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SpyIntel [72]
3 years ago
8

Almost done, three more left I think

Mathematics
1 answer:
Alexxandr [17]3 years ago
8 0

Answer:

Nice!!

Step-by-step explanation:

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2 years ago
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A Food Marketing Institute found that 39% of households spend more than $125 a week on groceries. Assume the population proporti
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Answer:

0.6210

Step-by-step explanation:

Given that a Food Marketing Institute found that 39% of households spend more than $125 a week on groceries

Sample size n =87

Sample proportion will follow a normal distribution with p =0.39

and standard error = \sqrt{\frac{0.39(1-0.39)}{87} } \\=0.0523

the probability that the sample proportion of households spending more than $125 a week is between 0.29 and 0.41

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There is 0.6210 probability that the sample proportion of households spending more than $125 a week is between 0.29 and 0.41

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84102 - 1052=?<br><br><br>pasagot po pls with solution ​
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83050

Step-by-step explanation:

that's the answer to the question

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2 years ago
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LiRa [457]

Answer:

1a

Step-by-step explanation:

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Find f(-4) for f(x)=9(3)^x
jeka94

Answer:

f(-4) = \dfrac{1}{9}

Step-by-step explanation:

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