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salantis [7]
3 years ago
8

Hi I need help can somone also give me and explain or how to work out the answer​

Mathematics
1 answer:
SIZIF [17.4K]3 years ago
5 0

Answer:

(A) looks right

Step-by-step explanation:

I am sorry if i am wrong

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Does anyone know the answer to this?
mr Goodwill [35]

Answer:

I think she has about 3 segments for sprinting and 6 segments for jogging

7 0
3 years ago
Five eighths of people you invite to a party show up, as do six people who were not directly invited. If a total of 41 people sh
denpristay [2]

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3 0
3 years ago
Kim and Melody are selling popcorn buckets for a school fundraiser. Customers can buy small buckets or large buckets. Kim sold 3
Papessa [141]

Answer:

The cost of

A small bucket = x = $8

A Larger bucket = y = $13

Step-by-step explanation:

Let the cost of

A small bucket = x

A Larger bucket = y

Kim sold 3 small buckets and 14 large buckets of popcorn for a total of $206.

3x + 14y = 206... Equation 1

Melody sold 11 small buckets and 11 large buckets of popcorn for a total of $231.

11x + 11y = 231.... Equation 2

What is the cost for a small bucket and a large bucket of popcorn?

3x + 14y = 206... Equation 1

11x + 11y = 231 ....... Equation 2

We solve using Elimination method

Multiply Equation 1 by 11 and Equation 2 by 3 to Eliminate x

33x + 154y = 2266.... Equation 3

33x + 33y = 693....... Equation 4

We subtract Equation 4 from Equation 3

121y = 1573

y = 1573/121

y = $13

Solving for x

3x + 14y = 206... Equation 1

3x + 14 × 13 = 206

3x + 182 = 206

3x = 206 - 182

3x = 24

x = 24/3

x = $8

Hence, the cost of

A small bucket = x = $8

A Larger bucket = y = $13

4 0
3 years ago
The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.9 minutes and a standard deviation of 2.9
Eva8 [605]

Answer:

a) 0.2981 = 29.81% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

b) 0.999 = 99.9% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes

c) 0.2971 = 29.71% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 8.9 minutes and a standard deviation of 2.9 minutes.

This means that \mu = 8.9, \sigma = 2.9

Sample of 37:

This means that n = 37, s = \frac{2.9}{\sqrt{37}}

(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?

320/37 = 8.64865

Sample mean below 8.64865, which is the p-value of Z when X = 8.64865. So

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

By the Central Limit Theorem

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

Z = \frac{8.64865 - 8.9}{\frac{2.9}{\sqrt{37}}}

Z = -0.53

Z = -0.53 has a p-value of 0.2981

0.2981 = 29.81% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes?

275/37 = 7.4324

Sample mean above 7.4324, which is 1 subtracted by the p-value of Z when X = 7.4324. So

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

Z = \frac{7.4324 - 8.9}{\frac{2.9}{\sqrt{37}}}

Z = -3.08

Z = -3.08 has a p-value of 0.001

1 - 0.001 = 0.999

0.999 = 99.9% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes?

Sample mean between 7.4324 minutes and 8.64865 minutes, which is the p-value of Z when X = 8.64865 subtracted by the p-value of Z when X = 7.4324. So

0.2981 - 0.0010 = 0.2971

0.2971 = 29.71% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes

7 0
2 years ago
To win a game, roger needs to score 2000 points. So far he has scored 837 points how many more points does roger need to score?
BartSMP [9]
Roger needs to score 1163 more points to win the game.


5 0
3 years ago
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