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Likurg_2 [28]
3 years ago
9

After its first day of life, a baby blue whale started growing. It grew 47.075 inches. If the average baby blue whale grows at a

rate of 1.5 inches a day, for how many days did the baby whale grow, to the nearest tenth of a day?
Mathematics
2 answers:
pantera1 [17]3 years ago
7 0

I believe its 31 days

I divided and got the answer

skad [1K]3 years ago
4 0

Answer:

After the first day, Length of baby blue whale=47.075 inches

It is given that, average baby blue whale grows at a rate of 1.5 inches a day.

⇒ To the nearest tenth of a day, increase in Height of baby blue whale will be

=47.075 \times \frac{1.5}{10}\\\\=7.06125  inches        

Number of Days for which the baby whale grow, to the nearest tenth of a day

    =\frac{47.075}{7.06125}\\\\=6.67

=Approximately 7 days

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RADIATION ON MARS
Darina [25.2K]

Answer:

0.71% probability that the daily average amount of radiation received by a rover will exceed 102

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 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 problem, we have that:

\mu = 100, \sigma = 20, n = 600, s = \frac{20}{\sqrt{600}} = 0.8165

What is the probability that the daily average amount of radiation received by a rover will exceed 102?

This is 1 subtracted by the pvalue of Z when X = 102. So

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

By the Central Limit Theorem

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

Z = \frac{102 - 100}{0.8165}

Z = 2.45

Z = 2.45 has a pvalue of 0.9929

1 - 0.9929 = 0.0071

0.71% probability that the daily average amount of radiation received by a rover will exceed 102

5 0
3 years ago
The average temperature in Fairbanks, Alaska, was —4°F one day and
Oksanka [162]

Answer:

  • 20°

Step-by-step explanation:

<u>We know that:</u>

  • -4 + x = 16 [x = difference between two temperatures]

<u>Solution:</u>

  • -4 + x = 16
  • => x = 16 + 4
  • => x = 20

Hence, the difference between the two temperatures is 20°.

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Answer:

R has to be 8 or above

Step-by-step explanation:

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3 years ago
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Zolol [24]
The answer is C because all you have to do is divide 70 in half.
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Suppose the total weight of the fish caught stayed the same but instead of 100 fish caught during the weeklong fishing trip only
Paha777 [63]
The weight of each fish would be multiplied by 10.

Assume the first time, the weight was 800 lb.  Since there were 100 fish caught, the weight of each fish would be 800/100 = 8 lb.

This time, however, only 10 fish were caught.  The weight would be 800/10 = 80 lb.  

80 = 8(10), so the weight is 10 times more.
4 0
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