The radius of a clock face is 8.5 centimeters. What is the area of the clock face? Approximate Π as 3.14.

A number cube has sides labeled 1 to 6. Gia rolls the number cube. What is the probbillty the she rolls 4?

kotegsom [21]

<span>A cube Has 6 sides.So,n(A)=6={1,2,3,4,5,6} Probability to getting number 4 in a cube is Event A. n(s)=1 Probability of event A that occurs : P(A) = 1/6. P(A)=0.167 Probability of event A that does not occur P(A') = 1 - 0.167 =0.833 Probability to getting number 4 in a cube is 0.167</span>

8 0

4 years ago

Read 2 more answers

The quotient of 9 times an unknown number and 16 is 81.

lys-0071 [83]

Represent the unknown number with x
9x / 16 = 81
Multiply both sides by 16
9x = 1296
Divide both sides by 9
x = 144

5 0

3 years ago

Which of the

alisha [4.7K]

Given:

The inequality is:

$x>111$

To find:

The x-values that are the solutions to the given inequality.

Solution:

We have,

$x>111$

It means all the real numbers greater than 111 are the solutions to the given inequality.

$97$ , so 97 is not a solution.

$101$ , so 101 is not a solution.

$119>111$ , so 119 is a solution.

Therefore, the correct option is C.

7 0

3 years ago

Let l(x) be the statement "x has an Internet connection" and C(x, y) be the statement "x and y have chatted over the Internet,"

Bas_tet [7]

The question is:

Let l(x) be the statement "x has an Internet connection" and C(x, y) be the statement "x and y have chatted over the Internet," where the domain for the variables x and y consists of all students in your class. Use quantifiers to express each of these statements.

a. Jerry does not have an Internet connection.

b. Rachel has not chatted over the Internet with Chelsea.

c. Jan and Sharon have never chatted over the Internet.

d. No one in the class has chatted with Bob.

e. Sanjay has chatted with everyone except Joseph.

f. Someone in your class does not have an Internet connection.

g. Not everyone in your class has an Internet connection.

h. Exactly one student in your class has an internet connection.

i. Everyone except one student in your class has an Internet connection.

j. Everyone in your class with an Internet connection has chatted over the Internet with at least one other student in your class.

k. Someone in your class has an Internet connection but has not chatted with anyone else in your class.

l. There are two students in your class who have not chatted with each other over the Internet.

m. There is a student in your class who has chatted with everyone in your class over the Internet.

n. There are at least two students in your class who have not chatted with the same person in your class.

o. There are two students in the class who between them have chatted with everyone else in the class.

Step-by-step explanation:

Note that: Because the arguments are variables, we write

Ix = I(x)

Cxy = C(x,y)

a. Jerry does not have an Internet connection.

~I(Jerry).

b. Rachel has not chatted over the internet with Chelsea.

~C(Rachel,Chelsea).

c. Jan and Sharon have never chatted over the internet.

~C(Jan,Sharon).

d. No one in the class has chatted with Bob.

~∃x C(x, Bob).

e. Sanjay has chatted with everyone except Joseph.

∀y (y ≠ Joseph → C(Sanjay, y)).

f. Someone in your class does not have an internet connection.

∃x ~Ix.

g. Not everyone in your class has an internet connection.

~∀x Ix.

h. Exactly one student in your class has an internet connection.

∃x (Ix ∧ ∀y (Iy → y = x)).

i. Everyone except one student in your class has an internet connection. This means exactly one student does not have a connection.

∃x (~Ix∧ ∀y (Iy → y = x)).

j. Everyone in your class with an internet connection has chatted over the internet with at least one other student in your class.

∀x (Ix → ∃y (Cxy ∧ x ≠ y)).

Assuming nobody can chat with him or herself, then

∀x (Ix → ∃y (Cxy))

k. Someone in your class has an internet connection but has not chatted with anyone else in your class.

∃x (Ix ∧ ∀y ~Cxy).

l. There are two students in your class who have not chatted with each other over the internet.

∃x ∃y (x ≠ y∧~Cxy).

m. There is a student in your class who has chatted with everyone in your class over the internet.

∃x ∀y Cxy.

n. There are at least two students in your class who have not chatted with the same person in your class. ∃x ∃y (x ≠ y ∧ ∃z (~Cxz ∧ ~Cyz)).

o. There are two students in the class who between them have chatted with everyone else in the class.

∃x ∃y (x ≠ y ∧ ∀z (Cxz ∨ Cyz)).

5 0

4 years ago

The average undergraduate cost for tuition, fees, and room and board for two-year institutions last year was $13,252. The follow

kipiarov [429]

Answer:

The p-value of the test is 0.0041 < 0.05, which means that there is sufficient evidence at the 0.05 significance level to conclude that the mean cost has increased.

Step-by-step explanation:

The average undergraduate cost for tuition, fees, and room and board for two-year institutions last year was $13,252. Test if the mean cost has increased.

At the null hypothesis, we test if the mean cost is still the same, that is:

$H_0: \mu = 13252$

At the alternative hypothesis, we test if the mean cost has increased, that is:

$H_1: \mu > 13252$

The test statistic is:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, s is the standard deviation and n is the size of the sample.

13252 is tested at the null hypothesis:

This means that $\mu = 13252$

The following year, a random sample of 20 two-year institutions had a mean of $15,560 and a standard deviation of $3500.

This means that $n = 20, X = 15560, s = 3500$

Value of the test statistic:

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{15560 - 13252}{\frac{3500}{\sqrt{20}}}$

$t = 2.95$

P-value of the test and decision:

The p-value of the test is found using a t-score calculator, with a right-tailed test, with 20-1 = 19 degrees of freedom and t = 2.95. Thus, the p-value of the test is 0.0041.

The p-value of the test is 0.0041 < 0.05, which means that there is sufficient evidence at the 0.05 significance level to conclude that the mean cost has increased.

7 0

3 years ago