All categories

3 years ago

HELP PLEASE!!!!!!

Mathematics

2 answers:

Westkost [7]3 years ago

8 0

False true true false true false

spayn [35]3 years ago

5 0

Answer:

The function is discontinuous at x = 1.

The value of the function is never negative.

The value of the function at x = -1 is 0

Step-by-step explanation:

Given is a graph of a function first a straight line, followed by a semi circle centred at the origin and another piece of straight line not connected to the other graph.

We can see that the funciton is discontinuous at x=1 since

right limit = 1 and left limit =0

x=1 is in the domain .

f(-1) =0 is right

But f(1) =0 is incorrect

Thus correct options are

The function is discontinuous at x = 1.

The value of the function is never negative.

The value of the function at x = -1 is 0

You might be interested in

What is the value of 4 in the number 546 210

Komok [63]

Answer:

4 has value of 10,000 which ten thousand place value.

Step-by-step explanation:

Given number 546210

We have to find the value of 4 in given number 546210.

We write the given number in value form as ,

546210 = 5×100000 + 4× 10000 + 6 ×1000 + 2×100 + 10

in words it is written as

Here, 4 has value of 10,000 which ten thousand place value.

3 0

4 years ago

Read 2 more answers

What decimal is represented by this expanded form?

Radda [10]

2*10000 is 20000.

Start your expanded form with 20000. Add 7000, add 400, add 9, add 0.3 and add 0.008:

27409.308 (answer)

6 0

4 years ago

I NEED HELP ASAP ⚠️⚠️ If 1 is a solution to g(x) = 0, what is the other solution?

bezimeni [28]

9514 1404 393

Answer:

A. -1

Step-by-step explanation:

The table shows that g(x) = 0 for x = -1. (third row)

The other solution is x = -1.

8 0

3 years ago

IF UR AN EXPERT IN MATH CAN U PLSPLSPLSPLSPLSPLS ANSWER THIS QUESTION, I WILL GIVE BRAINLIEST 5 STARS AND A THANKS :C DUE TODAY!

Lynna [10]

Answer:

The answer is $2 + \frac{1}{4}x < \frac{1}{3}x$

Step-by-step explanation:

<h2>PLEASE MARK AS BRAINLIEST!!!!!</h2>

3 0

3 years ago

Read 2 more answers

The amount of money spent on textbooks per year for students is approximately normal.

Contact [7]

Answer:

(A) A 95% confidence for the population mean is [$332.16, $447.84] .

(B) If the confidence level in part (a) changed from 95% to 99%, then the margin of error for the confidence interval would increase.

(C) If the sample size in part (a) changed from 19 to 22, then the margin of error for the confidence interval would decrease.

(D) A 99% confidence interval for the proportion of students who purchase used textbooks is [0.363, 0.477] .

Step-by-step explanation:

We are given that 19 students are randomly selected the sample mean was $390 and the standard deviation was $120.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

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

where, $\bar X$ = sample mean = $390

s = sample standard deviation = $120

n = sample of students = 19

$\mu$ = population mean

Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-2.101 < $t_1_8$ < 2.101) = 0.95 {As the critical value of t at 18 degrees of

freedom are -2.101 & 2.101 with P = 2.5%}

P(-2.101 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.101) = 0.95

P( $-2.101 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.101 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.101 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.101 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-2.101 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.101 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $\$390-2.101 \times {\frac{\$120}{\sqrt{19} } }$ , $\$390+2.101 \times {\frac{\$120}{\sqrt{19} } }$ ]

= [$332.16, $447.84]

(A) Therefore, a 95% confidence for the population mean is [$332.16, $447.84] .

(B) If the confidence level in part (a) changed from 95% to 99%, then the margin of error for the confidence interval which is $Z_(_\frac{\alpha}{2}_) \times \frac{s}{\sqrt{n} }$ would increase because of an increase in the z value.

(C) If the sample size in part (a) changed from 19 to 22, then the margin of error for the confidence interval which is $Z_(_\frac{\alpha}{2}_) \times \frac{s}{\sqrt{n} }$ would decrease because as denominator increases; the whole fraction decreases.

(D) We are given that to estimate the proportion of students who purchase their textbooks used, 500 students were sampled. 210 of these students purchased used textbooks.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion students who purchase their used textbooks = $\frac{210}{500}$ = 0.42

n = sample of students = 500

p = population proportion

Here for constructing a 99% confidence interval we have used a One-sample z-test statistics for proportions

So, 99% confidence interval for the population proportion, p is ;

P(-2.58 < N(0,1) < 2.58) = 0.99 {As the critical value of z at 0.5%

level of significance are -2.58 & 2.58}

P(-2.58 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 2.58) = 0.99

P( $-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

P( $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

99% confidence interval for p = [ $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.42 -2.58 \times {\sqrt{\frac{0.42(1-0.42)}{500} } }$ , $0.42 +2.58 \times {\sqrt{\frac{0.42(1-0.42)}{500} } }$ ]

= [0.363, 0.477]

Therefore, a 99% confidence interval for the proportion of students who purchase used textbooks is [0.363, 0.477] .

8 0

3 years ago