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alexandr1967 [171]
2 years ago
14

The mass of Jupiter is 1.898 × 10(27) kilograms, and the mass of Saturn is 5.68 × 10(26) kilograms. How much greater is Jupiter'

s mass than Saturn's mass?
Mathematics
1 answer:
grandymaker [24]2 years ago
6 0
Just subtract but take care of the power or else you can come up with wrong results !

so first make the power same ;

for Jupiter = 1.898× 10(27) = 18.98 × 10(26)
for Saturn = 5.68×10(26)

so now subtract = (18.98-5.68) × 10(26)
= 13.3 × 10(26)

so the answer is 13.3 ×10(26)
or, 1.33× 10(27)
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Assume that females have pulse rates that are normally distributed with a mean of μ=73.0 beats per minute and a standard deviati
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Answer:

a. the probability that her pulse rate is less than 76 beats per minute is 0.5948

b. If 25 adult females are randomly​ selected,  the probability that they have pulse rates with a mean less than 76 beats per minute is 0.8849

c.   D. Since the original population has a normal​ distribution, the distribution of sample means is a normal distribution for any sample size.

Step-by-step explanation:

Given that:

Mean μ =73.0

Standard deviation σ =12.5

a. If 1 adult female is randomly​ selected, find the probability that her pulse rate is less than 76 beats per minute.

Let X represent the random variable that is normally distributed with a mean of 73.0 beats per minute and a standard deviation of 12.5 beats per minute.

Then : X \sim N ( μ = 73.0 , σ = 12.5)

The probability that her pulse rate is less than 76 beats per minute can be computed as:

P(X < 76) = P(\dfrac{X-\mu}{\sigma}< \dfrac{X-\mu}{\sigma})

P(X < 76) = P(\dfrac{76-\mu}{\sigma}< \dfrac{76-73}{12.5})

P(X < 76) = P(Z< \dfrac{3}{12.5})

P(X < 76) = P(Z< 0.24)

From the standard normal distribution tables,

P(X < 76) = 0.5948

Therefore , the probability that her pulse rate is less than 76 beats per minute is 0.5948

b.  If 25 adult females are randomly​ selected, find the probability that they have pulse rates with a mean less than 76 beats per minute.

now; we have a sample size n = 25

The probability can now be calculated as follows:

P(\overline X < 76) = P(\dfrac{\overline X-\mu}{\dfrac{\sigma}{\sqrt{n}}}< \dfrac{ \overline X-\mu}{\dfrac{\sigma}{\sqrt{n}}})

P( \overline X < 76) = P(\dfrac{76-\mu}{\dfrac{\sigma}{\sqrt{n}}}< \dfrac{76-73}{\dfrac{12.5}{\sqrt{25}}})

P( \overline X < 76) = P(Z< \dfrac{3}{\dfrac{12.5}{5}})

P( \overline X < 76) = P(Z< 1.2)

From the standard normal distribution tables,

P(\overline X < 76) = 0.8849

c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

In order to determine the probability in part (b);  the  normal distribution is perfect to be used here even when the sample size does not exceed 30.

Therefore option D is correct.

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5 0
2 years ago
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oee [108]
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5 0
2 years ago
4x + 6y = 18 &amp; 4x – 2y = 10
kherson [118]

Answer:

x=4.875

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Step-by-step explanation:

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x=4.875

y=2(4.875)-10

y=-0.25

8 0
2 years ago
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