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expeople1 [14]
2 years ago
10

Susie has 20 pieces of candy in a bag: 6 mint sticks, 5 jelly treats, and 9 fruit tart chews. If she eats one piece every 9 minu

tes, what is the probability her first two pieces will both be jelly treats?
Mathematics
1 answer:
motikmotik2 years ago
4 0

Answer:

Step-by-step explanation:

7 fruit tart chews. If he eats one piece every 10 minutes, what is the probability his first two pieces will be a jelly treat and a mint stick? ... First you add all the candies together to get 20 in the bag 2+11+7=20 jelly treat: ... Paul has a bag with 6 mint sticks, 9 jelly treats, and 5 fruit tart chews. If he eats one ...

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Help! Find the coordinates of P so that P partitions the segment AB in the ratio 6:2 if A(−4,12) and B(9,−4).
LekaFEV [45]

(5.75,0)

Let A be (x1,y1) , B be (x2,y2) , P be (x,y) and the ratio be m:n

Formula is,

x=(x2*m+x1*n)/(m+n)

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7 0
1 year ago
Of 560 samples of seafood purchased from various kinds of food stores in different regions of a country and genetically compared
frutty [35]

Answer:

a) (0.5256,0.5944)  

c) Criticism is invalid

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 560

Proportion of mislabeled = 56%

\hat{p} = 0.56

a) 90% Confidence interval:

\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}

z_{critical}\text{ at}~\alpha_{0.05} = \pm 1.64

Putting the values, we get:

0.56\pm 1.64(\sqrt{\dfrac{0.56(1-0.56)}{560}}) = 0.56\pm 0.0344\\\\=(0.5256,0.5944)

b) Interpretation of confidence interval:

We are 90% confident that the true proportion of all seafood in the country that is mislabeled or misidentified is between 0.5256 and 0.5944 that is 52.56% and 59.44%.

c) Validity of criticism

Conditions for validity:

np > 10\\n(1-p)>10

Verification:

560\times 0.56 = 313.6>10\\560(1-0.56) = 246.4>10

Both the conditions are satisfied. This, the criticism is invalid.

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2 years ago
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I think the answer is 76 ...
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