All categories

3 years ago

Write n^4 • n^3 without exponents

Mathematics

1 answer:

Elis [28]3 years ago

8 0

Answer:

n*n*n*n*n*n*n

Step-by-step explanation:

n^4 * n^3 = n^7 = n*n*n*n*n*n*n

You might be interested in

I need help pls no wronge answers this is for a test

VashaNatasha [74]

Answer:

I think the answer is

Step-by-step explanation:

(D) OR (B)

4 0

3 years ago

The average weight of the class of 35 students was 45 KG. with the admission of a new student the average weight came down to 44

melamori03 [73]

The weight of the new student is 27 kg.

Average weight

= total weight ÷total number of students

<h3>1) Define variables</h3>

Let the total weight of the 35 students be y kg and the weight of the new student be x kg.

<h3>2) Find the total weight of the 35 students</h3>

y= 35(45)

y= 1575 kg

<h3>3) Write an expression for average weight of students after the addition of the new student</h3>

New total number of students

= 35 +1

= 36

Total weight

= total weight of 35 students +weight of new students

= y +x

$44.5 = \frac{y + x}{36}$

<h3>4) Substitute the value of y</h3>

$44.5 = \frac{1575 + x}{36}$

<h3>5) Solve for x</h3>

36(44.5)= 1575 +x

1602= x +1575

<em>Subtract 1575 from both sides:</em>

x= 1602 -1575

x= 27

Thus, the weight of the new student is 27 kg.

7 0

3 years ago

Question 3 Translate this sentence into an equation. The sum of 13 and Julie's age is 50. Use the variable to represent Julie's

LekaFEV [45]

Answer: 13 + x = 50

Step-by-step explanation: the sum is 50 , 13 is provided , julie’s age is the variable

3 0

3 years ago

Construct a 99% confidence interval for the population mean, mu. Assume the population has a normal distribution. A group of 1

Zarrin [17]

Answer:

99% confidence interval for the population mean is [19.891 , 24.909].

Step-by-step explanation:

We are given that a group of 19 randomly selected students has a mean age of 22.4 years with a standard deviation of 3.8 years.

Assuming the population has a normal distribution.

Firstly, the pivotal quantity for 99% confidence interval for the population mean is given by;

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

where, $\bar X$ = sample mean age of selected students = 22.4 years

s = sample standard deviation = 3.8 years

n = sample of students = 19

$\mu$ = population mean

<em>Here for constructing 99% confidence interval we have used t statistics because we don't know about population standard deviation.</em>

So, 99% confidence interval for the population mean, $\mu$ is ;

P(-2.878 < $t_1_8$ < 2.878) = 0.99 {As the critical value of t at 18 degree of

freedom are -2.878 & 2.878 with P = 0.5%}

P(-2.878 < $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ < 2.878) = 0.99

P( $-2.878 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.878 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

P( $\bar X -2.878 \times {\frac{s}{\sqrt{n} }$ < $\mu$ < $\bar X +2.878 \times {\frac{s}{\sqrt{n} }$ ) = 0.99

<u>99% confidence interval for</u> $\mu$ = [ $\bar X -2.878 \times {\frac{s}{\sqrt{n} }$ , $\bar X +2.878 \times {\frac{s}{\sqrt{n} }$ ]

= [ $22.4 -2.878 \times {\frac{3.8}{\sqrt{19} }$ , $22.4 +2.878 \times {\frac{3.8}{\sqrt{19} }$ ]

= [19.891 , 24.909]

Therefore, 99% confidence interval for the population mean is [19.891 , 24.909].

6 0

3 years ago

What is the inverse of the function fx) x +3?

xz_007 [3.2K]

<span>My answer -
f(x)=x^3+2 To find the inverse function f(x)^-1 , we will assume" y=x^3+2 y-2= x^3 ==> x= (y-2)^1/3 ==> f(x)-1= (x-2)^1/3</span>
P.S

Happy to help you have an AWESOME!!!!! day

6 0

3 years ago