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Shkiper50 [21]
3 years ago
12

The manager of a store that specializes in selling tea decides to experiment with a new blend.She will mix some Earl Grey tea th

at sells for $6 per pound with some Orange Pekoe tea that sells for $4 per pound to get 600 pounds of the new blend. the selling price of the new blend is $4.50 per pound and there was something to be no difference in revenue from selling the new blend versus selling the other types. how many pounds of the Earl Grey tea and orange Pekoe tea are required?
​
Mathematics
1 answer:
Marizza181 [45]3 years ago
3 0

Answer:

  • Earl Grey: 150 pounds
  • Orange Pekoe: 450 pounds

Step-by-step explanation:

Let "e" represent the number of pounds of Earl Grey tea required for the mix. Then (600-e) is the number of pounds of Orange Pekoe tea required. The total cost of the mix is ...

  6·e + 4·(600 -e) = 4.50·600

  2e = 600·(4.50 -4.00) . . . . . simplify, subtract 4·600

  e = 300/2 = 150 . . . . . . . . . . divide by the coefficient of x

Required are 150 pounds of Earl Grey and 450 pounds of Orange Pekoe.

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Ganezh [65]

Answer:

1. -3/8     2. 1/2     3. -3/2

Step-by-step explanation:

You can use Y2-Y1/X2-X1 to find the answers

Ask me if you want it more in detail

6 0
3 years ago
(43 points) In the US, 85% of the population has Rh positive blood. Suppose we take a random sample of 6 persons and let Y denot
VladimirAG [237]

Answer:

a) Binomial distribution with parameters p=0.85 q=0.15 n=6

b) 62.29%

c) 2.38%

d) See explanation below

Step-by-step explanation:

a)

We could model this situation with a binomial distribution

P(6;k)=\binom{6}{k}p^kq^{6-k}

where P(6;k) is the probability of finding exactly k people out of 6 with Rh positive, p is the probability of finding one person with Rh positive and q=(1-p) the probability of finding a person with no Rh.

So

\bf P(Y=k)=\binom{6}{k}(0.85)^k(0.15)^{6-k}

b)  

The probability that Y is less than 6 is

P(Y=0)+P(Y=1)+...+P(Y=5)

Let's compute each of these terms

P(Y=0)=P(6;0)=\binom{6}{0}(0.85)^0(0.15)^{6}=1.139*10^{-5}

P(Y=1)=P(6;1)=\binom{6}{1}(0.85)^1(0.15)^{5}=0.0000387281

P(Y=2)=P(6;2)=\binom{6}{2}(0.85)^2(0.15)^{4}=0.005486484

P(Y=3)=P(6;3)=\binom{6}{3}(0.85)^3(0.15)^{3}=0.041453438

P(Y=4)=P(6;4)=\binom{6}{4}(0.85)^4(0.15)^{2}=0.176177109

P(Y=5)=P(6;5)=\binom{6}{5}(0.85)^5(0.15)^{1}=0.399334781

and adding up these values we have that the probability that Y is less than 6 is

\sum_{i=1}^{5}P(Y=i)=0.622850484\approx 0.6229=62.29\%

c)

In this case is a binomial distribution with n=200 instead of 6.

p and q remain the same.

The mean of this sample would be 85% of 200 = 170.  

In a binomial distribution, the standard deviation is  

s = \sqrt{npq}

In this case  

\sqrt{200(0.85)(0.15)}=5.05

<em>Let's approximate the distribution with a normal distribution with mean 170 and standard deviation 5.05</em>

So, the approximate probability that there are fewer than 160 persons with Rh positive blood in a sample of 200 would be the area under the normal curve to the left of 160

(see picture attached)

We can compute that area with a computer and find it is  

0.0238 or 2.38%

d)<em> In order to approximate a binomial distribution with a normal distribution we need a large sample like the one taken in c).</em>

In general, we can do this if the sample of size n the following inequalities hold:

np\geq 5 \;and\;nq \geq 5

in our case np = 200*0.85 = 170 and nq = 200*0.15 = 30

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3 years ago
Complete the statement.
Ne4ueva [31]
I am pretty sure that the answer is 3
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Nick bought 6 tickets to an air show if he spent 156 dollars how much did the tickets cost
marysya [2.9K]

Answer:

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Step-by-step explanation: when you divide 156 divided by 6 you get 26

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(7 + i) - (6 - 2i) = 7 + i - 6 + 2i = 1 + 3i
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