Pls help me out lol. I’m desperate

Here is the feasible region.

8090 [49]

Answer:

A(8, 10)

Step-by-step explanation:

Vertex A is on the vertical line x=8, so we can find the y-coordinate by solving the boundary equation ...

5x +8y = 120

5(8) +8y = 120 . . . use the x-value of the vertical line

5 + y = 15 . . . . . . . divide by 8

y = 10 . . . . . . . . . . subtract 5

Point A is (8, 10).

6 0

4 years ago

The exact value of sec(-60) equals

MaRussiya [10]

This looks hard... good luck ! :D

4 0

3 years ago

Use the compound interest formula to determine the final value of the following amount. $1900 at 10.4% compounded monthly for 4.

nikklg [1K]

Answer:

$3027.80

Explanation:

The compound interest formula is the following.

$A=P(1+\frac{r}{n})^{nt}$

where

A = final amount

P = principle amount

r = interest rate / 100

n = number of compounds per interval

t = time interval

Now in our case,

A = unknown

P = $1900

r = 10.4/100

n = 12 months / year ( because the interest is compounded monthly)

t = 4.5 yrs

Therefore, the compound interest formula gives

$A=1900(1+\frac{10.4/100}{12})^{12*4.5}$

Using a calculator, we evaluate the above to get

$\boxed{A=\$3027.80}$

which is our answer!

8 0

1 year ago

In the data set below, what are the lower quartile, the median, and the upper quartile?

Studentka2010 [4]

The median, upper and lower quartiles of the data set are:

Lower Quartile = 46

Median = 49

Upper Quartile = 90

<h3>What is the Median?</h3>

The center value or middle value of a data set is the median.

<h3>What are the Upper and Lower Quartiles?</h3>

Upper quartile (Q3) is the center or middle data point of the second half of a data set.

Order the data set given as:

32, 46, 49, 49, 77, 90, 96

Lower Quartile = 32, (46), 49, 49, 77, 90, 96 = 46

Median = 32, 46, 49, (49,) 77, 90, 96 = 49

Upper Quartile = 32, 46, 49, 49, 77, (90,) 96 = 90

Learn more about the Median, Upper and Lower Quartiles on:

brainly.com/question/15572643

3 0

2 years ago

Suppose that the probability that any particle emitted by a radioactive material will penetrate a certain shield is 0.01. If 10

AnnyKZ [126]

Answer:

a)P=0.42

b) $n\geq 297$

Step-by-step explanation:

We have a binomial distribution, since the result of each experiment admits only two categories (success and failure) and the value of both possibilities is constant in all experiments. The probability of getting k successes in n trials is given by:

$P=\begin{pmatrix}n\\ k\end{pmatrix} p^k(1-p)^{n-k}=\frac{n!}{k!(n!-k!)}p^k(1-p)^{n-k}$

a) we have k=2, n=10 and p=0.01:

$P=\frac{10!}{2!(10!-2!)}0.01^2(1-0.01)^{10-2}\\P=\frac{10!}{2!*8!}0.01^2(0.99)^{8}\\P=45*0.01^2(0.99)^8=0.42$

b) We have, $1-(1-p)^n=P$ , Here P is the probability that at least one particle will penetrate the shield, this probabity has to be equal or greater than 0.95. Therefore, this will be equal to subtract from the total probability, the probability that the particles do not penetrate raised to the total number of particles.

$1-0.99^n\geq 0.95\\0.99^n\leq 1-0.95\\0.99^n\leq 0.05\\n\geq 297$

4 0

3 years ago