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3 years ago

8

What is the equation of the line whose Y intercept of three and So what is the equation of the line whose Y-intercept is 3 and s

lope is 1?

1 answer:

viva [34]3 years ago

3 0

Answer:

y = x + 3

Step-by-step explanation:

the equation ofa line in slope-intercept form is

y = mx + c ( m is the slope and c the y-intercept )

here m = 1 and c = 3, hence

y = x + 3 ← is the equation of the line

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Jack opens a bank account with $80 and decides to deposit $25 each month into the account. Which equation can be used to find th

kupik [55]

<span>405=25n+80

Hope this helped!</span>

7 0

4 years ago

The New York Rangers hockey team won 3/4 Of their games last season. If they lost 21 games, how many games did they play in the

bija089 [108]

We know that the won 3/4. So that means that the remaining 1/4 is the games lost, which is 21.
To find the total games played, we simply multiply 21 by 4 to get 4/4, or the total games.
$21 \times 4 = 84$

3 0

3 years ago

An article reported that, in a study of a particular wafer inspection process, 356 dies were examined by an inspection probe and

Lady bird [3.3K]

Answer:

95% (two-sided) confidence interval for the proportion of all dies that pass the probe is between 0.470 and 0.574

Step-by-step explanation:

Confidence interval can be calculated as p±ME where

p is the sample proportion of dies that pass the probe
ME is the margin of error

and margin of error (ME) around the mean can be found using the formula

ME= $\frac{z*\sqrt{p*(1-p)}}{\sqrt{N} }$ where

z is the corresponding statistic in 95% confidence level (1.96)
p is the sample proportion ( $\frac{186}{356}=0.522$ )

N is the sample size (356)

Using the numbers in the formula we get

ME= $\frac{1.96*\sqrt{0.522*0.478}}{\sqrt{356} }$ ≈0.052

then the 95% confidence interval would be 0.522±0.052

6 0

4 years ago

19) Given that f(x)x² - 8x+ 15x² - 25find the horizontal and vertical asymptotes using the limits of the function.A) No Vertical

Tems11 [23]

EXPLANATION

Since we have the function:

$f(x)=\frac{x^2-8x+15}{x^2}$

Vertical asymptotes:

$For\:rational\:functions,\:the\:vertical\:asymptotes\:are\:the\:undefined\:points,\:also\:known\:as\:the\:zeros\:of\:the\:denominator,\:of\:the\:simplified\:function.$

Taking the denominator and comparing to zero:

$x+5=0$

The following points are undefined:

$x=-5$

Therefore, the vertical asymptote is at x=-5

Horizontal asymptotes:

$\mathrm{If\:denominator's\:degree\:>\:numerator's\:degree,\:the\:horizontal\:asymptote\:is\:the\:x-axis:}\:y=0.$ $If\:numerator's\:degree\:=\:1\:+\:denominator's\:degree,\:the\:asymptote\:is\:a\:slant\:asymptote\:of\:the\:form:\:y=mx+b.$ $If\:the\:degrees\:are\:equal,\:the\:asymptote\:is:\:y=\frac{numerator's\:leading\:coefficient}{denominator's\:leading\:coefficient}$ $\mathrm{If\:numerator's\:degree\:>\:1\:+\:denominator's\:degree,\:there\:is\:no\:horizontal\:asymptote.}$ $\mathrm{The\:degree\:of\:the\:numerator}=1.\:\mathrm{The\:degree\:of\:the\:denominator}=1$ $\mathrm{The\:degrees\:are\:equal,\:the\:asymptote\:is:}\:y=\frac{\mathrm{numerator's\:leading\:coefficient}}{\mathrm{denominator's\:leading\:coefficient}}$ $\mathrm{Numerator's\:leading\:coefficient}=1,\:\mathrm{Denominator's\:leading\:coefficient}=1$ $y=\frac{1}{1}$ $\mathrm{The\:horizontal\:asymptote\:is:}$ $y=1$

In conclusion:

$\mathrm{Vertical}\text{ asymptotes}:\:x=-5,\:\mathrm{Horizontal}\text{ asymptotes}:\:y=1$

4 0

2 years ago

What is the median of the following data set?<br> (3, 4, 2, 8, 5)<br> 02<br> 03<br> 04<br> 06

serious [3.7K]

Answer:

4

Step-by-step explanation:

The median is the middle number when the list is sorted from least to greatest or greatest to least.

We make this list to get 2,3,4,5,8 or 8,5,4,3,2 and the middle number in both lists is 4.

7 0

2 years ago