16. A sweater is marked 30% off during an

Which graph represents the inequality y>3−x?

nekit [7.7K]

Answer:

B

Step-by-step explanation:

The first step is to draw the line y = 3 - x to see what the line itself looks like. Is it going from left to right as in A and C is going up as you go from right to left as in B and D? The graph on the left (below) gives you the answer. It is going up as you progress from right to left.

The next step is to answer which is it: B or D.

y has to be above the line. it is B. See the graph below on the right.

4 0

3 years ago

Zen deposited some money in a savings account. The graph below shows the value of Zen's investment y, in dollars, after x months

kifflom [539]

The y-intercept of 1000 in the graph is at the point where x is equal to 0.

At $x=0$ , it means no month have passed. It is the initial point where Zen deposited the money.

What is the amount (y axis)? 1000!

So y represents the initial deposit of Zen, which is 1000.

ANSWER: Answer choice D (Amount of money Zen deposited in the savings account)

6 0

3 years ago

Read 2 more answers

Item 3 The Hampton family used 17,158 kilowatts of electricity last year. They used about the same amount of electricity each we

katen-ka-za [31]

17,158/52= 329.96 rounded to the nearest tenth is 330

7 0

3 years ago

An article contained the following observations on degree of polymerization for paper specimens for which viscosity times concen

devlian [24]

Answer:

(a) A 95% confidence interval for the population mean is [433.36 , 448.64].

(b) A 95% upper confidence bound for the population mean is 448.64.

Step-by-step explanation:

We are given that article contained the following observations on degrees of polymerization for paper specimens for which viscosity times concentration fell in a certain middle range:

420, 425, 427, 427, 432, 433, 434, 437, 439, 446, 447, 448, 453, 454, 465, 469.

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 = $\frac{\sum X}{n}$ = 441

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 14.34

n = sample size = 16

$\mu$ = population mean

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

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

P(-2.131 < $t_1_5$ < 2.131) = 0.95 {As the critical value of t at 15 degrees of

freedom are -2.131 & 2.131 with P = 2.5%}

P(-2.131 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.131) = 0.95

P( $-2.131 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.131 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.131 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.131 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X -2.131 \times {\frac{s}{\sqrt{n} } }$ , $\bar X +2.131 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $441-2.131 \times {\frac{14.34}{\sqrt{16} } }$ , $441+2.131 \times {\frac{14.34}{\sqrt{16} } }$ ]

= [433.36 , 448.64]

(a) Therefore, a 95% confidence interval for the population mean is [433.36 , 448.64].

The interpretation of the above interval is that we are 95% confident that the population mean will lie between 433.36 and 448.64.

(b) A 95% upper confidence bound for the population mean is 448.64 which means that we are 95% confident that the population mean will not be more than 448.64.

6 0

3 years ago

Given the functions f(x)= 1/x−4 +3 and g(x)=1/x+1 +6 .

melomori [17]

ANSWER

The graph shifts 5 units left and 3 units up.

EXPLANATION

The given functions are:

$f(x) = \frac{1}{x - 4} + 3$