7 0

Answer:

C

Step-by-step explanation:

120 *.3 = 36

120-36 = 84

84*.3 = 25.2

84-25.2= 59.8

120*.6 = 72

120-72 = 48

59.8-48 = 10.8 C

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Answer:

Option C, the graph does not flip because 2 is positive

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What is the probability that a customer ordered a hot drink given that he or she ordered a large?
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Read 2 more answers
153 is .9% of what number?
victus00 [196]
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3 years ago
For about $1 billion in new space shuttle expenditures, NASA has proposed to install new heat pumps, power heads, heat exchanger
11111nata11111 [884]

Answer:

The probability of one or more catastrophes in:

(1) Two mission is 0.0166.

(2) Five mission is 0.0410.

(3) Ten mission is 0.0803.

(4) Fifty mission is 0.3419.

Step-by-step explanation:

Let <em>X</em> = number of catastrophes in the missions.

The probability of a catastrophe in a mission is, P (X) = p=\frac{1}{120}.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The probability mass function of <em>X </em>is:

P(X=x)={n\choose x}\frac{1}{120}^{x}(1-\frac{1}{120})^{n-x};\x=0,1,2,3...

In this case we need to compute the probability of 1 or more than 1 catastrophes in <em>n</em> missions.

Then the value of P (X ≥ 1) is:

P (X ≥ 1) = 1 - P (X < 1)

             = 1 - P (X = 0)

             =1-{n\choose 0}\frac{1}{120}^{0}(1-\frac{1}{120})^{n-0}\\=1-(1\times1\times(1-\frac{1}{120})^{n-0})\\=1-(1-\frac{1}{120})^{n-0}

(1)

Compute the compute the probability of 1 or more than 1 catastrophes in 2 missions as follows:

P(X\geq 1)=1-(1-\frac{1}{120})^{2-0}=1-0.9834=0.0166

(2)

Compute the compute the probability of 1 or more than 1 catastrophes in 5 missions as follows:

P(X\geq 1)=1-(1-\frac{1}{120})^{5-0}=1-0.9590=0.0410

(3)

Compute the compute the probability of 1 or more than 1 catastrophes in 10 missions as follows:

P(X\geq 1)=1-(1-\frac{1}{120})^{10-0}=1-0.9197=0.0803

(4)

Compute the compute the probability of 1 or more than 1 catastrophes in 50 missions as follows:

P(X\geq 1)=1-(1-\frac{1}{120})^{50-0}=1-0.6581=0.3419

6 0
3 years ago
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skelet666 [1.2K]

Step-by-step explanation:

3h+11/4h+16=19/26

78h+286=76h+304

12h=18

h=3/2

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2 years ago
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