Evaluate 2p - 3r + t for p = 10, r = -4 and t = 5. 8 13 37

Evaluate The following expression . in e^e

erik [133]

$\bf \textit{Logarithm Cancellation Rules} \\\\ \stackrel{\stackrel{\textit{let's use this one}}{\downarrow }}{log_a a^x = x}\qquad \qquad a^{log_a x}=x \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ln(e^e)\implies log_e(e^e)\implies e$

3 0

3 years ago

What is 2/9 simplified?

zavuch27 [327]

2/9 is already simplified but it’s decimal form is
_
0.2

4 0

3 years ago

The National Center for Education Statistics surveyed a random sample of 4400 college graduates about the lengths of time requir

Paha777 [63]

Answer:

95% confidence interval for the mean time required to earn a bachelor’s degree by all college students is [5.10 years , 5.20 years].

Step-by-step explanation:

We are given that the National Center for Education Statistics surveyed a random sample of 4400 college graduates about the lengths of time required to earn their bachelor’s degrees. The mean was 5.15 years and the standard deviation was 1.68 years respectively.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean time = 5.15 years

$\sigma$ = sample standard deviation = 1.68 years

n = sample of college graduates = 4400

$\mu$ = population mean time

Here for constructing 95% confidence interval we have used One-sample z test statistics although we are given sample standard deviation because the sample size is very large so at large sample values t distribution also follows normal.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5%

level of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $5.15-1.96 \times {\frac{1.68}{\sqrt{4400} } }$ , $5.15+1.96 \times {\frac{1.68}{\sqrt{4400} } }$ ]

= [5.10 , 5.20]

Therefore, 95% confidence interval for the mean time required to earn a bachelor’s degree by all college students is [5.10 years , 5.20 years].

8 0

3 years ago

I need help SOMEONE PLZ HELP ME OUT WITH THIS !! THANK UU

Mama L [17]

Answer:

since they are alternate exteriors, we put a equal sign

2x+78=5x+15

next we subtract 78 on both sides

2x+78-78=5x+15-78

2x=5x-63

then we subtract 5x on both sides

2x-5x=5x-5x-63

-3x=-63

after that we divide -3 on both sides

-3x/-3=-63/-3

and the answer is

x=21

Step-by-step explanation:

8 0

2 years ago

USE A MODEL Olivia and her brother William had a bicycle race. Olivia rode at a speed

Anni [7]

Answer:

Let the total distance Aileen ran be 'x'

Let the distance ran by Andrew be 'y'

Refer to graph attached for graphical representation of distance travelled by both.

Applying the formula,

distance=speed*time

x=20*t Equation 1

y= 15*t equation 2

Since it can be easily understood by the graph attached, that

x=150+y equation 3

Now putting values of x and y in equation 3

20*t =150 +15*t.

5t= 150

t=30 seconds.

putting the values of t in both the equation 1 and 2

x=20t

=20*30

x = 600 feet

So the distance ran by Aileen is 600 feet, whereas Andrew ran 450 feet.

Step-by-step explanation:

Do not mind the names just change them. I had the same question before with different names so that is why. Hope this helps:)

7 0

2 years ago