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3 years ago

There+are+no+rational+numbers+that+are+also+whole+numbers

2 answers:

meriva3 years ago

8 0

Actually all the whole numbers are kind of rational numbers like : 1/1 , 2/1 , 3/1 , ...
so there are some rational numbers that are whole numbers too :))
i hope this be helpful
have a nice day

Ilya [14]3 years ago

4 0

No. ALL whole numbers are rational. And what's with all the plus signs ? ? They really make your questions hard to read.

You might be interested in

The sum of two numbers is 39. The sum of twice the larger number and three times the smaller number is 93. Find the smaller numb

ExtremeBDS [4]

Answer:3 * (n + 4) = 93

3n + 12 = 93

Subtract 12 to both sides:

3n = 81

Divide 3 to both sides:

n = 27

Step-by-step explanation:

6 0

3 years ago

Hey can some one help me, Define a variable and write each phrase as an algebraic expression:

Elena L [17]

Answer:

1. d - 6 ft

where d is the width

2. t + 6 h/week

where t is the amount of time per week Theodore studies

3. t’ - 6 a

where t’ is Tracy’s age

4. (k/3) - 2

where k is the points that the panthers scored

Step-by-step explanation:

Write the variable and do the operations as described!

“less than” means subtraction, “more […] than” means addition and “one-third of” means division by 3.

8 0

3 years ago

Plz plz help

viktelen [127]

X^2 - 10 = 3x
X^2 - 3x - 10 = 0
(x - 5) (x + 2) = 0
x = 5 and x = -2
Answer: x = -2

6 0

3 years ago

Historically, a certain region has experienced 92 thunder days annually. (A "thunder day" is day on which at least one instance

DiKsa [7]

Answer:

We conclude that the mean number of thunder days is less than 92.

Step-by-step explanation:

We are given that Historically, a certain region has experienced 92 thunder days annually.

Over the past fifteen years, the mean number of thunder days is 72 with a standard deviation of 38.

Let $\mu$ = mean number of thunder days.

So, Null Hypothesis, $H_0$ : $\mu \geq$ 92 days {means that the mean number of thunder days is more than or equal to 92}

Alternate Hypothesis, $H_A$ : $\mu$ < 92 days {means that the mean number of thunder days is less than 92}

The test statistics that would be used here One-sample t test statistics as we don't know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean number of thunder days = 72

s = sample standard deviation = 38

n = sample of years = 15

So, test statistics = $\frac{72-92}{\frac{38}{\sqrt{15} } }$ ~ $t_1_4$

= -2.038

The value of z test statistics is -2.038.

Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the t table gives critical value of -1.761 at 14 degree of freedom for left-tailed test.

Since our test statistics is less than the critical value of t as -2.038 < -1.761, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the mean number of thunder days is less than 92.

7 0

3 years ago

The number of students who attended FPM in 2015 was 1380. In 2016, 1405 students attended FPM. what was the percent of change in

Fittoniya [83]

The percent of change in student enrollment is 1.81 % increase

Solution:

Given that, The number of students who attended FPM in 2015 was 1380

In 2016, 1405 students attended FPM

To find: Percent change

The percent change is given by formula:

$\text { percent change }=\frac{\text { final value - initial value}}{\text { initial value }} \times 100$

If the result is positive, it is percent increase

If the result is negative, it is percent decrease

From given,

Initial value in 2015 = 1380

Final value in 2016 = 1405

Substituting the values in formula, we get

$\text { percent change }=\frac{1405-1380}{1380} \times 100\\\\\text { percent change }=\frac{25}{1380} \times 100\\\\\text { percent change }= 0.01811 \times 100\\\\\text { percent change }= 1.81$

Thus percent of change in student enrollment is 1.81 % increase

6 0

3 years ago