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dsp73
2 years ago
5

What is the area, in square centimeters, of the shape below? 7.4 cm 9.4 cm​

Mathematics
1 answer:
HACTEHA [7]2 years ago
7 0

Answer:

34.78

Step-by-step explanation:

A=h×b/2

A=7.4×9.4/2

A=34.78

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A researcher was interested in seeing how many names a class of 38 students could remember after playing a name game After playi
Andreas93 [3]

Answer:

Proportion of the students recalled more than 15 names is 91.77%.

Step-by-step explanation:

We are given that a researcher was interested in seeing how many names a class of 38 students could remember after playing a name game After playing the name game, the students were asked to recall as many first names of fellow students as possible.

The mean number of names recalled was 19.41 with a standard deviation of 3.17.

<em>Let X = number of names recalled</em>

SO, X ~ N(\mu = 19.41,\sigma^{2} = 3.17^{2})

The z-score probability distribution is given by ;

                  Z = \frac{X-\mu}{\sigma} } } ~ N(0,1)

where, \mu = mean number of names recalled = 19.41

            \sigma = standard deviation = 3.17

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, proportion of the students recalled more than 15 names is given by = P(X > 15 names)

     P(X > 15) = P( \frac{X-\mu}{{\sigma} } } > \frac{15-19.41}{3.17}  } ) = P(Z > -1.39)

                                                      = P(Z < 1.39) = 0.9177  {using z table}

<em>Therefore, proportion of the students recalled more than 15 names is </em><em>91.77%.</em>

4 0
3 years ago
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
Please help me answer these. Its all about upper and lower bounds.
sergeinik [125]

Answer:

red question= 8.5 yellow question=4.25

Step-by-step explanation:

5 0
3 years ago
If f(x)= 0, what is x?
crimeas [40]

Answer:

X is the input to the function ... it appears that it is unrestricted...

it can be anything from -∞ to + ∞ (the domain) ..

in every case the output (range) is "0"

Step-by-step explanation:

7 0
3 years ago
2/5x + 4 = 1/5x + 8 what does x equal
uysha [10]

Answer:

x=20

Step-by-step explanation:

6 0
3 years ago
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