Abox contains red apples and green apples .if an apple is picked at random from the box the probability of picking a red apple is 3/8 .which of these best describes the likelihood of picking a green apple?

Question says that there are red apples and green apples in the box.

And the probability of picking a red apple is 3/8

Now as there are only two types of apples

So probability of picking a green apple = 1 - 3/8 = 5/8

Now as we can clearly see that the probability of picking a green apple is more than the probability of picking a red apple.

Hence likelihood of picking a green apple is more LIKELY , then red apple.

a

step-by-step explanation:

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d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;