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masha68 [24]
3 years ago
7

To stay on budget, Kurt must spend less than $50. Yesterday, he spent $7 for lunch. Today he wants to buy sports cards, which co

st $1 each.
Let s represent the number of sports cards Kurt can buy.

Which statement determines how many sports cards Kurt can buy to stay on budget?

To stay on budget, Kurt can buy fewer than 7 cards.

To stay on budget, Kurt can buy fewer than 43 cards.

To stay on budget, Kurt can buy fewer than 50 cards.

To stay on budget, Kurt can buy fewer than 58 cards.
Mathematics
2 answers:
bagirrra123 [75]3 years ago
8 0
A. To stay on budget, Kurt can buy few 7 cards because it says that he has to spend less that $50 and he already spent 10 so he's left with $43 if he gets 43 cards then that's $43 and he would've spent $50 dollars altogether but he said he wanted to spend less that $50 dollars so A is your answer.
Hope this helps!!!
Please mark as Brainleist 
zheka24 [161]3 years ago
7 0
A. would be your answer
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The mean life of a television set is 119 months with a standard deviation of 14 months. If a sample of 74 televisions is randoml
irina [24]

Answer:

50.34% probability that the sample mean would differ from the true mean by less than 1.1 months

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 normally distributed samples are 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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 119, \sigma = 14, n = 74, s = \frac{14}{\sqrt{74}} = 1.63

If a sample of 74 televisions is randomly selected, what is the probability that the sample mean would differ from the true mean by less than 1.1 months

This is the pvalue of Z when X = 119 + 1.1 = 120.1 subtracted by the pvalue of Z when X = 119 - 1.1 = 117.9. So

X = 120.1

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

By the Central Limit Theorem

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

Z = \frac{120.1 - 119}{1.63}

Z = 0.68

Z = 0.68 has a pvalue of 0.7517

X = 117.9

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

Z = \frac{117.9 - 119}{1.63}

Z = -0.68

Z = -0.68 has a pvalue of 0.2483

0.7517 - 0.2483 = 0.5034

50.34% probability that the sample mean would differ from the true mean by less than 1.1 months

8 0
3 years ago
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