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MAXImum [283]
3 years ago
15

What is the range of y = -5sin(x)?

Mathematics
1 answer:
klemol [59]3 years ago
8 0

Answer:

[-5,5]

Step-by-step explanation:

The minimum value of just sin(x) would be -1, while the max would be 1.

Now we just have to multiply these by the -5 that is there. That gives us a min of -5 and a max of 5

You might be interested in
Kayla squared a number x and added this result to -6.5. This gave her an answer of 42.5.
ladessa [460]

Answer:

7, -7

Step-by-step explanation:

x^2-6.5= 42.5 add 6.5 to both sides

x^2= 49

The square root of 49 is 7 and -7.

7 0
3 years ago
Find the equations of the following lines
White raven [17]

You need these two basic solutions and facts to find the equation of a line:

  • If you know the gradient m and one point (x_0,y_0):

y-y_0=m(x-x_0)

  • If you know the gradient two points (x_1,y_1),\ (x_2,y_2):

\dfrac{x-x_2}{x_1-x_2}=\dfrac{y-y_2}{y_1-y_2}

  • The slope of a line is the coefficient m when you write it in the y=mx+q form
  • Parallel lines have the same slope
  • The slopes of perpendicular lines give -1 when multiplied

We can use this list to solve all the exercises:

b)

Use the first equation to get

y-1=-4(x-2) \iff y=-4x+9

c)

Use the second equation to get

\dfrac{x-4}{2-4}=\dfrac{y-2}{-1-2} \iff \dfrac{x-4}{-2}=\dfrac{y-2}{-3}\iff 3(x-4)=2(y-2) \iff 3x-12=2y-4 \iff 2y = 3x-8 \iff y = \frac{3}{2}x-4

d) same as c)

e) We derive the slope of the line by writing it as

5y = -x-15 \iff y = -\dfrac{1}{5}x-3

So, the slope is -1/5. From here, it's the same as b)

f) same as e)

g) Again we find the slope as

3x+y+2=0\iff y=-3x-2

so the slope is -3, and a perpendicular line has slope 1/3. From there, it's the same as b).

6 0
4 years ago
Hi everyone, I’m currently trying to dive into some lessons before school starts and I’m taking algebra 2 this year and the less
3241004551 [841]
An imaginary number is a complex number that can be written as a real number multiplied by the imaginary unit i, which is defined by its property i² = −1. The square of an imaginary number bi is −b². For example, 5i is an imaginary number, and its square is −25
5 0
3 years ago
For the hypothesis test H0: μ = 10 against H1: μ >10 and variance known, calculate the Pvalue for each of the following test
Basile [38]

Answer:

a) p_v =P(Z>2.05)=1-P(z

b) p_v =P(Z>-1.84)=1-P(z

c) p_v =P(Z>0.4)=1-P(z

Step-by-step explanation:

Some previous concepts

The p-value is the probability of obtaining the observed results of a test, assuming that the null hypothesis is correct.

A z-test for one mean "is a hypothesis test that attempts to make a claim about the population mean(μ)".

The null hypothesis attempts "to show that no variation exists between variables or that a single variable is no different than its mean"

The alternative hypothesis "is the hypothesis used in hypothesis testing that is contrary to the null hypothesis"

Hypothesis

Null hypothesis: \mu=10

Alternative hypothesis: \mu >10

If the random variable is distributed like this: X \sim N(\mu,\sigma)

We assume that the variance is known so the correct test to apply here is the z test to compare means, the statistic is given by the following formula:

z_o=\frac{\bar X -\mu}{\sigma}

Since we have the values for the statistic already calculated we can calculate the p value using the following formulas:

Part a

p_v =P(Z>2.05)=1-P(z

And in order to find the answer using excel we can use the following code:

"=1-NORM.DIST(2.05,0,1,TRUE)"

Part b

p_v =P(Z>-1.84)=1-P(z

And in order to find the answer using excel we can use the following code:

"=1-NORM.DIST(-1.84,0,1,TRUE)"

Part c

p_v =P(Z>0.4)=1-P(z

And in order to find the answer using excel we can use the following code:

"=1-NORM.DIST(0.4,0,1,TRUE)"

Conclusions

If we use a reference value for the significance, let's say \alpha=0.05. For part a the p_v so then we can reject the null hypothesis at this significance level.

For part b the p_v>\alpha so then we FAIL to reject the null hypothesis at this significance level.

For part c the p_v>\alpha so again we FAIL to reject the null hypothesis at this significance level.

3 0
3 years ago
Part A: The average rate of change between the 2nd and 3rd point is Select a Value Part A: The average rate of change between th
Fofino [41]

Answer:

Rate = 6

Rate = 10

Rate = 18

Rate = 162

Step-by-step explanation:

Given

See attachment for table

Solving (a): Rate of change between 2nd and 3rd point on A

The rate of change is calculated as:

Rate = \frac{y_2 - y_1}{x_2 - x_1}

In table A, the 2nd and 3rd point is:

(x_1,y_1) =(1,7)

(x_2,y_2) =(2,13)

So, the average rate of change is:

Rate = \frac{13 - 7}{2 - 1}

Rate = \frac{6}{1}

Rate = 6

Solving (b): Rate of change between 3rd and 4th point on A

In table A, the 3rd and 4th point is:

(x_1,y_1) =(2,13)

(x_2,y_2) =(3,23)

So, the average rate of change is:

Rate = \frac{23 - 13}{3 - 2}

Rate = \frac{10}{1}

Rate = 10

Solving (c): Rate of change between 2nd and 3rd point on B

In table B, the 2nd and 3rd point is:

(x_1,y_1) =(2,11)

(x_2,y_2) =(3,29)

So, the average rate of change is:

Rate = \frac{29 - 11}{3 - 2}

Rate = \frac{18}{1}

Rate = 18

Solving (d): Rate of change between 4th and 5th point on B

In table B, the 4th and 5th point is:

(x_1,y_1) =(4,83)

(x_2,y_2) =(5,245)

So, the average rate of change is:

Rate = \frac{245 - 83}{5 - 4}

Rate = \frac{162}{1}

Rate = 162

7 0
3 years ago
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