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2 years ago

Is this graph shown below a proportional or non proportional linear relationship

Mathematics

1 answer:

mr Goodwill [35]2 years ago

8 0

Step-by-step explanation:

where is the graph

complete the question

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Nora and Eli are making homemade spring

yawa3891 [41]

Answer:

eli

Step-by-step explanation:

3 0

3 years ago

In a random sample of 5 residents of the state of Texas, the mean waste recycled per person per day was 1.9 pounds with a standa

sammy [17]

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

The confidence interval for population mean is given by :-

$\overline{x} \pm\ t_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

Given : Sample size : n= 5, since n<30 , so the test we use here is t-test.

Sample mean : $\overline{x}=1.9\text{ pounds}$

Standard deviation: $\sigma=0.89\text{ pounds}$

Significance level : $1-\alpha:1-0.99=0.01$

By using the standard normal distribution table , the critical value corresponds to the given significance level will be :-

$t_{n-1,\alpha/2}=t_{5-1,0.01/2}=t_{4,0.005}=4.604$

Now, the 99% confidence interval for the mean waste recycled per person per day for the population of Texas will be :-

$1.9\pm\ (4.604)\dfrac{0.89}{\sqrt{5}}\\\\\approx1.9\pm1.832\\\\=(1.9-1.832,1.9+1.832)=(0.068,\ 3.732)$

Hence, the 99% confidence interval for the mean waste recycled per person per day for the population of Texas = (0.068, 3.732)

5 0

3 years ago

Two problems here I need solved! I need every step, so please have that with your answers!!

Free_Kalibri [48]

QUESTION 1

We want to solve,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-6x+8}$

We factor the denominator of the fraction on the right hand side to get,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-4x - 2x+8}.$

This implies
$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x(x-4) - 2(x - 4)}.$

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{(x-4)(x - 2)}$

We multiply through by LCM of
$(x-4)(x - 2)$

$(x - 2) + x(x-4) = 2$

We expand to get,

$x - 2 + {x}^{2} - 4x= 2$

We group like terms and equate everything to zero,

${x}^{2} + x - 4x - 2 - 2 = 0$

We split the middle term,

${x}^{2} + - 3x - 4 = 0$

We factor to get,

${x}^{2} + x - 4x- 4 = 0$

$x(x + 1) - 4(x + 1) = 0$

$(x + 1)(x - 4) = 0$

$x + 1 = 0 \: or \: x - 4 = 0$

$x = - 1 \: or \: x = 4$

But
$x = 4$
is not in the domain of the given equation.

It is an extraneous solution.

$\therefore \: x = - 1$
is the only solution.

QUESTION 2

$\sqrt{x+11} -x=-1$

We add x to both sides,

$\sqrt{x+11} =x-1$

We square both sides,

$x + 11 = (x - 1)^{2}$

We expand to get,

$x + 11 = {x}^{2} - 2x + 1$

This implies,

${x}^{2} - 3x - 10 = 0$

We solve this quadratic equation by factorization,

${x}^{2} - 5x + 2x - 10 = 0$

$x(x - 5) + 2(x - 5) = 0$

$(x + 2)(x - 5) = 0$

$x + 2 = 0 \: or \: x - 5 = 0$

$x = - 2 \: or \: x = 5$

But
$x = - 2$
is an extraneous solution

$\therefore \: x = 5$