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sertanlavr [38]
3 years ago
15

The measure of central angle XYZ is 1.25 radians.

Mathematics
2 answers:
jekas [21]3 years ago
6 0

Answer:

The answer is C

I hope this helps

Nataly [62]3 years ago
4 0

Answer:

40 units2 or c

Step-by-step explanation:

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Answer:

14 is the answer.

Step-by-step explanation:

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Wat is 7x² + 4x - 5= 0
frosja888 [35]
Answer and workings in the attachment below.

4 0
3 years ago
Read 2 more answers
The pregnancy length in days for a population of new mothers can be approximated by a normal distribution with a mean of days an
solniwko [45]

Answer:

(a) 283 days

(b) 248 days

Step-by-step explanation:

The complete question is:

The pregnancy length in days for a population of new mothers can be approximated by a normal distribution with a mean of 268 days and a standard deviation of 12 days. ​(a) What is the minimum pregnancy length that can be in the top 11​% of pregnancy​ lengths? ​(b) What is the maximum pregnancy length that can be in the bottom ​5% of pregnancy​ lengths?

Solution:

The random variable <em>X</em> can be defined as the pregnancy length in days.

Then, from the provided information X\sim N(\mu=268, \sigma^{2}=12^{2}).

(a)

The minimum pregnancy length that can be in the top 11​% of pregnancy​ lengths implies that:

P (X > x) = 0.11

⇒ P (Z > z) = 0.11

⇒ <em>z</em> = 1.23

Compute the value of <em>x</em> as follows:

z=\frac{x-\mu}{\sigma}\\\\1.23=\frac{x-268}{12}\\\\x=268+(12\times 1.23)\\\\x=282.76\\\\x\approx 283

Thus, the minimum pregnancy length that can be in the top 11​% of pregnancy​ lengths is 283 days.

(b)

The maximum pregnancy length that can be in the bottom ​5% of pregnancy​ lengths implies that:

P (X < x) = 0.05

⇒ P (Z < z) = 0.05

⇒ <em>z</em> = -1.645

Compute the value of <em>x</em> as follows:

z=\frac{x-\mu}{\sigma}\\\\-1.645=\frac{x-268}{12}\\\\x=268-(12\times 1.645)\\\\x=248.26\\\\x\approx 248

Thus, the maximum pregnancy length that can be in the bottom ​5% of pregnancy​ lengths is 248 days.

8 0
3 years ago
Depth of swimming pool
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8 0
3 years ago
The test scores on a 100-point test were recorded for 20 students:71 93 91 86 7573 86 82 76 5784 89 67 62 7277 68 65 75 84a. Can
Dafna11 [192]

Answer: a. Yes

              b. mean = 76.65

                  standard deviation = 10.04

              c. 76.65 ± 4.4

Step-by-step explanation:

a. <u>Stem</u> <u>and</u> <u>leaf</u> <u>Plot</u> shows the frequencies with which classes of value occur. To create this plot, we divide the set of numbers into 2 columns: <u>stem</u>, the left column, which contains the tens digits; <u>leaf</u>, the right column, which contains the unit digits.

<u>Normal</u> <u>distribution</u> is a type of distribution: it's a bell-shaped, symmetrical, unimodal distribution.

A stem and leaf plot displays the main features of the distribution. If turned on its side, we can see the shape of the data.

The figure below shows the stem and leaf plot of the 100-point test score. As we can see, when turned, the plot resembles bell-shaped distribution. So, this test scores were selected from a normal population.

b. <u>Mean</u> is the average number of a data set. It is calculated as the sum of all the data divided by the quantity the sample has:

mean = \frac{\Sigma x}{n}

For the 100-point test score:

mean = \frac{71+93+91+...+65+75+84}{20}

mean = 76.65

<u>Standard</u> <u>Deviation</u> determines how much the data is dispersed from the mean. It is calculated as:

s=\sqrt{\frac{\Sigma (x-mean)^{2}}{n-1} }

For the 100-point test score:

s=\sqrt{\frac{[(71-76.65)+(93-76.65)+...+(84-76.65)]^{2}}{20-1} }

s = 10.04

The mean and standard deviation of the scores are 76.65 and 10.04, respectively.

c. <u>Confidence</u> <u>Interval</u> is a range of values we are confident the real mean lies.

The calculations for the confidence interval is

mean ± z\frac{s}{\sqrt{n} }

where

z is the z-score for the 95% confidence interval, which is equal 1.96

Calculating interval

76.65 ± 1.96.\frac{10.04}{\sqrt{20} }

76.65 ± 4.4

The 95% confidence interval for the average test score in the population of students is between 72.25 and 81.05.

7 0
3 years ago
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