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3 years ago

What is the maximum value of Sinx+cosx

Mathematics

1 answer:

Harrizon [31]3 years ago

6 0

The answer is 2 since both

$\sin(x),\cos(x)\in[0,1]$

The maximum value of $\sin(x)$ is 1 and the same for cosine. The sum becomes 1 + 1 which evaluates to 2.

Hope this helps.

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GIVING 100 POINTS FOR THIS QUESTION PLS PL SPL SPLS PLS HELPPP!!!!! !!

sergiy2304 [10]

Answer: 65000$ worth of things

This is edited, because I got it backwards last time.

We need to make 2500, but she only makes 4% of whatever she sells.

That means the equation is

2500 = 0.04x

Now, we just need to divide both sides by 0.04 and we get 65000.

Sorry for the confusion.

7 0

3 years ago

A jewelry store offers its own brand of customizable charm bracelets. You are able to choose from 9

Elis [28]

Answer:

By the way that it is worded, I believe that you are only allowed to put 1 charm on the bracelet, so there are 9 ways to pick a charm.

Hope this helps(and also if you have a more refined wording please put it in the comments)!

4 0

3 years ago

Which expression is not a polynomial

polet [3.4K]

4-x I know this because my teacher just went over this problem

6 0

3 years ago

Imagine that you would like to purchase a $275,000 home. Using 20% as

vfiekz [6]

Answer:

The mortgage chosen is option A;

15-year mortgage term with a 3% interest rate because it has the lowest total amount paid over the loan term of $270,470

Step-by-step explanation:

The details of the home purchase are;

The price of the home = $275,000

The mode of purchase of the home = Mortgage

The percentage of the loan amount payed as down payment = 20%

The amount used as down payment for the loan = $55,000

The principal of the mortgage borrowed, P = The price of the house - The down payment

∴ P = $275,000 - 20/100 × $275,000 = $275,000 - $55,000 = $220,000

The principal of the mortgage, P = $220,000

The formula for the total amount paid which is the cost of the loan is given as follows;

$Outstanding \ Loan \ Balance = \dfrac{P \cdot \left[\left(1+\dfrac{r}{12} \right)^n - \left(1+\dfrac{r}{12} \right)^m \right] }{1 - \left(1+\dfrac{r}{12} \right)^n }$

The formula for monthly payment on a mortgage, 'M', is given as follows;

$M = \dfrac{P \cdot \left(\dfrac{r}{12} \right) \cdot \left(1+\dfrac{r}{12} \right)^n }{\left(1+\dfrac{r}{12} \right)^n - 1}$

A. When the mortgage term, t = 15-years,

The interest rate, r = 3%

The number of months over which the loan is payed, n = 12·t

∴ n = 12 months/year × 15 years = 180 months

n = 180 months

The monthly payment, 'M', is given as follows;

M =

The total amount paid over the loan term = Cost of the mortgage

Therefore, we have;

220,000*0.05/12*((1 + 0.05/12)^360/( (1 + 0.05/12)^(360) - 1)

$M = \dfrac{220,000 \cdot \left(\dfrac{0.03}{12} \right) \cdot \left(1+\dfrac{0.03}{12} \right)^{180} }{\left(1+\dfrac{0.03}{12} \right)^{180} - 1} \approx 1,519.28$

The minimum monthly payment for the loan, M ≈ $1,519.28

The total amount paid over loan term, A = n × M

∴ A ≈ 180 × $1,519.28 = $273,470

The total amount paid over loan term, A ≈ $270,470

B. When t = 20 year and r = 6%, we have;

n = 12 × 20 = 240

$\therefore M = \dfrac{220,000 \cdot \left(\dfrac{0.06}{12} \right) \cdot \left(1+\dfrac{0.06}{12} \right)^{240} }{\left(1+\dfrac{0.06}{12} \right)^{240} - 1} \approx 1,576.15$

The total amount paid over loan term, A = 240 × $1,576.15 ≈ $378.276

The monthly payment, M = $1,576.15

C. When t = 30 year and r = 5%, we have;

n = 12 × 30 = 360

$\therefore M = \dfrac{220,000 \cdot \left(\dfrac{0.05}{12} \right) \cdot \left(1+\dfrac{0.05}{12} \right)^{360} }{\left(1+\dfrac{0.05}{12} \right)^{360} - 1} \approx 1,181.01$

The total amount paid over loan term, A = 360 × $1,181.01 ≈ $425,163

The monthly payment, M ≈ $1,181.01

The mortgage to be chosen is the mortgage with the least total amount paid over the loan term so as to reduce the liability

Therefore;

The mortgage chosen is option A which is a 15-year mortgage term with a 3% interest rate;

The total amount paid over the loan term = $270,470

8 0

3 years ago

A gambler has a coin which is either fair (equal probability heads or tails) or is biased with a probability of heads equal to 0

yawa3891 [41]

Answer:

(a) 0.1719

(b) 0.3504

Step-by-step explanation:

For every coin the number of heads follows a Binomial distribution and the probability that x of the 10 times are heads is equal to:

$P(x)=\frac{n!}{x!(n-x)!}*p^x*(1-p)^{10-x}$

Where n is 10 and p is the probability to get head. it means that p is equal to 0.5 for the fair coin and 0.3 for the biased coin

So, for the fair coin, the probability that the number of heads is less than 4 is:

$P(x$

Where, for example, P(0) and P(1) are calculated as:

$P(0)=\frac{10!}{0!(10-0)!}*0.5^0*(1-0.5)^{10-0}=0.0009\\P(1)=\frac{10!}{1!(10-1)!}*0.5^1*(1-0.5)^{10-1}=0.0098$

Then, $P(x$ , so there is a probability of 0.1719 that you conclude that the coin is biased given that the coin is fair.

At the same way, for the biased coin, the probability that the number of heads is at least 4 is:

$P(x\geq4 )=P(4)+P(5)+P(6)+...+P(10)$

Where, for example, P(4) is calculated as:

$P(4)=\frac{10!}{4!(10-4)!}*0.3^4*(1-0.3)^{10-4}=0.2001$

Then, $P(x\geq4 )=0.3504$ , so there is a probability of 0.3504 that you conclude that the coin is fair given that the coin is biased.

7 0

4 years ago