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3 years ago

I need help with the following if possible

Mathematics

2 answers:

scoundrel [369]3 years ago

5 0

1. -1
2. 1
3. 5
4. -1
5.-1
6. -3

kenny6666 [7]3 years ago

3 0

Answer:

2+(-3)= 2-3= -1
-3+4=1
-4+9=5
7+(-8)= 7-8= -1
-2+1= 1
6+ -9= -3

You might be interested in

Prove that 1/sin^2A -1/tan^2A= 1

Vika [28.1K]

Answer:

Step-by-step explanation:

$LHS =\dfrac{1}{Sin^{2} \ A }-\dfrac{1}{Tan^{2} \ A }\\\\\\ = \dfrac{1}{sin^{2} \ A}- \dfrac{1}{\dfrac{Sin^{2} \ A}{Cos^{2} \ A}}\\\\\\= \dfrac{1}{sin^{2} \ A } - \dfrac{Cos^{2} \ A}{Sin^{2} \ A}\\\\\\= \dfrac{1-Cos^{2} \ A}{Sin^{2} \ A}\\\\\\= \dfrac{Sin^{2} \ A}{Sin^{2} \ A}\\\\\\= 1 = \ RHS$

Hint: 1 - Cos² A = Sin² A

4 0

3 years ago

In a population of 10,000, there are 5000 nonsmokers, 2500 smokers of one pack or less per day, and 2500 smokers of more than on

Kazeer [188]

Answer:

In one month, we will have 4,950 non-smokers, 2,650 smokers of one pack and 2,400 smokers of more than one pack.

In two months, we will have 4,912 non-smokers, 2,756 smokers of one pack and 2,332 smokers of more than one pack.

In a year, we will have 4,793 non-smokers, 3,005 smokers of one pack and 2,202 smokers of more than one pack.

Step-by-step explanation:

We have to write the transition matrix M for the population.

We have three states (nonsmokers, smokers of one pack and smokers of more than one pack), so we will have a 3x3 transition matrix.

We can write the transition matrix, in which the rows are the actual state and the columns are the future state.

- There is an 8% probability that a nonsmoker will begin smoking a pack or less per day, and a 2% probability that a nonsmoker will begin smoking more than a pack per day. Then, the probability of staying in the same state is 90%.

- For smokers who smoke a pack or less per day, there is a 10% probability of quitting and a 10% probability of increasing to more than a pack per day. Then, the probability of staying in the same state is 80%.

- For smokers who smoke more than a pack per day, there is an 8% probability of quitting and a 10% probability of dropping to a pack or less per day. Then, the probability of staying in the same state is 82%.

The transition matrix becomes:

$\begin{vmatrix} &NS&P1&PM\\NS& 0.90&0.08&0.02 \\ P1&0.10&0.80 &0.10 \\ PM& 0.08 &0.10&0.82 \end{vmatrix}$

The actual state matrix is

$\left[\begin{array}{ccc}5,000&2,500&2,500\end{array}\right]$

We can calculate the next month state by multupling the actual state matrix and the transition matrix:

$\left[\begin{array}{ccc}5000&2500&2500\end{array}\right] * \left[\begin{array}{ccc}0.90&0.08&0.02\\0.10&0.80 &0.10\\0.08 &0.10&0.82\end{array}\right] =\left[\begin{array}{ccc}4950&2650&2400\end{array}\right]$

In one month, we will have 4,950 non-smokers, 2,650 smokers of one pack and 2,400 smokers of more than one pack.

To calculate the the state for the second month, we us the state of the first of the month and multiply it one time by the transition matrix:

$\left[\begin{array}{ccc}4950&2650&2400\end{array}\right] * \left[\begin{array}{ccc}0.90&0.08&0.02\\0.10&0.80 &0.10\\0.08 &0.10&0.82\end{array}\right] =\left[\begin{array}{ccc}4912&2756&2332\end{array}\right]$

In two months, we will have 4,912 non-smokers, 2,756 smokers of one pack and 2,332 smokers of more than one pack.

If we repeat this multiplication 12 times from the actual state (or 10 times from the two-months state), we will get the state a year from now:

$\left( \left[\begin{array}{ccc}5000&2500&2500\end{array}\right] * \left[\begin{array}{ccc}0.90&0.08&0.02\\0.10&0.80 &0.10\\0.08 &0.10&0.82\end{array}\right] \right)^{12} =\left[\begin{array}{ccc}4792.63&3005.44&2201.93\end{array}\right]$

In a year, we will have 4,793 non-smokers, 3,005 smokers of one pack and 2,202 smokers of more than one pack.

3 0

3 years ago

Is this right? IDK I need help!!!

True [87]

Answer:

um... you are wrong

Step-by-step explanation:

its the the last one

you can't add when its division

you can change exponents from 1/x^-2

to x^2

<h2>I nweed brainliest</h2>

5 0

3 years ago

The hours of daylight, y, in Utica in days x from January 1, 2013 can be modeled by the equation y = 2.89sin(0.0145x-1.40) + 10.

Pie

Answer:

9.3 hours

Step-by-step explanation:

Given

$y = 2.89\sin(0.0145x-1.40) + 10.99.$

Required

Hours of sunlight on Feb 21, 2013

First, calculate the number of days from Jan 1, 2013 to Feb 21, 2013

$days = 52$

So:

$x = 52$

So, we have:

$y = 2.89\sin(0.0145x-1.40) + 10.99.$

$y = 2.89\sin(0.0145*52-1.40) + 10.99.$

$y = 2.89\sin(0.754-1.40) + 10.99.$

$y = 2.89\sin(-0.646) + 10.99.$

$y = 2.89*-0.6020 + 10.99.$

$y = -1.740 + 10.99.$

$y = 9.25$

$y = 9.3$ --- approximated

8 0

3 years ago

A pharmacist is trying to stock the pharmacy with cheaper alternatives for customers. She will stop carrying the more expensive

Leno4ka [110]

Answer:

c. The caplet form is the most expensive

Step-by-step explanation:

A pharmacist is trying to determine which of the form is most expensive. These tablets form comes in 50 mg tablets and these tablet cost is 75.00 for 100 tablets. Capsule comes in 75 mg capsule and cost is 63. 75 for bottle of 85 capsules caplet comes in 100 mg.and 40 caplets are cost of $ 70.00.

Based on these information caplet form is most expensive

8 0

3 years ago

Read 2 more answers