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Oksanka [162]
2 years ago
5

A grocer purchased a 100-pound mixture of nuts for $515.The mixture contains 70 pounds of peanuts that cost $3.50 per pound.The

remaining nuts in the mixture are almonds.What is the cost per pound of the almonds in the mixture.??
Mathematics
1 answer:
alexdok [17]2 years ago
3 0
Answer $9.00 per pound for almonds
Reason
First, 70# x $3.50 = $245 of peanuts

Second, $515 total cost - $245 of peanuts = $270 of almonds

Third, we know 100# - 70# peanuts = 30# of almonds

Last, $270 cost of almonds divided by 30# of almonds = $9.00 per pound for almonds

You might be interested in
35 points!! How much does a fish weigh if its tail weighs 4 kilograms, it's head weighs as much as it's tail and half its body,
NeX [460]

Let t, h, b represent the weighs of tail, head, and body, respectively.

t = 4 . . . . given

h = t + b/2 . . . . the head weighs as much as the tail and half the body

b/2 = h + t . . . . half the body weighs as much as the head and tail

_____

Substituting for b/2 in the second equation using the expression in the third equation, we have

... h = t + (h + t)

Subtracting h from both sides gives

... 0 = 2t . . . . . . in contradiction to the initial statement about tail weight.

Conclusion: there's no solution to the problem given here.

4 0
3 years ago
What two numbers multiply to 54 and add up to 7
k0ka [10]
Xy = 54
x + y = 7

xy = 54
\frac{xy}{x} = \frac{54}{x}
y = \frac{54}{x}

x + y = 7
x + \frac{54}{x} = 7
\frac{x^{2}}{x} + \frac{54}{x} = 7
\frac{x^{2} + 54}{x} = 7
x^{2} + 54 = 7x
x^{2} - 7x + 54 = 0
x = \frac{-(-7) \± \sqrt{(-7)^{2} - 4(1)(54)}}{2(1)}
x = \frac{7 \± \sqrt{49 - 216}}{2}
x = \frac{7 \± \sqrt{-167}}{2}
x = \frac{7 \± i\sqrt{167}}{2}
x = \frac{7 \± 12.9i}{2}
x = 3.5 \± 6.45i
x = 3.5 + 6.45i    or    3.5 - 6.45i

                  x + y = 7
   3.5 + 6.45i + y = 7
- (3.5 + 6.45i)       - (3.5 + 6.45i)
                        y = 3.5 - 6.45i
                  (x, y) = (3.5 + 6.45i, (3.5 - 6.45i)

                          or

                  x + y = 7
   3.5 - 6.45i + y = 7
- (3.5 - 6.45i)     - (3.5 - 6.45i)
                       y = 3.5 + 6.45i
                 (x, y) = (3.5 - 6.45i, 3.5 + 6.45i)

The two numbers that add up to 7 and can multiply to 54 is 3.5 ± 6.45i.
5 0
3 years ago
- What is the range of the following set of points: {(3, 4), (7, - 2), (11, – 8)}?
konstantin123 [22]

Answer:

B) {-8,-2,3,4,7,11}

Step-by-step explanation:

range is y

3 0
3 years ago
Kwamee created a coffee blend for his cafe by mixing Kona beans and Fuji beans. Kona beans cost $11 per pound and Fuji beans cos
yKpoI14uk [10]
Let's use K for Kona and F for Fuji. The system of equations has to be a balanced system. For example, you can't mix the number of pounds of beans with the cost for each because pounds and dollars are different and you can only combine like terms...pounds with pounds and dollars with dollars. So let's start with the number of pounds. Since we don't know how much of each he bought we have the 2 unknowns, F and K, but we DO know that he bought 23 pounds total. So the first equation is
K + F = 23
Now let's see what we can do with the dollars. Again, we don't know how much he bought of each kind of coffee, but we do know that Kona beans cost $11 per pound and that Fuji beans cost $7.50 per pound, and we know that he spent a total of $197. So let's set that up:
11K + 7.50F = 197
Those are your 2 equations. It doesn't say you need to solve them, so you're done.
7 0
3 years ago
Each of a group of 20 intermediate tennis players is given two rackets, one having nylon strings and the other synthetic gut str
sergeinik [125]

Answer:

Find answers below

Step-by-step explanation:

H0: P <= 0.5

Ha: P > 0.5

, the number who prefer gut strings is <= a number or the test tends towards the  left-tailed.

{0,1,2,3,4,5} ;  

{15,16,17,18,19,20} is a right-tailed test and not appropriate for H0:

{ 0,1,2,3,17,18,19,20} is two-tailed and not appropriate for H0:

b)

Does the region specify a level .05 test? No

 

P = proportion who prefer gut strings to nylon

P = X /20

Assume alpha = 0.05

z(alpha) = -1.645

Reject if (x/20 - 0.5) / sqrt[ (0.5)(0.5)/20 ] < -1.645

Reject if (x/20 - 0.5) <  < (-1.645) sqrt ( (0.5)(0.5)/20 )

Reject if x/20   < (-1.645) sqrt ( (0.5)(0.5)/20 ) + 0.5

Reject if x/20   < 0.316

Reject if x   < (0.316)(20) = 6.32

{0,1,2,3,4,5,6} is the region for the best level 0.05 test

c)

According to (a),  reject H0 if x <= 5

P( Type II error) = P( do not reject H0/ when Ha is true)

P( Type II error) = P( x > 5/ P=0.6)

x ---p(x)

6  0.004854  0.998388  

7  0.014563    

8  0.035497  

9  0.070995  

10  0.117142  

11  0.159738  

12  0.179706  

13  0.165882  

14  0.124412  

15  0.074647  

16  0.034991  

17  0.012350  

18  0.003087  

19  0.000487  

20  0.000037  

add: 0.9984 --  proba bility of a type II error

Assuming P=0.8

P( Type II error) = P( x > 5/ P=0.8)

6  0.000002  1.000000  

7  0.000013  

8  0.000087  

9  0.000462  

10  0.002031  

11  0.007387  

12  0.022161  

13  0.054550  

14  0.109100  

15  0.174560  

16  0.218199  

17  0.205364  

18  0.136909  

19  0.057646  

20  0.011529  

add: 1.0000  probability of a type II error

d)

P( x <= 13) =  

0  0.000001  

1  0.000019  

2  0.000181  

3  0.001087  

4  0.004621  

5  0.014786  

6  0.036964  

7  0.073929  

8  0.120134  

9  0.160179  

10  0.176197  

11  0.160179  

12  0.120134  

13  0.073929  

add: 0.9423 < 0.10 ,  H0 cannot be rejected

5 0
3 years ago
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