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Svetach [21]
2 years ago
10

Tomorrow, Mrs. Wendel's class will be using toothpicks for a science project. Each student must use at least 5 toothpicks for th

e project.
Mrs. Wendel knows that there is already a bag of 50 toothpicks in the class storage room. She plans to buy x additional toothpicks this afternoon to make sure her class will have enough.


If her class has 29 students, which number sentence represents this situation?

A.

1/29x+1 21/29 ≥ 5


B.

1/29+25/29 ≥ 5

C.

1/29+25/29 ≤ 5


D.

1/29x+1 21/29 ≤ 5
Mathematics
1 answer:
Amiraneli [1.4K]2 years ago
7 0

Answer:

I'm pretty sure A but let me know if I'm wrong

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Artist 52 [7]

Answer:

0% probability that at least 1,500 will agree to respond

Step-by-step explanation:

I am going to use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

n = 15000, p = 0.09

So

\mu = E(X) = np = 15000*0.09 = 1350

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{15000*0.09*0.91} = 35.05

What is the probability that at least 1,500 will agree to respond

This is 1 subtracted by the pvalue of Z when X = 1500-1 = 1499. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{1499 - 1350}{35.05}

Z = 4.25

Z = 4.25 has a pvalue of 1.

1 - 1 = 0

0% probability that at least 1,500 will agree to respond

6 0
2 years ago
PLEASE HELP! I SUCK AT MATHHH, THANK YOUUU!!!
vfiekz [6]

Answer: 21/400

Step-by-step explanation:

Multivitamin:

8 cherry, 5 grape, and 7 orange

Total fruits in Multivitamin container :

(8 + 5 + 7) = 20

P(orange) = Total required outcome / Total possible outcomes)

P(orange) = 7/20

Calcium vitamin :

11 berry, 3 lemon, and 6 pineapple

Total fruits in calcium vitamin container:

(11 + 3 + 6) = 20

Probability of picking a lemon:

P(lemon) = Total required outcome / Total possible outcomes)

P(lemon) = 3/20

Therefore, probability of picking an orange and a lemon equals:

P(orange) × p(lemon) = (7/20) × (3/20) = 21/400

3 0
2 years ago
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
motikmotik

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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