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Anna [14]
3 years ago
15

How many hours did lucille drive if she drove 258 miles at 65 mph? round to the nearest tenth hour?

Mathematics
2 answers:
joja [24]3 years ago
7 0
258/65 = 3.9692
= 4.0 hours rounded to the nearest tenth hour
lyudmila [28]3 years ago
5 0
The answer would be 4.0 rounded

hope this helps


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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
Gennadij [26K]

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.

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

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2 years ago
Least common denominator in 1/30 + 1/24
aev [14]

Answer:

The LCD in 1/30 + 1/24 is <u>90</u>.

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2 years ago
Find the distance d(A,B) between points A and B.
Misha Larkins [42]

Answer:

A(4,-5) ; B(4,3)

distance d(A,B) = \sqrt{( 4 - 4)^{2} + (3 -(-5))^{2} } = 8

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15 pts!!! will mark BRAINLIEST Sharon found a couch for $1,770.00. The sales person said it was on sale for 25% off. How much di
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Answer:

$1407.15

Step-by-step explanation:

$1,770.00*0.25= $442.5

$1.770-$442.50 = $1327.5

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360 divided by 7 = 51

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