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otez555 [7]
4 years ago
7

If a cow has a mass of 9×1029×102 kilograms, and a blue whale has a mass of 1.8×1051.8×105 kilograms, which of these statements

is true?
A.) The blue whale has about 200200 times more mass.
B.)The blue whale has about 500500 times more mass.
C.)The cow has about 200200 times more mass.
D.)There is no way to compare the masses.
Mathematics
2 answers:
STatiana [176]4 years ago
3 0
Cow : 9 x 10^2 = 9 * 100 = 900
whale : 1.8 x 10^5 = 1.8 * 100,000 = 180,000

180,000/900 = 200

The blue whale has about 200 times more mass <==


Natalija [7]4 years ago
3 0

Answer:

The blue whale has about 200 times more mass.

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Yes, the ratios do form a proportion.

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Please help me solve this !<br> will mark brainliest !<br> (spammers reported)
kherson [118]

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it is five

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Aleks04 [339]

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243

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7 0
3 years ago
127.5 is 250% of what?
Inga [223]
Your answer is 127.5 is 250% of 51
4 0
3 years ago
The weight of an adult swan is normally distributed with a mean of 26 pounds and a standard deviation of 7.2 pounds. A farmer ra
Snezhnost [94]
Let X denote the random variable for the weight of a swan. Then each swan in the sample of 36 selected by the farmer can be assigned a weight denoted by X_1,\ldots,X_{36}, each independently and identically distributed with distribution X_i\sim\mathcal N(26,7.2).

You want to find

\mathbb P(X_1+\cdots+X_{36}>1000)=\mathbb P\left(\displaystyle\sum_{i=1}^{36}X_i>1000\right)

Note that the left side is 36 times the average of the weights of the swans in the sample, i.e. the probability above is equivalent to

\mathbb P\left(36\displaystyle\sum_{i=1}^{36}\frac{X_i}{36}>1000\right)=\mathbb P\left(\overline X>\dfrac{1000}{36}\right)

Recall that if X\sim\mathcal N(\mu,\sigma), then the sampling distribution \overline X=\displaystyle\sum_{i=1}^n\frac{X_i}n\sim\mathcal N\left(\mu,\dfrac\sigma{\sqrt n}\right) with n being the size of the sample.

Transforming to the standard normal distribution, you have

Z=\dfrac{\overline X-\mu_{\overline X}}{\sigma_{\overline X}}=\sqrt n\dfrac{\overline X-\mu}{\sigma}

so that in this case,

Z=6\dfrac{\overline X-26}{7.2}

and the probability is equivalent to

\mathbb P\left(\overline X>\dfrac{1000}{36}\right)=\mathbb P\left(6\dfrac{\overline X-26}{7.2}>6\dfrac{\frac{1000}{36}-26}{7.2}\right)
=\mathbb P(Z>1.481)\approx0.0693
5 0
3 years ago
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