Find the area of the semicircle.

Mathematics

1 answer:

n200080 [17]2 years ago

6 0

Answer:

Step-by-step explanation:

semi circle: pi(r^2)/ 2

r=3

AREA: 14.14

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Consider the ellipse 25xsquared plus 4y squared equals 1 in the xy-plane. a. If this ellipse is revolved about the x-axis, wha

BaLLatris [955]

Answer:

about x: 25x² +4y² + 4z² = 1
about y: 25x² +4y² +25z² = 1

Step-by-step explanation:

When the figure is revolved around the x-axis, its extent in the z-direction will match its extent in the y-direction. So, the z-term will have the same coefficient as the y-term. 25x² +4y² +4z² = 1

Similarly, when the figure is revolved around the y-axis, its extent in the z-direction will match its extent in the x-direction. Then the z-term will have the same coefficient as the x-term. 25x² +4y² +25z² = 1

5 0

3 years ago

The mean gross annual incomes of certified welders are normally distributed with the mean of $20,000 and a standard deviation of

masya89 [10]

Answer:

Z=±1.65

Step-by-step explanation:

Some previous concepts

The p-value is the probability of obtaining the observed results of a test, assuming that the null hypothesis is correct.

A z-test for one mean "is a hypothesis test that attempts to make a claim about the population mean(μ)".

The null hypothesis attempts "to show that no variation exists between variables or that a single variable is no different than its mean"

The alternative hypothesis "is the hypothesis used in hypothesis testing that is contrary to the null hypothesis"

System of hypothesis

Null hypothesis: $\mu=20000$

Alternative hypothesis: $\mu \neq 20000$

If the random variable is distributed like this: $X \sim N(\mu=20000,\sigma=2000)$

The significance level provided is $\alpha=0.1$ and the value of $\alpha/2= 0.05$ .

Since we are conducting a bilateral test, then we have two critical values. We need to find two quantiles that accumulates 0.05 of the area on the tails of the normal standard distribution. And in order to find it we can use the following excel codes:

"=NORM.INV(0.05,0,1)" , "=NORM.INV(1-0.05,0,1)"

And we got this:

$z_{\alpha/2}=-1.65$ and $z_{1-\alpha/2}=1.65$