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3 years ago

14

Tell whether the function

title="y - 4x {}^{2} = 3 x" alt="y - 4x {}^{2} = 3 x" align="absmiddle" class="latex-formula">
linear or nonlinear

1 answer:

ivanzaharov [21]3 years ago

3 0

Non linear because the degree of variable y is 1 and the degree of variable x is 2

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Pls help me with this

OlgaM077 [116]

Answer:

1/(x^(17/12)y^(5/3))

Step-by-step explanation:

We assume you want to simplify the expression. The applicable rules of exponents are ...

(a^b)(a^c) = a^(b+c)

(a^b)^c = a^(bc)

1/a^b = a^-b

Using these "all at once", we can add up the exponents of each of the variables:

x^(-2/3 -2·0 -(1/2)(3/2))·y^(-2/3-(1/2)·2) = x^(-2/3-3/4)y^(-2/3-1)

= x^(-17/12)y^(-5/3)

= $\dfrac{1}{x^{\frac{17}{12}}y^{\frac{5}{3}}}$

4 0

4 years ago

G+3>6<br> A (3)<br> B (4)<br> C (-3)<br> D (-4)

aleksley [76]

G>6-3
G>3 is definitely the answer

7 0

3 years ago

Camila Cabello wants to order cheesecakes over the internet. Each cheesecake costs $15.99 and shipping for the entire order is

ladessa [460]

Answer:

she can purchase 5 cheesecakes

Step-by-step explanation:

15.99*5=79.95

if we add 9.99 for the shipping of thw whole order

79.95+9.99=89.94

6 0

3 years ago

What are the solution(s) to the quadratic equation 9x2 = 4?

Lunna [17]

Multiply 9 by 2

18 = 4

Then subtract 4 from both sides

18 - 4 = 0 will become

14 = 0

Since 14 ≠ 0, there are no solutions

3 0

4 years ago

Measurements of the sodium content in samples of two brands of chocolate bar yield the following results (in grams):

Tpy6a [65]

Answer:

98% confidence interval for the difference μX−μY = [ 0.697 , 7.303 ] .

Step-by-step explanation:

We are give the data of Measurements of the sodium content in samples of two brands of chocolate bar (in grams) below;

Brand A : 34.36, 31.26, 37.36, 28.52, 33.14, 32.74, 34.34, 34.33, 29.95

Brand B : 41.08, 38.22, 39.59, 38.82, 36.24, 37.73, 35.03, 39.22, 34.13, 34.33, 34.98, 29.64, 40.60

Also, $\mu_X$ represent the population mean for Brand B and let $\mu_Y$ represent the population mean for Brand A.

Since, we know nothing about the population standard deviation so the pivotal quantity used here for finding confidence interval is;

P.Q. = $\frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ~ $t_n__1+n_2-2$

where, $Xbar$ = Sample mean for Brand B data = 36.9

$Ybar$ = Sample mean for Brand A data = 32.9

$n_1$ = Sample size for Brand B data = 13

$n_2$ = Sample size for Brand A data = 9

$s_p = \sqrt{\frac{(n_1-1)s_X^{2}+(n_2-1)s_Y^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(13-1)*10.4+(9-1)*7.1 }{13+9-2} }$ = 3.013

Here, $s^{2}_X$ and $s^{2} _Y$ are sample variance of Brand B and Brand A data respectively.

So, 98% confidence interval for the difference μX−μY is given by;

P(-2.528 < $t_2_0$ < 2.528) = 0.98

P(-2.528 < $\frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < 2.528) = 0.98

P(-2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ < $(Xbar -Ybar) -(\mu_X-\mu_Y)$ < 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ ) = 0.98

P( (Xbar - Ybar) - 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ < $(\mu_X-\mu_Y)$ < (Xbar - Ybar) + 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ ) = 0.98

98% Confidence interval for μX−μY =

[ (Xbar - Ybar) - 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ , (Xbar - Ybar) + 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ ]

[ $(36.9 - 32.9)-2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9}$ , $(36.9 - 32.9)+2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9}$ ]

[ 0.697 , 7.303 ]

Therefore, 98% confidence interval for the difference μX−μY is [ 0.697 , 7.303 ] .

4 0

3 years ago