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mestny [16]
3 years ago
10

3 coffees and 4 donuts cost $10.05. In the same cafeteria, 5 coffees and 7 donuts cost $17.15. How much do you have to pay for 4

coffees and 6 donut
Mathematics
1 answer:
MakcuM [25]3 years ago
5 0

Answer

You need to pay $14.20 to get 4 coffees and 6 donuts.

Explanation

Let's say that the price of one coffee and x and the price of one donut is y.

In the first instance, 3x+4y=10.05.

In the second instance, 5x+7y=17.15.

You can use these equations to find the value of a coffee and the value of a donut.

We can find x and y using elimination. To do this, you should add or subtract one equation from another to "get rid" of a variable (I'll "get rid" of x). We can't just add or subtract the equations right now, since that wouldn't lead to 0x.

Multiply the first equation by 5, and the second equation by 3. After this, both equations will have 15x. Make sure to multiply each term by 5 and 3.

15x+20y=50.25. This means that 15 coffees and 20 donuts is $50.25.

15x+21y=51.45. This means that 15 coffees and 21 donuts is 51.45.

Now since there are an equal number of coffees, we can subtract these equations.

(15x+21y=51.45)-(15x+20y=50.25) equals y=1.20 (a donut costs $1.20).

Plug the price of the donut into y to find x; 3x+4(1.20)=10.05. The value of x is 1.75. The price of a coffee is 1.75.

You can multiply 1.75 by 4 to find the price of 4 coffees and 1.20 by 6 to find the price of 6 donuts.

1.75*4 is 7.00 and 1.20*6 is 7.20.

You can add those to find the total price; 7.00+7.20=14.20.

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The U.S. government has devoted considerable funding to missile defense research over the past 20 years. The latest development
Bad White [126]

Answer:

a) Let the random variable X= "number of these tracks where SBIRS detects the object." in order to use the binomial probability distribution we need to satisfy some conditions:

1) Independence between the trials (satisfied)

2) A value of n fixed , for this case is 20 (satisfied)

3) Probability of success p =0.2 fixed (Satisfied)

So then we have all the conditions and we can assume that:

X \sim Bin(n =20, p=0.8)

b) X \sim Bin(n =20, p=0.8)

c) P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456

d) P(X \geq 15) = P(X=15)+ .....+P(X=20)

P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456

P(X=16)=(20C16)(0.8)^{16} (1-0.8)^{20-16}=0.218

P(X=17)=(20C17)(0.8)^{17} (1-0.8)^{20-17}=0.205

P(X=18)=(20C18)(0.8)^{18} (1-0.8)^{20-18}=0.137

P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.058

P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.012

P(X\geq 15)=0.804208

e) E(X) = np = 20*0.8 = 16

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

Part a

Let the random variable X= "number of these tracks where SBIRS detects the object." in order to use the binomial probability distribution we need to satisfy some conditions:

1) Independence between the trials (satisfied)

2) A value of n fixed , for this case is 20 (satisfied)

3) Probability of success p =0.2 fixed (Satisfied)

So then we have all the conditions and we can assume that:

X \sim Bin(n =20, p=0.8)

Part b

X \sim Bin(n =20, p=0.8)

Part c

For this case we just need to replace into the mass function and we got:

P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456

Part d

For this case we want this probability: P(X\geq 15)

And we can solve this using the complement rule:

P(X \geq 15) = P(X=15)+ .....+P(X=20)

P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456

P(X=16)=(20C16)(0.8)^{16} (1-0.8)^{20-16}=0.218

P(X=17)=(20C17)(0.8)^{17} (1-0.8)^{20-17}=0.205

P(X=18)=(20C18)(0.8)^{18} (1-0.8)^{20-18}=0.137

P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.058

P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.012

P(X\geq 15)=0.804208

Part e

The expected value is given by:

E(X) = np = 20*0.8 = 16

5 0
3 years ago
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