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Inessa [10]
4 years ago
12

a sack of potatoes weighs 14 pounds 9 ounces. After Wendy makes potato salad for a picnic, the sack weighs 9 pounds 14 ounces. w

hat is the weight of the potatoes Wendy used for the potato salad
Mathematics
1 answer:
Mamont248 [21]4 years ago
5 0
You have to convert the pound to ounces and ounces to pounds


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x = 4y^2 - 2
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The graph of these two is below.

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A package of ground beef costs $6.98. The price per pound is $1.86. How many pounds of ground beef are in the package? Round to
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3 years ago
Use the given information to determine which segments must be parallel. If there are no segments, write none.
densk [106]

Answer:

11. AW║BY

12. DX║YZ

13. AW║YZ

14. AW║DZ

Step-by-step explanation:

11. If alternate interior angles are congruent, the lines the transversal crosses are parallel.

12. The sum ∠5+∠6 is an alternate interior angle with ∠10. The reasoning of problem 11 applies.

13. If same-side interior angles are suppplementary, the lines the transversal crosses are parallel.

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6 0
3 years ago
Two student council representatives are chosen at random from a group of 7 girls and 3 boys. Let G be the random variable denoti
Archy [21]

This question is incomplete, the complete question is;

Two student council representatives are chosen at random from a group of 7 girls and 3 boys. Let G be the random variable denoting the number of girls chosen.

a) What is E[G] or range of G

b) Give the distribution over the random variable G

Answer:

a) E[G] is [ 0, 1 , 2 ]

b)

the distribution over the random variable G.

x     P(x)

0     0.0667

1      0.4667

2     0.4667

Step-by-step explanation:

Given the data in the question;

Number to be chosen is 2

from 7 girls and 3 boys

a) range of G

number of girls can be; [ 0, 1 , 2 ]

Therefore, E[G] is [ 0, 1 , 2 ]

b) he distribution over the random variable G.

Distribution of the random variable G is Hypergeometric;

so

with P( G=g ) = P( getting g girls from 7 and 2-g boys from 3)

= ((^7C_g) × (^3C_{2-g)) / ^{10}C_2

now since, the range of G is [ 0, 1 , 2 ]

P( G=0 ) = ((^7C_0) × (^3C_{2)) / ^{10}C_2

= [(7!/(0!(7-0)!)) × (3!/(2!(3-2)!))] / (10!/(2!(10-2)!))

= [1 × 3] / 45 = 3/45 = 0.0667

P( G=1 ) = ((^7C_1) × (^3C_{1)) / ^{10}C_2

= [(7!/(1!(7-1)!)) × (3!/(1!(3-1)!))] / (10!/(2!(10-2)!))

= [ 7 × 3 ] / 45 = 21/45 = 0.4667

P( G=2 ) = ((^7C_2) × (^3C_{0)) / ^{10}C_2

= [(7!/(2!(7-2)!)) × (3!/(0!(3-0)!))] / (10!/(2!(10-2)!))

= [ 21 × 1 ] / 45 = 21/45 = 0.4667

Therefore, the distribution over the random variable G.

x     P(x)

0     0.0667

1      0.4667

2     0.4667

3 0
3 years ago
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