All categories

3 years ago

Is 3 sin(alpha)+4 cos(beta)=8 possible

Mathematics

2 answers:

sdas [7]3 years ago

8 0

Answer:

No, it's not possible

Step-by-step explanation:

We know that for various values of x , $-1\leq \sin x\leq 1$

and $-1\leq \cos x \leq 1$

For values of $\alpha$ , $-1\leq \sin \alpha\leq 1$

On multiplying all sides by 3, we get

$-3\leq 3\sin \alpha\leq 3$

For values of $\beta$ , $-1\leq\cos \beta\leq 1$

On multiplying all sides by 4, we get

$-4\leq 4\cos \beta\leq 4$

On adding all sides of $-3\leq 3\sin \alpha\leq 3\,\,,\,\,-4\leq 4\cos \beta\leq 4$ , we get

$-3-4\leq 3\sin \alpha+ 4\cos \beta\leq 3+4\\-7\leq 3\sin \alpha+ 4\cos \beta\leq 7$

Therefore, maximum value of $3\sin \alpha+ 4\cos \beta$ is 7.

So, it's not possible that $3\sin \alpha+ 4\cos \beta=8$

babunello [35]3 years ago

3 0

No because the maximum amount for sin and cos is 1.if alpha is 90 and beta is 0 the answer is 7 so it's not right.

You might be interested in

Which angles are corresponding angles?

Zolol [24]

Its ONL and PQS
theyre congruent

8 0

3 years ago

Read 2 more answers

Ms. Chu needs to buy hot dogs and buns for the school picnic. She needs at least one hot dog and bun for each of the 40 students

PtichkaEL [24]

Heksndurnfndjfndjwjfjrnwkfh

6 0

3 years ago

When an electric current passes through two resistors with resistance r1 and r2, connected in parallel, the combined resistance,

kondaur [170]

Answer:

The combined resistance of a circuit consisting of two resistors in parallel is given by:

$\frac{1}{R}=\frac{1}{r_1}+\frac{1}{r_2}$

where

R is the combined resistance

$r_1, r_2$ are the two resistors

We can re-write the expression as follows:

$\frac{1}{R}=\frac{r_1+r_2}{r_1r_2}$

$R=\frac{r_1 r_2}{r_1+r_2}$

In order to see if the function is increasing in r1, we calculate the derivative with respect to r1: if the derivative if > 0, then the function is increasing.

The derivative of R with respect to r1 is:

$\frac{dR}{dr_1}=\frac{r_2(r_1+r_2)-1(r_1r_2)}{(r_1+r_2)^2}=\frac{r_2^2}{(r_1+r_2)^2}$

We notice that the derivative is a fraction of two squared terms: therefore, both factors are positive, so the derivative is always positive, and this means that R is an increasing function of r1.

To solve this part, we use again the expression for R written in part a:

$R=\frac{r_1 r_2}{r_1+r_2}$

We start by noticing that there is a limit on the allowed values for r1: in fact, r1 must be strictly positive,

$r_1>0$

So the interval of allowed values for r1 is

$0$

From part a), we also said that the function is increasing versus r1 over the whole domain. This means that if we consider a certain interval

a ≤ r1 ≤ b

The maximum of the function (R) will occur at the maximum value of r1 in this interval: so, at

$r_1=b$

6 0

3 years ago

GRE Test: The Graduate Record Examination (GRE) is a test required for admission to many US graduate schools. Student’s scores o

Zepler [3.9K]

Answer:

a) $\bar X=144.3$

b) The 95% confidence interval would be given by (138.846;149.754)

c) n=34

d) n=298

e) n=1190

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

We can calculate the sample mean with this formula:

$\bar X = \frac{\sum_{i=1}^n x_i}{n}$

$\bar X=144.3$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=8.8$ represent the population standard deviation

n=10 represent the sample size

a) Find a point estimate of the population mean

We can calculate the sample mean with this formula:

$\bar X = \frac{\sum_{i=1}^n x_i}{n}$

$\bar X=144.3$

b) Construct a 95% confidence interval for the true mean score for this population.

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $t_{\alpha/2}=1.96$

Now we have everything in order to replace into formula (1):

$144.3-1.96\frac{8.8}{\sqrt{10}}=138.846$

$144.3+1.96\frac{8.8}{\sqrt{10}}=149.754$

So on this case the 95% confidence interval would be given by (138.846;149.754)

Part c

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

And on this case we have that ME =+20 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got $z_{\alpha/2}=1.960$ , replacing into formula (5) we got: