Can a tangent ratio be greater than one

There are 27 students in Ms. Langley's

Yuliya22 [10]

Answer:

405 Problems

Step-by-step explanation:

One student in class has 15 problems.

There are 27 students in the class.

Let's multiply 15 by 27.

$15*27=405$

In all, there would be 405 problems assigned to the whole class.

8 0

3 years ago

Night Owl Motel rents rooms with 2 beds for $68 per night and rooms with 1 bed for $57 per night. One evening Night Owl Motel re

mixer [17]

1 bed room = 12
2 bed room = 15

68 x 15 = 1020
57 x 12 = 684
-----------
1020 + 684 = $1704 (total made)

15 two-bed room + 12 one-bed room = 27 rooms in all

5 0

3 years ago

10 times The sum of half a numberand six is eight

GuDViN [60]

The numerical form of this worded expression is:

10(1/2n + 6)=8

4 0

3 years ago

An object is heated to 100°. It is left to cool in a room that

stepladder [879]

Answer:

Step-by-step explanation:

Use Newton's Law of Cooling for this one. It involves natural logs and being able to solve equations that require natural logs. The formula is as follows:

$T(t)=T_{1}+(T_{0}-T_{1})e^{kt}$ where

T(t) is the temp at time t

T₁ is the enviornmental temp

T₀ is the initial temp

k is the cooling constant which is different for everything, and

t is the time (here, it's in minutes)

If we are looking first for the temp after 20 minutes, we have to solve for the k value. That's what we will do first, given the info that we have:

T(t) = 80

T₁ = 30

T₀ = 100

t = 5

k = ?

Filling in to solve for k:

$80=30+(100-30)e^{5k}$ which simplifies to

$50=70e^{5k}$ Divide both sides by 70 to get

$\frac{50}{70}=e^{5k}$ and take the natural log of both sides:

$ln(\frac{5}{7})=ln(e^{5k})$

Since you're learning logs, I'm assuming that you know that a natural log and Euler's number, e, "undo" each other (just like taking the square root of something squared). That gives us:

$-.3364722366=5k$

Divide both sides by 5 to get that

k = -.0672944473

Now that we have a value for k, we can sub that in to solve for T(20):

$T(20)=30+(100-30)e^{-.0672944473(20)}$ which simplifies to

$T(20)=30+70e^{-1.345888946}$

On your calculator, raise e to that power and multiply that number by 70:

T(20)= 30 + 70(.260308205) and

T(20) = 30 + 18.22157435 so

T(20) = 48.2°

Now we can use that k value to find out when (time) the temp of the object cools to 35°:

T(t) = 35

T₁ = 30

T₀ = 100

k = -.0672944473

t = ?

$35=30+100-30)e^{-.0672944473t}$ which simplifies to

$5=70e^{-.0672944473t}$

Now divide both sides by 70 and take the natural log of both sides:

$ln(\frac{5}{70})=ln(e^{-.0672944473t})$ which simplifies to

-2.63905733 = -.0672944473t

Divide to get

t = 39.2 minutes

3 0

3 years ago

An article contained the following observations on degree of polymerization for paper specimens for which viscosity times concen

devlian [24]

Answer:

(a) A 95% confidence interval for the population mean is [433.36 , 448.64].

(b) A 95% upper confidence bound for the population mean is 448.64.

Step-by-step explanation:

We are given that article contained the following observations on degrees of polymerization for paper specimens for which viscosity times concentration fell in a certain middle range:

420, 425, 427, 427, 432, 433, 434, 437, 439, 446, 447, 448, 453, 454, 465, 469.

Firstly, the pivotal quantity for finding the 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 = $\frac{\sum X}{n}$ = 441

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 14.34

n = sample size = 16

$\mu$ = population mean

Here for constructing a 95% confidence interval we have used One-sample t-test statistics as we don't know about population standard deviation.

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

P(-2.131 < $t_1_5$ < 2.131) = 0.95 {As the critical value of t at 15 degrees of

freedom are -2.131 & 2.131 with P = 2.5%}

P(-2.131 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.131) = 0.95

P( $-2.131 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.131 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.131 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.131 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X -2.131 \times {\frac{s}{\sqrt{n} } }$ , $\bar X +2.131 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $441-2.131 \times {\frac{14.34}{\sqrt{16} } }$ , $441+2.131 \times {\frac{14.34}{\sqrt{16} } }$ ]

= [433.36 , 448.64]

(a) Therefore, a 95% confidence interval for the population mean is [433.36 , 448.64].

The interpretation of the above interval is that we are 95% confident that the population mean will lie between 433.36 and 448.64.

(b) A 95% upper confidence bound for the population mean is 448.64 which means that we are 95% confident that the population mean will not be more than 448.64.

6 0

3 years ago