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3 years ago

How do you convert set notation to interval notation?

Mathematics

1 answer:

Semenov [28]3 years ago

6 0

Answer:

Imagine we are to express [ a , b ] in set notation

A = [ a , b ] , then A = { x ∈ R / a ≤ x ≤ b }

In this notation we define the characteristics of all x belonging to this set A ....x must be greater or equal to a and simultaneously smaller or equal to b...

Interval notation is other way to say the same but assuming that means the extreme a is IN the interval and means extreme a is not.

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Is it a cylinder or a sphere?

Hunter-Best [27]

Answer: Sphere

Step-by-step explanation:

7 0

2 years ago

What is an equation of the line that passes through the points (-3,8) and (-7,8) help pleaseeee

Bingel [31]

Answer:

y = 8

Step-by-step explanation:

find slope first

(8-8)/(-7-(-3)) = 0/-4, the slope is 0

You know that 8 is y intercept

Equation: y = 0x + 8, or y = 8

3 0

2 years ago

The following are the ages (years) of 5 people in a room:

strojnjashka [21]

Answer:

67 years old. (you only gave 4 people and missed out one person)

Step-by-step explanation:

take all the ages first and put the unknown (age of the person who entered the room) as x.

14, 16, 16, 13, x

all of these numbers added to together would be:

59 + x

Because the question is talking about the mean age, to find mean:

add all the values in the data (ages) and divide by the total number of values on the list. (in your case would be 6 people (following your question) but you gave only 5 people including the x).

then (59 + x) / 6 = 21

21 x 6 = 126

126 = 59 + x

x = 67

7 0

2 years ago

Im absolutely clueless with part b and c

Elden [556K]

Answer:

19.8 and 8/(sin(theta))+ 6/(cos(theta))

Step-by-step explanation:

If you make a table you can see that L_1 + L_2 gets larger as you increase theta prom pi/2 and decrease theta from pi/2. This means that pi/2 is the theta that will yield the smallest length for the ladder. Plugging this into L_1 + L_2 you get 19.8 (rounded to the nearest hundred)

C = 8/(sin(theta))+ 6/(cos(theta))

7 0

3 years ago

A study of 4,430 U.S. adults found that 2,913 were obese or overweight. The 95% confidence interval for the proportion of adults

IrinaK [193]

Answer:

$0.658 - 1.96\sqrt{\frac{0.658(1-0.658)}{4430}}=0.644$

$0.658 + 1.96\sqrt{\frac{0.658(1-0.658)}{4430}}=0.672$

We are 95% confident that the true proportion of adults overweight is between 0.644 and 0.672

Step-by-step explanation:

We can begin founding the estimated proportion of people overweight:

$\hat p =\frac{2913}{4430}= 0.658$

We need to find a critical value for the confidence level using the normal standard distribution. We know that 95% is the confidence level, then the significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The confidence interval for the true population proportion of interest is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

Replacing the data provided we got:

$0.658 - 1.96\sqrt{\frac{0.658(1-0.658)}{4430}}=0.644$

$0.658 + 1.96\sqrt{\frac{0.658(1-0.658)}{4430}}=0.672$

We are 95% confident that the true proportion of adults overweight is between 0.644 and 0.672

6 0

2 years ago