Please help there is a attachment to this i think.

In a woodworking class, you need to make five triangles out of plywood. Each triangle has a height of 6 inches and a base of 14

Bogdan [553]

Answer:5 * (1/2 * 6 * 14)= x

Step-by-step explanation: :D

3 0

3 years ago

A 50-gallon rain barrel is filled to capacity. It drains at a rate of 10 gallons per minute. Write an equation to show how much

makvit [3.9K]

Answer:

The quantity of water drain after x min is 50 $(0.9)^{x}$

Step-by-step explanation:

Given as :

Total capacity of rain barrel = 50 gallon

The rate of drain = 10 gallon per minutes

Let The quantity of water drain after x min = y

Now, according to question

The quantity of water drain after x min = Initial quantity of water × $(1-\dfrac{\textrm rate}{100})^{\textrm time}$

I.e The quantity of water drain after x min = 50 gallon × $(1-\dfrac{\textrm 10}{100})^{\textrm x}$

or, The quantity of water drain after x min = 50 gallon × $(0.9)^{x}$

Hence the quantity of water drain after x min is 50 $(0.9)^{x}$ Answer

4 0

3 years ago

Read 2 more answers

Solve the equations and match them with their solutions.

Artyom0805 [142]

1 is x=13.5

2 is x=-260

3 is x=38

4 is x=-37^38

i think this is correct

7 0

3 years ago

A sample of size 6 will be drawn from a normal population with mean 61 and standard deviation 14. Use the TI-84 Plus calculator.

AVprozaik [17]

Answer:

$X \sim N(61,14)$

Where $\mu=61$ and $\sigma=14$

Since the distribution for X is normal then the distribution for the sample mean is also normal and given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

$\mu_{\bar X} = 61$

$\sigma_{\bar X}= \frac{14}{\sqrt{6}}= 5.715$

So then is appropiate use the normal distribution to find the probabilities for $\bar X$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean". The letter $\phi(b)$ is used to denote the cumulative area for a b quantile on the normal standard distribution, or in other words: $\phi(b)=P(z$

Solution to the problem

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

$X \sim N(61,14)$

Where $\mu=61$ and $\sigma=14$

Since the distribution for X is normal then the distribution for the sample mean $\bar X$ is also normal and given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

$\mu_{\bar X} = 61$

$\sigma_{\bar X}= \frac{14}{\sqrt{6}}= 5.715$

So then is appropiate use the normal distribution to find the probabilities for $\bar X$

8 0

3 years ago

Data is collected to compare two different types of batteries. We measure the time to failure of 12 batteries of each type. Batt

-Dominant- [34]

Using the t-distribution, it is found that since the test statistic is greater than the critical value for the right-tailed test, it is found that there is enough evidence to conclude that Battery B outlasts Battery A by more than 2 hours.

At the null hypothesis, it is tested if it does not outlast by more than 2 hours, that is, the subtraction is not more than 2:

$H_0: \mu_B - mu_A \leq 2$

At the alternative hypothesis, it is tested if it outlasts by more than 2 hours, that is:

$H_1: \mu_B - \mu_A > 2$

The sample means are: $\mu_A = 8.65, \mu_B = 11.23$

The standard deviations for the samples are $s_A = s_B = 0.67$

Hence, the standard errors are:

$s_{Ea} = S_{Eb} = \frac{0.67}{\sqrt{12}} = 0.1934$

The distribution of the difference has mean and standard deviation given by:

$\overline{x} = \mu_B - \mu_A = 11.23 - 8.65 = 2.58$

$s = \sqrt{s_{Ea}^2 + s_{Eb}^2} = \sqrt{0.1934^2 + 0.1934^2} = 0.2735$

The test statistic is given by:

$t = \frac{\overline{x} - \mu}{s}$

In which $\mu = 2$ is the value tested at the hypothesis.

Hence:

$t = \frac{\overline{x} - \mu}{s}$

$t = \frac{2.58 - 2}{0.2735}$

$t = 2.12$

The critical value for a right-tailed test, as we are testing if the subtraction is greater than a value, with a 0.05 significance level and 12 + 12 - 2 = 22 df is given by $t^{\ast} = 1.71$

Since the test statistic is greater than the critical value for the right-tailed test, it is found that there is enough evidence to conclude that Battery B outlasts Battery A by more than 2 hours.

A similar problem is given at brainly.com/question/13873630

7 0

2 years ago