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3 years ago

Rewrite the following equation in slope intercept form 14x+13y=12

Mathematics

1 answer:

Alenkinab [10]3 years ago

8 0

Answer:

$y = -\dfrac{14}{13}x + \dfrac{12}{13}$

Step-by-step explanation:

The slope-intercept form of the equation of a line is

y = mx + b

To write the equation of a line in the slope-intercept form, solve the equation for y.

14x + 13y = 12

Subtract 14x from both sides.

13y = -14x + 12

Divide both sides by 13.

$y = -\dfrac{14}{13}x + \dfrac{12}{13}$

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If the equation of a line has a slope of 0 and passes through the point (3, 3), which form would be used to write the equation o

3241004551 [841]

Y=mx+b

3=0(3)+b

3+b

y=0x+3

3 0

4 years ago

Which of these graphs represents a function?

mash [69]

Answer:

A, C, and D all do because if you were to hold a pencil up vertically, the line would only intersect it once, if it intersected more, it is not a function

Step-by-step explanation:

5 0

3 years ago

X+19=13<br><br><br> Find the value of x.

Oliga [24]

To find the value of x we need to use the INVERSE OPERATION of addition which is subtraction. Then we will check our work. lets do it:-

x + 19 = 13
x = ?
13 - 19 = x
<span>13 - 19 = -6</span>
x = - 6

CHECK OUR WORK:-

x + 19 = 13
x = -6
-6 + 19 = 13
We were RIGHT!!

So, x = -6

Hope I helped ya!! xD

4 0

4 years ago

Read 2 more answers

The LB Company has long manufactured a light bulb with an average life of 5400 hours. Company researchers have recently develope

elena55 [62]

<h2><u>Answer with explanation</u>:</h2>

Let $\mu$ be the average life of light bulbs.

As per given , we have

Null hypothesis : $H_0 : \mu =5400$

Alternative hypothesis : $H_a : \mu >5400$

Since $H_a$ is right-tailed and population standard deviation is also known, so we perform right-tailed z-test.

Formula for Test statistic : $z=\dfrac{\overlien{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

where, n= sample size

$\overline{x}$ = sample mean

$\mu$ = Population mean

$\sigma$ =population standard deviation

For $n=95,\ \overline{x}=5483\ \&\ \sigma=500$ , we have

$z=\dfrac{5483-5400}{\dfrac{500}{\sqrt{95}}}=1.61796786124\approx1.6180$

Using z-value table , Critical one-tailed test value for 0.06 significance level :

$z_{0.06}=1.5548$

Decision : Since critical z value (1.5548) < test statistic (1.6180), so we reject the null hypothesis .

[We reject the null hypothesis when critical value is less than the test statistic value .]

Conclusion : We have enough evidence at 0.06 significance level to support the claim that the new filament yields a longer bulb life

8 0

4 years ago

The probability that a randomly selected 3-year-old male chipmunk will live to be 4 years old is 0.96516.

mezya [45]

Using the binomial distribution, it is found that there is a:

a) The probability that two randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.93153 = 93.153%.

b) The probability that six randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.80834 = 80.834%.

c) The probability that at least one of six randomly selected 3-year-old male chipmunks will not live to be 4 years old is 0.19166 = 19.166%. This probability is not unusual, as it is greater than 5%.

-----------

For each chipmunk, there are only two possible outcomes. Either they will live to be 4 years old, or they will not. The probability of a chipmunk living is independent of any other chipmunk, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

0.96516 probability of a chipmunk living through the year, thus $p = 0.96516$

Item a:

Two is P(X = 2) when n = 2, thus:

$P(X = 2) = C_{2,2}(0.96516)^2(1-0.96516)^{0} = 0.9315$

The probability that two randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.93153 = 93.153%.

Item b:

Six is P(X = 6) when n = 6, then:

$P(X = 6) = C_{6,6}(0.96516)^6(1-0.96516)^{0} = 0.80834$

The probability that six randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.80834 = 80.834%.

Item c:

At least one not living is:

$P(X < 6) = 1 - P(X = 6) = 1 - 0.80834 = 0.19166$

The probability that at least one of six randomly selected 3-year-old male chipmunks will not live to be 4 years old is 0.19166 = 19.166%. This probability is not unusual, as it is greater than 5%.

A similar problem is given at brainly.com/question/24756209

6 0

3 years ago