Solve the following -1/2x(8x +4)

Find the value of cos W rounded to the nearest hundredth, if necessary,

Aliun [14]

Answer:

cos W = 7/15

Step-by-step explanation:

Mathematically the cosine of an angle is the ratio of the adjacent to the hypotenuse side

From the diagram given, 14 is adjacent to vertex w and 30 faces the right angle which makes it the hypotenuse

So, we have it that;

cos W = 14/30

cos W = 7/15

4 0

3 years ago

HELP 100 POINTS The population of a city is modeled with the function P=250,000e^0.013t, where t is the number of years since 2

erma4kov [3.2K]

Pretty difficult problem, but that’s why I’m here.
First we need to identify what we’re looking for, which is t. So now plug 450k into equation and solve for t.
450000 = 250000e^0.013t
Now to solve this, we need to remember this rule: if you take natural log of e you can remove x from exponent. And natural log of e is 1.
Basically ln(e^x) = xln(e) = 1*x
So knowing this first we need to isolate e
450000/250000 = e^0.013t
1.8 = e^0.013t
Now take natural log of both
Ln(1.8) = ln(e^0.013t)
Ln(1.8) = 0.013t*ln(e)
Ln(1.8) = 0.013t * 1
Now solve for t
Ln(1.8)/0.013 = t
T= 45.21435 years
Now just to check our work plug that into original equation which we get:
449999.94 which is basically 500k (just with an error caused by lack of decimals)
So our final solution will be in the 45th year after about 2 and a half months it will reach 450k people.

6 0

2 years ago

Please help!<br><br>4^x*5^y=20, 4^y*5^x=20^2, x+y=?

photoshop1234 [79]

Answer:

x + y are either 2 or 4

Step-by-step explanation:

conclusion x + y = 2 OR 4

7 0

3 years ago

A pen company averages 1.2 defective pens per carton produced (200 pens). The number of defects per carton is Poisson distribute

nlexa [21]

Answer:

a. P(x = 0 | λ = 1.2) = 0.301

b. P(x ≥ 8 | λ = 1.2) = 0.000

c. P(x > 5 | λ = 1.2) = 0.002

Step-by-step explanation:

If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:

$P(k)=\frac{\lambda^{k}e^{-\lambda}}{k!}= \frac{1.2^{k}\cdot e^{-1.2}}{k!}$

a. What is the probability of selecting a carton and finding no defective pens?

This happens for k=0, so the probability is:

$P(0)=\frac{1.2^{0}\cdot e^{-1.2}}{0!}=e^{-1.2}=0.301$

b. What is the probability of finding eight or more defective pens in a carton?

This can be calculated as one minus the probablity of having 7 or less defective pens.

$P(k\geq8)=1-P(k$

$P(0)=1.2^{0} \cdot e^{-1.2}/0!=1*0.3012/1=0.301\\\\P(1)=1.2^{1} \cdot e^{-1.2}/1!=1*0.3012/1=0.361\\\\P(2)=1.2^{2} \cdot e^{-1.2}/2!=1*0.3012/2=0.217\\\\P(3)=1.2^{3} \cdot e^{-1.2}/3!=2*0.3012/6=0.087\\\\P(4)=1.2^{4} \cdot e^{-1.2}/4!=2*0.3012/24=0.026\\\\P(5)=1.2^{5} \cdot e^{-1.2}/5!=2*0.3012/120=0.006\\\\P(6)=1.2^{6} \cdot e^{-1.2}/6!=3*0.3012/720=0.001\\\\P(7)=1.2^{7} \cdot e^{-1.2}/7!=4*0.3012/5040=0\\\\$

$P(k$

c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?

We can calculate this as we did the previous question, but for k=5.

$P(k>5)=1-P(k\leq5)=1-\sum_{k=0}^5P(k)\\\\P(k>5)=1-(0.301+0.361+0.217+0.087+0.026+0.006)\\\\P(k>5)=1-0.998=0.002$

5 0

3 years ago

Can somebody plz answer this math problem correctly thanks!<br><br> WILL MARK BRAINLIEST :D

Soloha48 [4]

Answer:

Ellen and Blake answered 20 questions correctly since Ellen had 0.8 correct and that is 80% same with Blake

Stephen scored a .84 or an 84%

I hope this helps :)

Step-by-step explanation:

plz mark B R A I N L I E S T

3 0

3 years ago

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