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mixas84 [53]
2 years ago
9

How can i solve them

Mathematics
1 answer:
Mariulka [41]2 years ago
4 0

Answer:

2x²+x-15=0

Step-by-step explanation:

A=L×W

12cm²=(2x+3)(x-1)

12=2x²-2x+3x-3

12=2x²+x-3

2x²+x-3-12=0

2x²+x-15=0

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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.

5 0
2 years ago
2. A vine called the mile-a-minute weed is known for growing at a very fast rate. It can grow up to 0.5 ft per day. How fast in
gayaneshka [121]

Answer:

.25 inches per hour

Step-by-step explanation:

.5 ft

-----------

day

We need to get to inches

1 ft = 12 inches

.5 ft          12 inches      

----------- * ------------

day             1 ft  

We also need to convert days to minutes

1 day = 24 hours

.5 ft           12 inches             1 day          

----------- * ------------   * -------------    

day             1 ft       24 hours      

6 inches

-------------

24 hours

.25 inches per hour

8 0
2 years ago
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What is the following sum? 4(5√x^2 y) +3(5√x^2 y)
Levart [38]

the answer is 35 x √ 2 y

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3 years ago
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Greatest to least 3.008 3.825 3.09 3.18
brilliants [131]
3.825,3.18,3.09,3.008
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Another name for the term ____ is trade off
WITCHER [35]
I think the answer is an auction!
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