MATHEMATICS COLLEGE

Susan reads a book at a rate of 1 page every 3 minutes. If her reading rate remains the same, which method could be used to determine the number of minutes for her to read 18 page? *

Answers

Answer 1

Answer:

Answer:

54 minutes

Step-by-step explanation:

Hope I helped!

I can almost guarantee this answer is correct.

Have a nice day!

Answer 2

Answer:

Answer:

54 minutes

Step-by-step explanation:

Every 3 minutes she reads 1

18x3=54

Hope it helps

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I know the answer is 36 but I need to know the working out and I am unsure how to work the question out.Plz explain

Thanks in advance!! :)

Answers

Answer:

Step-by-step explanation:

Since the 3 numbers have a ratio of 2:3:7, that means it simplifies to that. So, there must be a common factor, let’s say x, for each of the numbers. Thus, the numbers are 2x, 3x, and 7x. To find the mean, we add up all of the numbers and divide by the number of numbers: (2x + 3x + 7x)/3 = 12x/3 = 4x = 48. Dividing by 4 on both sides gets x = 12. The median of the numbers is the number in the middle which is 3x. Substituting x = 12, we get: 3(12) = 36.

I hope this helps!!! :)

What is three and sixteen hundredths written in expanded form

Answers

Answer:

3.16

Step-by-step explanation:

Three is equal to well, 3.

Sixteen hundredths gives you 0.16.

Three and sixteenth hundredths is the same as saying 3 and 0.16, which is just 3.16.

Three and sixteen hundredths in expanded form is 3.16

Simplify –2x 3 y × xy 2 . A. –3x 4y 3 B. –12x 4y 3 C. –2x 3y 4

Answers

You will simplify if there are fractions!

Can someone help me solve this please?

Answers

h=2A/b

Hope this helps!

It is believed that the average amount of money spent per U.S. household per week on food is about $98, with standard deviation $11. A random sample of 36 households in a certain affluent community yields a mean weekly food budget of $100. We want to test the hypothesis that the mean weekly food budget for all households in this community is higher than the national average. State the null and alternative hypotheses for this test, the test statistic and determine if the results significant at the 5% level.

Answers

Answer:

The null hypothesis is

$H_o : \mu =$ $98

The alternative hypothesis is

$H_a : \mu >$ $98

test statistics $t_s = 1.091$

The the result of the test statistics is significant

Step-by-step explanation:

From the question we are told that

The population mean is $\mu =$ $98

The standard deviation is $\sigma =$ $11

The sample size is $n = 36$

The sample mean is $\= x =$ $100

The level of significance is $\alpha = 5$ % = 0.05

The null hypothesis is

$H_o : \mu =$ $98

The alternative hypothesis is

$H_a : \mu >$ $98

Now the critical values for this level of significance obtained from critical value for z-value table is $z_\alpha = 1.645$

The test statistics is mathematically evaluated as

$t_s = (\= x - \mu)/( (\sigma )/( √(n) ) )$

substituting values

$t_s = (100 - 98)/( (11 )/( √(36) ) )$

$t_s = 1.091$

Looking at $z_\alpha \ and \ t_s$ we see that $z_\alpha \ > t_s$ hence the we fail to reject the null hypothesis

hence there is no sufficient evidence to conclude that the mean weekly food budget for all households in this community is higher than the national average.

Thus the the result is significant

At a college, 69% of the courses have final exams and 42% of courses require research papers. Suppose that 29% of courses have a research paper and a final exam. Let F be the even that a course has a final exam. Let R be the event that a course requires a research paper. (a) Find the probability that a course has a final exam or a research paper. Your answer is : (b) Find the probability that a course has NEITHER of these two requirements. Your answer is :

Answers

Answer:

a) 0.82

b) 0.18

Step-by-step explanation:

We are given that

P(F)=0.69

P(R)=0.42

P(F and R)=0.29.

P(course has a final exam or a research paper)=P(F or R)=?

P(F or R)=P(F)+P(R)- P(F and R)

P(F or R)=0.69+0.42-0.29

P(F or R)=1.11-0.29

P(F or R)=0.82.

Thus, the the probability that a course has a final exam or a research paper is 0.82.

P( NEITHER of two requirements)=P(F' and R')=?

According to De Morgan's law

P(A' and B')=[P(A or B)]'

P(A' and B')=1-P(A or B)

P(A' and B')=1-0.82

P(A' and B')=0.18

Thus, the probability that a course has NEITHER of these two requirements is 0.18.

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Stephanie claims that 26540is greater than 50. Do you agree or disagree? Explain.

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