All categories

3 years ago

What property of an even function do you see in this graph?

Mathematics

1 answer:

Brut [27]3 years ago

6 0

Answer:

Step-by-step explanation:

I'm confused about what you want me to do but I know for a faction it is a function.

You might be interested in

what is the largest of three consecutive integers if the sum of the two larger integers is 3 more than two times the smallest in

Alina [70]

Answer:

Ambitious Ambitios Ambitious Ambitions

5 0

2 years ago

A line passes through (9, –9) and (10, –5).Write an equation for the line in point-slope form.

CaHeK987 [17]

Y - y1 = [-9-(-5)]/[(9-10] * (x - x1)

y - (-9) = [-9 + 5] / [-1] * (x - 9)

y + 9 = (-4) / (-1) * (x - 9)

y + 9 = 4(x - 9)

y + 9 = 4x - 36

y = 4x - 36 - 9

y = 4x - 45 This is the slope - intercept form.

y - 4x + 45 = 0 This is the standard form

5 0

2 years ago

83 random samples were selected from a normally distributed population and were found to have a mean of 32.1 and a standard devi

arlik [135]

Answer:

$\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}$

$4.525 \leq \sigma^2 \leq 7.602$

Now we just take square root on both sides of the interval and we got:

$2.127 \leq \sigma \leq 2.757$

Step-by-step explanation:

Information given

$\bar X=32.1$ represent the sample mean

$\mu$ population mean (variable of interest)

s=2.4 represent the sample standard deviation

n=83 represent the sample size

Confidence interval

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

The degrees of freedom given by:

$df=n-1=8-1=7$

The confidence level is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical values.

The excel commands would be: "=CHISQ.INV(0.05,82)" "=CHISQ.INV(0.95,82)". so for this case the critical values are:

$\chi^2_{\alpha/2}=104.139$

$\chi^2_{1- \alpha/2}=62.132$

The confidence interval is given by:

$\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}$

$4.525 \leq \sigma^2 \leq 7.602$

Now we just take square root on both sides of the interval and we got:

$2.127 \leq \sigma \leq 2.757$

5 0

3 years ago

Trigonometry help!! - double angle formulae

ivolga24 [154]

Answer:

The two rules we need to use are:

Sin(a + b) = sin(a)*cos(b) + sin(b)*cos(a)

cos(a + b) = cos(a)*cos(b) - sin(a)*sin(b)

And we also know that:

sin^2(a) + cos^2(a) = 1

To solve the relations, we start with the left side and try to construct the right side.

a) Sin(3*A) = sin (2*A + A) = sin(2*A)*cos(A) + sin(A)*cos(2*A)

sin(A + A)*cos(A) + sin(A)*cos(A + A)

(sin(A)*cos(A) + sin(A)*cos(A))*cos(A) + sin(A)*(cos(A)*cos(A) - sin(A)*sin(A))

sin(A)*cos^2(A) + sin(A)*cos^2(A) + sin(A)*cos^2(A) - sin^3(A)

3*sin(A)*cos^2(A) - sin(A)*sin^2(A)

sin(A)*(3*cos^2(A) - sin^2(A))

Now we can add and subtract 4*sin^3(A)

sin(A)*(3*cos^2(A) - sin^2(A)) + 4*sin^3(A) - 4*sin^3(A)

sin(A)*(3*cos^2(A) + 3*sin^2(A)) - 4*sin^3(A)

sin(A)*3*(cos^2(A) + sin^2(A)) - 4*sin^3(A)

3*sin(A) - 4*sin^3(A)

b) Here we do the same as before:

cos(3*A) = 4*cos^3(A) - 3*cos(A)

We start with:

Cos(2*A + A) = cos(2*A)*cos(A) - sin(2*A)*sin(A)

= cos(A + A)*cos(A) - sin(A + A)*sin(A)

= (cos(A)*cos(A) - sin(A)*sin(A))*cos(A) - ( sin(A)*cos(A) + sin(A)*cos(A))*sin(A)

= (cos^2(A) - sin^2(A))*cos(A) - sin^2(A)*cos(A) - sin^2(A)*cos(A)

= cos^3(A) - 3*sin^2(A)*cos(A)

= cos(A)*(cos^2(A) - 3*sin^2(A))

now we subtract and add 4*cos^3(A)

= cos(A)*(cos^2(A) - 3*sin^2(A)) + 4*cos^3(A) - 4*cos^3(A)

= cos(A)*(-3*cos^2(A) - 3*sin^2(A)) + 4*cos^3(A)

= cos(A)*(-3)*(cos^2(A) + sin^2(A)) + 4*cos^3(A)

= -3*cos(A) + 4*cos^3(A)

8 0

3 years ago

30 POINTS!!!!!!!<br><br> IT WILL TELL ME IF YOU ARE WRONG!!!!

Lady bird [3.3K]

Answer:

40 unit squared

Step-by-step explanation:

10*8=80

80/2=40

7 0

3 years ago