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2 years ago

Can anyone help me with this question

Mathematics

1 answer:

lesya692 [45]2 years ago

7 0

Answer:

False

Step-by-step explanation:

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2 years ago

How to solve 6x+25=10x-47

Irina-Kira [14]

Answer:

Step-by-step explanation:

6x+25=10x-47

16x -22

(6x + 10x) + (-47 +25)

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The results of a survey indicate that the true proportion of households who want a park in their neighborhood is likely in the i

GrogVix [38]

Answer:

0.66

Step-by-step explanation:

What are you giving there is a confidence interval. You can obtain a confidence interval based on a sample you got. The length of the confidence interval is determined on how much confidence do you want for your interval (the probability of the real value being inseide the interval) and how big is the sample: the bigger the sample, the smaller the length of the confidence interval. Independently of the sample length, all intervals are centered on the average value you got for the sample, and that is your estimate. In this case, the center of the interval is 0.52+0.8/2 = 0.66.

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2 years ago

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A bag contains different colored candies. There are 50 candies in the bag, 28 are red, 10 are blue, 8 are green and 4 are yellow

Mrrafil [7]

Answer:

$\displaystyle \frac{54}{5405}$ .

Step-by-step explanation:

How many unique combinations are possible in total?

This question takes 5 objects randomly out of a bag of 50 objects. The order in which these objects come out doesn't matter. Therefore, the number of unique choices possible will the sames as the combination

$\displaystyle \left(50\atop 5\right) = 2,118,760$ .

How many out of that 2,118,760 combinations will satisfy the request?

Number of ways to choose 2 red candies out a batch of 28:

$\displaystyle \left( 28\atop 2\right) = 378$ .

Number of ways to choose 3 green candies out of a batch of 8:

$\displaystyle \left(8\atop 3\right)=56$ .

However, choosing two red candies out of a batch of 28 red candies does not influence the number of ways of choosing three green candies out of a batch of 8 green candies. The number of ways of choosing 2 red candies and 3 green candies will be the product of the two numbers of ways of choosing

$\displaystyle \left( 28\atop 2\right) \cdot \left(8\atop 3\right) = 378\times 56 = 21,168$ .

The probability that the 5 candies chosen out of the 50 contain 2 red and 3 green will be:

$\displaystyle \frac{21,168}{2,118,760} = \frac{54}{5405}$ .

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2 years ago