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zhannawk [14.2K]
3 years ago
12

IO.7(4.2)+15.9= 51.25+43.85+14=

Mathematics
1 answer:
katovenus [111]3 years ago
7 0

Answer:

10.7(4.2)+15.9 equals 60.84

51.25+43.85+14 equals 109.1

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Perform the following mathematical operation, and report the answer to the correct umber of significant figures 7.125*8.00
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Answer:23☺️

Step-by-step explanation:

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Write a real world situation for 15x-20<130 (greater than or equal to is the symbol)
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15x<130+20, 15x<150 divided both sides by 15 ,the answer is 10
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19 POINTS AND ILL GIVE BRAINLIEST!!
Archy [21]

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3.104 (option c)

Step-by-step explanation:

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Read 2 more answers
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.

5 0
3 years ago
What is the slope of the line that passes through the points (-3, 5) and (1, 7)?
Stells [14]

Answer: The slope of the line is \frac{1}{2}

Step-by-step explanation:

The slope of a line can be calculated with the following formula:

m=\frac{y_2-y_1}{x_2-x_1}

Then, given the points (-3, 5) and (1, 7), you can identify that:

y_2=5\\y_1=7\\\\x_2=-3\\x_1=1

Knowing these values you can substitute them into the formula for calculate the slope:

m=\frac{5-7}{-3-1}

Finally, evaluating, you get that the slope of that line is:

m=\frac{-2}{-4}\\\\m=\frac{1}{2}

8 0
3 years ago
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