Which pairs of angles in the figure below are vertical angles? Check all that apply.

PLS HELP ME ASAP FOR 18!! (MUST SHOW WORK/ STEPS!!) + LOTS OF POINTS!! *brainliest*

Tasya [4]

I'd say -2.75
or 2 3/4

4 0

3 years ago

Use f(x) = 1/2 x and f -1(x) = 2x to solve the problems. f(2) = 1 f−1(1) = 2 f−1(f(2)) = 2 f−1(−2) = f(−4) = f(f−1(−2)) =

vampirchik [111]

Answer:

In this problem, we are given the following functions:

$f(x)=\frac{1}{2}x$

and its inverse function:

$f^{-1}(x)=2x$

First of all, we want to calculate $f(2)$ . This can be obtained by substituting

x = 2

into f(x). Doing so, we find:

$f(2)=\frac{1}{2}\cdot 2 = 1$

Then we want to calculate $f^{-1}(1)$ . We can do it by substituting

x = 1

into $f^{-1}(x)$ . Doing so,

$f^{-1}(1)=2\cdot 1 = 2$

Then we want to calculate $f^{-1}(f(2))$ , which can be found by calculating f(2) and then using it as input for $f^{-1}(x).$ We know that

f(2) = 1

Therefore,

$f^{-1}(f(2))=f^{-1}(1)=2$

Then we want to calculate $f^{-1}(-2)$ , which can be calculated by plugging

x = -2

into $f^{-1}(x)$ . Doing so,

$f^{-1}(-2)=2\cdot (-2)=-4$

Then we want to calculate $f(-4)$ ; by substituting

x = 4

into f(x), we find

$f(-4)=\frac{1}{2}\cdot (-4)=-2$

Finally, we want to find $f(f^{-1}(-2))$

We know already that

$f^{-1}(-2)=-4$

So we have:

$f(f^{-1}(-2))=f(-4)=-2$

7 0

3 years ago

Read 2 more answers

I need help on this

Dvinal [7]

Answer:

1/12, 1/10

Step-by-step explanation:

P(Bead bracelet and red shoes) - 3/3 x 2 x 2

3/12=1/4 for bead bracelet

Red shoes - 1/3

Total probability for P(Bead bracelet and red shoes=1/4 x 1/3=1/12

P(Elastic bracelet and black shoes)

2/3 x 2 x 2

2/12=1/6 for elastic bracelet

Black shoes - 3/5

Total probablility for P(Elastic bracelet and black shoes)=1/6 x 3/5=3/10=1/10

Hope I helped!

5 0

3 years ago

Probability of getting 2 hears and 1 diamond in the first draw of 52 cards

iogann1982 [59]

Answer: 26/52

Step-by-step explanation:

5 0

3 years ago

The thickness of metal wires used in the manufacture of silicon wafers is assumed to be normallydistributed with meanμ. To monit

yanalaym [24]

Answer:

(a) Null Hypothesis, $H_0$ : $\mu$ = 10

(b) Alternate Hypothesis, $H_A$ : $\mu\neq$ 10

(c) One-sample t test statistics distribution : T.S. = $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

(d) The value of the test statistic is 1.054.

(e) The p-value = 0.1493

(f) We conclude that the mean equals the target value of 10 which means that the output can be considered acceptable as it doesn't differs from the target value of 10.

Step-by-step explanation:

We are given that the thickness of metal wires used in the manufacture of silicon wafers is assumed to be normally distributed with mean μ. To monitor the production process, the thickness of 40 wires is taken.

The output is considered unacceptable if the mean differs from the target value of 10. The 40 measurements yield a sample mean of 10.2 and sample standard deviation of 1.2.

Let $\mu$ = mean thickness of metal wires used in the manufacture of silicon wafers.

(a) Null Hypothesis, $H_0$ : $\mu$ = 10 {means that the mean equals the target value of 10}

(b) Alternate Hypothesis, $H_A$ : $\mu\neq$ 10 {means that the mean differs from the target value of 10}

(c) The test statistics that will be used here is One-sample t test statistics as we don't know about the population standard deviation;

T.S. = $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\mu$ = sample mean = 10.2

s = sample standard deviation = 1.2

n = sample of yields = 40

(d) So, test statistics = $\frac{10.2-10}{\frac{1.2}{\sqrt{40} } }$ ~ $t_3_9$

= 1.054

The value of the test statistic is 1.054.

(e) Now, P-value of the test statistics is given by;

P-value = P( $t_3_9$ > 1.054) = 0.1493

Now at 0.05 significance level, the t table gives critical values between -2.0225 and 2.0225 at 39 degree of freedom for two-tailed test. Since our test statistics lies within the critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

(e) Therefore, we conclude that the mean equals the target value of 10 which means that the output can be considered acceptable as it doesn't differs from the target value of 10.

7 0

3 years ago