PLEASE HELP I WILL GIVE BRAINLIEST!!!!

If 4x plus 7 equals 18, what is the value of 12x plus 21

8_murik_8 [283]

Answer: 54

--------------------

6 0

3 years ago

Given the relationship x2 + 3y2 =12, with y > 0 and dx, dt = 2 units/min., find the value of dy, dt at the instant y = 1 unit

matrenka [14]

$\bf x^2+3y^2=12\implies \stackrel{chain~rule}{2x\cfrac{dx}{dt}}+3 \stackrel{chain~rule}{\left(2y\cfrac{dy}{dt}\right)}=0
\\\\\\
2x\cfrac{dx}{dt}+6y\cfrac{dy}{dt}=0
\implies 
6y\cfrac{dy}{dt}=-2x\cfrac{dx}{dt}\implies \cfrac{dy}{dt}=\cfrac{-2x\frac{dx}{dt}}{
6y\frac{dy}{dt}}
\\\\\\
\boxed{\cfrac{dy}{dt}=-\cfrac{x\frac{dx}{dt}}{3y}}\\\\
-------------------------------$

$\bf \textit{now, when y = 1, what is \underline{x}?}\qquad x^2+3y^2=12\implies x^2+3(1)^2=12
\\\\\\
x^2=9\implies x=\sqrt{9}\implies x=3\\\\
-------------------------------\\\\
\begin{cases}
y=1\\
x=3\\
\frac{dx}{dt}=2
\end{cases}\implies \cfrac{dy}{dt}=-\cfrac{3\cdot 2}{3\cdot 1}\implies \cfrac{dy}{dt}=-2$

7 0

3 years ago

Simplify the expression 7.8n²+0.5+9n-.0.4-7n+n²

Natali [406]

The final answer for this question would be.....

8.8n^2+16n+0.1

Hope this helps!!!!

4 0

3 years ago

Read 2 more answers

A plane flies 640 mi with the wind in the same time that it takes to fly 440 mi against the wind. If this plane flies at the rat

777dan777 [17]

Answer:

what that this answer ⁉️

6 0

3 years ago

Assume a simple random sample of 10 BMIs with a standard deviation of 1.186 is selected from a normally distributed population o

kirza4 [7]

Answer:

a) H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

b) $df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

c) $t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

d) For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

Step-by-step explanation:

Information provided

n = 10 sample size

s= 1.186 the sample deviation

$\sigma_o =1.34$ the value that we want to test

$p_v$ represent the p value for the test

t represent the statistic (chi square test)

$\alpha=0.01$ significance level

Part a

On this case we want to test if the true deviation is 1,34 or no, so the system of hypothesis are:

H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

The statistic is given by:

$t=(n-1) [\frac{s}{\sigma_o}]^2$

Part b

The degrees of freedom are given by:

$df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

Part c

Replacing the info we got:

$t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

Part d

For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

5 0

3 years ago