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Greeley [361]
3 years ago
12

Penelope earned $50 in May. She earned $60 in June.

Mathematics
1 answer:
strojnjashka [21]3 years ago
3 0

b)25%

espero ter ajudado

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Round 0.9998 to 3 decimal places
mojhsa [17]
The answer would be one because  .9998 would round the 9 to a 10 which would round the second 9 and then the third nine to make 1

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What is this question?? (needing help)
Free_Kalibri [48]

468 x 0.001 = 0.468

46.8 x 0.1 = 4.68

4.68 x 10^3 = 4680

0.468 x 10^2 = 46.8

Hope this helps!

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I need help i attached this image
LenKa [72]

Answer:

k = -7

Step-by-step explanation:

Find the difference of the y-values of both vertexes:

f(x): y = 1

g(x): y = -6

g(x) - f(x)

-6 - 1 = -7

8 0
2 years ago
Which of the following correctly identifies the set of outputs?
ElenaW [278]
C is the correct answer
5 0
3 years ago
A particular fruit's weights are normally distributed, with a mean of 212 grams and a standard deviation of 20 grams.
djyliett [7]

Answer:

220 grams.

Step-by-step explanation:

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

Normal probability distribution

When the distribution is normal, we use 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 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.

In this question:

\mu = 212, \sigma = 20, n = 22, s = \frac{20}{\sqrt{22}} = 4.264

If you pick 22 fruits at random, then 3% of the time, their mean weight will be greater than how many grams?

We have to find the 100 - 3 = 97th percentile, which is X when Z has a pvalue of 0.97. So X when Z = 1.88.

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

By the Central Limit Theorem

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

1.88 = \frac{X - 212}{4.264}

X - 212 = 1.88*4.264

X = 220

The answer is 220 grams.

4 0
3 years ago
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