Which view is not correct for the figure above?

2. Determine if the expressions are equivalent by comparing the models.

Ksju [112]

Answer:

answer 4 y +rdcmeint or 67456

Step-by-step explanation:

8 0

2 years ago

5⁸ – = 5ⁿ 5⁴ work out the value of n.

Hitman42 [59]

Answer:

5 power 2 is the answer

Step-by-step explanation:

7 0

4 years ago

Your parents have decided to buy a new Toyota 4Runner for 25,635. According to the dealer, the car will depreciate in value appr

chubhunter [2.5K]

Let the value of the car be represented by V and the amount of years by y.
This gives us the following formula:
V = 25,635 - 3000y
(This is because we start with a value of $25,635 and the value decreases by $3,000 every year 'y')

Now, we want to know when the car is worth $3,135, so we know V = 3,135
Now we can make up our equation:
25,365 - 3,000y = 3,135

Collecting terms gives us:
-3,000y = -22,500

Finally we divide by -3,000 to find 'y'
y = -22,500 / -3,000 = 7.5

Hence, the car will be worth $3,135 after 7.5 years.

8 0

4 years ago

Read 2 more answers

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

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

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

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

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

= [19.891 , 24.909]

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

6 0

4 years ago

What is square root of 167?

enot [183]

The square root of 167 would be 12.922847. If you need it rounded to the nearest tenth it would be 12.9

If you need t rounded to the nearest hundredth, it would be 12.92.

Hope this helps :)

7 0

3 years ago