A gourment shop wants to mix coffe beans that cost $3.00 per pound with coffe beans that cost $4.25 per pound to create 25 pounds of a new blend that cost $3.50 per pound. Find the number of pounds of each nerded to produce the new blend

Answers

Answer 1

Answer: If the gourmet shopkeeper mix 15 pounds of coffee beans which costs 3.0$ per pound with 10 pounds of coffee beans costing 4.25$ per pound.
The mixture of 25 pounds will cost 3.5$ per pound.

It can be found using
3*x + 4.25*(25 -x) = 25*3.50

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When you round 6 5/11 what's the nearest whole number

Answers

6.

5 is less than half of eleven, so 6 5/11 would round down to 6.

If its easier you could multiply it by 2 to make it a larger #, which is 10/22, which means that the # would be rounded down, and stay as 6, hope I was helpful

What is the equation of a circle with (2,-5) and radius 4

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it is
(x-2)^2+(y+5)^2=16

Help me please. I don't understand this section

Answers

What you do is that you use your algebraic magic to find your c value. First, you will divide both sides by 5. Now you have 2b= the square root of c +2. Now subtract 2 from both sides and you will have 2b-2= the square root of c. To undo the square root of c, you will need to square it, so square both sides and you have your answer. Good luck!

What is the slope of 6x-3y=18??

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$the \ slope \ intercept \ form \ is : \n \n y= mx +b \n \n6x-3y=18 \n \n-3y=-6x+18 \ \ /:(-3)\n \ny=(-6)/(-3)x + (18)/(-3) \n \ny=2x-6 \n \n m = 2$

Can someone help me and explain step by step?

Answers

I'm not a master of probabilities, but I reckon the answer is 3/16.

The probability that the first spin lands on blue is $(6)/(8)$ (6 blue sections of all possible 8)
The probability that the second spin lands on grey is $(2)/(8)$ (2 gray sections of all possible 8)

So, the probability that the first spin lands on blue and the second spin lands on grey is $(6)/(8)\cdot(2)/(8)=(3)/(4)\cdot(1)/(4)=\boxed{(3)/(16)}$

Among the 10 most popular sports, men include competition-type sports - pool and billiards, basketball, and softball - whereas women include aerobics, running, hiking, and calisthenics. However, the top recreational activity for men was still the relaxing sport of fishing, with 41% of those surveyed indicating that they had fished during the year. Suppose 180 randomly selected men are asked whether they had fished in the past year. Suppose 180 randomly selected men are asked whether they has fished in the past year.a. What is the probability that fewer than 50 had fished?
b. What is the probability that between 50 and 75 (inclusive) had fished?
c. If the 180 men selected for the interview were selected by the marketing department of a sporting-goods company based on information obtained from their mailing lists, what would you conclude about the reliability of their survey results?

Answers

Answer:

(a) The probability that fewer than 50 had fished is 0.0002.

(b) The probability that between 50 and 75 (inclusive) had fished is 0.6026.

Step-by-step explanation:

Let X = number of men who had fished during the year.

The probability of the random variable X is, p = 0.41.

A random sample of n = 180 men are selected.

The random variable X follows a Binomial distribution with parameters n = 180 and p = 0.41.

A Normal approximation to Binomial is applied when the following conditions are met,

np ≥ 10
n(1 - p) ≥ 10

Check:

$np=180*0.41=73.8>10\nn(1-p)=180* (1-0.41)=63.72>10$

Thus, the distribution of x can be approximated by a Normal distribution with:

Mean = $np=180*0.41=73.8$

Standard deviation = $√(np(1-p))=√(180*0.41*(1-0.41))=6.599$

(a)

Compute the probability that fewer than 50 had fished as follows:

$P(X<50)=P((X-\mu)/(\sigma)>(50-73.8)/(6.599))\n=P(Z<-3.61)\n=1-P(Z<3.61)\n=1-0.9998\n=0.0002$

Thus, the probability that fewer than 50 had fished is 0.0002.

(b)

Compute the probability that between 50 and 75 (inclusive) had fished as follows:

$P(50\leq X\leq 75)=P(49.5<X<75.5)\n=P((49.5-73.8)/(6.599)<(X-\mu)/(\sigma)<(75.5-73.8)/(6.599))\n=P(-3.68<Z<0.26)\n=P(Z<0.26)-P(Z<-3.68)\n=0.6026-0\n=0.6026$

Thus, the probability that between 50 and 75 (inclusive) had fished is 0.6026.

(c)

If the sample of 810 men are selected from mailing list then it is highly probable that the sample is not a representative of the true population, i.e. sports men.

Because the people interested in sports are less likely to be interested in fishing.

Thus, it could be concluded that the survey results are not reliable.

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