All categories

3 years ago

G( x)=sqrt(3x) find g'(x)

1 answer:

4 0

Well, g'(x) is just a fancy way of saying "the slope of". In calculus, it's also called "the derivative".
So our original function is
sqrt3x (I'm leaving out the G(x))
First, we'll need to rewrite this in exponential form:
(3x)^1/2
Now, to take a derivative, move the exponent to the front of the expression:
1/2(3x)^1/2
And add 1 to the exponent:
g'(x)=1/2(3x)^3/2
g'(x)=1/2(sqrt27)x^3/2
g'(x)=((sqrt27)/2)x^3/2
I hope it didn't get confusing with all the parentheses! Sorry if it did!
Hope this helps!

You might be interested in

What is the equation of the graph ?

mezya [45]

5 over and 4 ^. not sure if this is correct but that’s what i got

8 0

3 years ago

Which question is statistical?

Stella [2.4K]

I thinks its B

p.s. spread le cheese ⌂

4 0

3 years ago

What is the correct answer?

vovikov84 [41]

Answer:

84 units2

Step-by-step explanation:

First, you multiply 24 *7 = 168

Second, you divide 168 / 2 = 84

Lastly, have a great day!

5 0

3 years ago

Help Me please11111111

Y_Kistochka [10]

Answer:

The first option im pretty sure

Step-by-step explanation:

6 0

2 years ago

After once again losing a football game to the college'ss arch rival, the alumni association conducted a survey to see if alumni

antoniya [11.8K]

Answer:

a) The 90% confidence interval would be given (0.561;0.719).

b) $p_v =P(z>2.8)=1-P(z$

c) Using the significance level assumed $\alpha=0.01$ we see that $p_v$ so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion is higher than 0.5 or 50%.

Step-by-step explanation:

1) Data given and notation

n=100 represent the random sample taken

X=64 represent were in favor of firing the coach

$\hat p=\frac{64}{100}=0.64$ estimated proportion for were in favor of firing the coach

$p_o=0.5$ is the value that we want to test since the problem says majority

$\alpha$ represent the significance level (no given, but is assumed)

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

p= population proportion of Americans for were in favor of firing the coach

Part a

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 90% confidence interval the value of $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.64$

And replacing into the confidence interval formula we got:

$0.64 - 1.64 \sqrt{\frac{0.64(1-0.64)}{100}}=0.561$

$0.64 + 1.64 \sqrt{\frac{0.64(1-0.64)}{100}}=0.719$

And the 90% confidence interval would be given (0.561;0.719).

Part b

We need to conduct a hypothesis in order to test the claim that the proportion exceeds 50%(Majority). :

Null Hypothesis: $p \leq 0.5$

Alternative Hypothesis: $p >0.5$

We assume that the proportion follows a normal distribution.

This is a one tail upper test for the proportion of union membership.

The One-Sample Proportion Test is "used to assess whether a population proportion $\hat p$ is significantly (different,higher or less) from a hypothesized value $p_o$ ".

Check for the assumptions that he sample must satisfy in order to apply the test

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

$np_o =100*0.64=64>10$

$n(1-p_o)=100*(1-0.64)=36>10$

Calculate the statistic

The statistic is calculated with the following formula:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o(1-p_o)}{n}}}$

On this case the value of $p_o=0.5$ is the value that we are testing and n = 100.

$z=\frac{0.64 -0.5}{\sqrt{\frac{0.5(1-0.5)}{100}}}=2.8$

The p value for the test would be:

$p_v =P(z>2.8)=1-P(z$

Part c

Using the significance level assumed $\alpha=0.01$ we see that $p_v$ so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion is higher than 0.5 or 50%.

6 0

3 years ago