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nignag [31]
3 years ago
7

s Chegg uppose you have purchased a filling machine for candy bags that is supposed to fill each bag with 16 oz of candy. Assume

that the weights of filled bags are approximately normally distributed. A random sample of 10 bags yields the following data (in oz): 15.87, 16.02, 15.78, 15.83, 15.69, 15.81, 16.04, 15.81, 15.92, 16.10 On the basis of this data, can you conclude that the mean fill weight is actually less than 16 oz? Let LaTeX: \alpha=5\%α = 5 %. What would be the appropriate null and alternative hypothesis?
Mathematics
1 answer:
Novosadov [1.4K]3 years ago
6 0

Answer:

it 68.9

Step-by-step explanation:

because u add all that

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Make sure your answer is in
romanna [79]

Answer:

210° , 330°

Step-by-step explanation:

using the sides of the 30- 60- 90 triangle

with legs 1, \sqrt{3} and hypotenuse 2 , then

sin^{-1} \frac{1}{2} = 30° ← related acute angle

since sin^{-1} - \frac{1}{2}

then angle is in third / fourth quadrants , then required angles are

180° + 30° = 210° ← in third quadrant

360° - 30° = 330° ← in fourth quadrant

6 0
2 years ago
(a) Omar changed 800 rands into dollars when the rate was $1 = 6.25 rands.
fenix001 [56]

Answer:$128

Step-by-step explanation:

If $1=6.25 rands then we must divide 800 by 6.25

5 0
3 years ago
X^2+6x-19=0 pls helpppp
USPshnik [31]

Answer:

Step-by-step explanation:

Since this is not factorable, let's just complete the square:

x^2+6x=19\\x^2+6x+9=28\\(x+3)^2=28\\x+3=\sqrt{28}\\x=2\sqrt{7}-3

8 0
3 years ago
What is the mode for the data set? <br> 59,57,56,50,58,51,54,59,55,52,53 (multiple choice)
jok3333 [9.3K]
The mode for this is 59

Other stuff you may need:
<span><span>Range - 9
</span><span>Median - 55</span><span>
</span><span>Mean - 54.909</span></span>
4 0
4 years ago
Read 2 more answers
What is not a requirement of the binomial probability​ distribution?
Dafna1 [17]

Answer:

B. the trials must be dependent

Step-by-step explanation:

The image for the question is attached.

The missing options are;

A. The probability of a success remains the same in all trials.

B. The trials must be dependent.

C. Each trial must have all outcomes classified into two categories.

D. The procedure has a fixed number of trials.

For a binomial distribution probability, the event must be repeated for a number of time (n) and the probability of success (q) must be the same during each trial.

Binomial distribution probability has two possible outcome: success (p) or failure (q). The event are also independent of each other.

The formula for a binomial distribution probability is given as:

P(X=x) = nCx (p ^ x) (q ^ (n-x))

From the above, we can see that the answer is:

B. the trials must be dependent

7 0
3 years ago
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