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3 years ago

Let f be the function defined by f(x) = −ln(x) for 0 < x ≤ 1. R is the region between the graph of f and the x-axis.

Mathematics

1 answer:

aivan3 [116]3 years ago

3 0

Answer:

0.089

Step-by-step explanation:

f(x) = -ln(x) and g(x) = x²

Start by graphing the region. The two curves intersect at about (0.653, 0.426), with g(x) on the left and f(x) on the right. The region is the triangular area between the curves and above the x-axis.

If we were to cut the region horizontally (perpendicular to the y-axis), the resulting line is the width of the square cross section. The thickness of this square is dy. So the volume of the square is:

dV = A dy

dV = s² dy

dV = (x₂ − x₁)² dy

dV = (e⁻ʸ − √y)² dy

The total volume is the sum of all the squares from y=0 to y=0.426.

V = ∫ dV

V = ∫₀⁰'⁴²⁶ (e⁻ʸ − √y)² dy

Evaluate with a calculator:

V ≈ 0.089

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Given the following information about a hypothesis test of the difference between two means based on independent random samples,

HACTEHA [7]

Answer:

$S^2_p =\frac{(13-1)(5)^2 +(10 -1)(3)^2}{13 +10 -2}=18.143$

And the deviation would be just the square root of the variance:

$S_p=4.259$

Then the statistic is given by:

$t=\frac{(12 -9)-(0)}{4.259\sqrt{\frac{1}{13}+\frac{1}{10}}}=1.674$

And the correct option would be:

t = 1.674

Step-by-step explanation:

Data given:

$n_1 =13$ represent the sample size for group 1

$n_2 =10$ represent the sample size for group 2

$\bar X_1 =12$ represent the sample mean for the group 1

$\bar X_2 =9$ represent the sample mean for the group 2

$s_1=5$ represent the sample standard deviation for group 1

$s_2=3$ represent the sample standard deviation for group 2

We are assuming two independent samples from two normal distributions with equal variances we are assuming that

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

And the statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

Where t follows a t distribution with $n_1+n_2 -2$ degrees of freedom and the pooled variance $S^2_p$ is given by this formula:

$\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 \leq \mu_2$

Alternative hypothesis: $\mu_1 > \mu_2$

The pooled variance is given by:

$S^2_p =\frac{(13-1)(5)^2 +(10 -1)(3)^2}{13 +10 -2}=18.143$

And the deviation would be just the square root of the variance:

$S_p=4.259$

Then the statistic is given by:

$t=\frac{(12 -9)-(0)}{4.259\sqrt{\frac{1}{13}+\frac{1}{10}}}=1.674$

And the correct option would be:

t = 1.674

4 0

3 years ago

Expand 3(5y-3) can someone answer this please

docker41 [41]

Answer:

15y -9

Step-by-step explanation:

3(5y-3)

Distribute

3*5y -3*3

15y -9

7 0

3 years ago

Read 2 more answers

The graph below shows the solution to which system of inequalities?

Anon25 [30]

I can’t see it’s staticy and stuff but uh

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3 years ago

At a local high school, enrollment in the drama club increases by 5% each year. What is the common ratio in this situation? 0. 0

Crazy boy [7]

The common ratio in the geometric progression is the ratio of any one phrase in the series divided by the preceding term. The common ratio in this situation is 1.05.

<h3>What is the common ratio?</h3>

The common ratio in the geometric progression is the ratio of any one phrase in the series divided by the preceding term. It is usually symbolized by the letter "r."

The given data in the problem will be;

Enrollment in the drama club increases by 5% each year

If the value of old enrollment is x and the value of new enrollment is y then

$\rm y=x+5 \% \ of\ x \\\\ y=x+0.05x= 1.05 x$