Which ordered pairs are a solution to the equation:

Students in Ms. Smiths Earth Science class collected data on three metamorphic rocks. What could the students perform to identif

kicyunya [14]

Answer:

Metamorphic rocks are rocks that have undergone a change from their original form due to changes in temperature, pressure or chemical alteration. The classification of metamorphic rocks is based on the minerals that are present and the temperature and pressure at which these minerals form.

Step-by-step explanation:

8 0

3 years ago

Plz help!! On question 7 thnx ✌️✨

Luba_88 [7]

Answer:

10 pieces

Step-by-step explanation:

(5 1/4)/(1/2) = 10.5

10 pieces with 1/4 ft left over

6 0

3 years ago

A random sample of BYU-Idaho students was surveyed and asked if they were in favor of retaining the penny as a form of currency

castortr0y [4]

Answer:

Step-by-step explanation:

Hello!

The objective is to compare the proportion of male and female students that are in favor of retaining the penny as a form of currency. Two groups of students were surveyed, the study variables are:

X₁: number of female students of the BYU-Idaho out of 116, that are in favor of retaining the penny as a form of currency.

X₂: number of male students of the BYU-Idaho out of 137, that is in favor of retaining the penny as a form of currency.

These two variables are discrete, and they follow the binomial criteria:

The number of observations of the trial is fixed and there are only two possible observations: success: the student is on the favor, failure: the student is against it.

Each observation in the trial is independent, this means that none of the trials will affect the probability of the next trial.

The probability of success in the same from one trial to another.

So we can say that

X₁~Bi(n₁;p₁)

X₂~Bi(n₂;p₂)

To study the difference between both population proportions through a hypothesis test or a Confidence Interval you theed to apply the Central Limit theorem and approximate the distribution of both sample proportions to normal. For that the sample sizes have to be big enough, as a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.

n₁ and n₂ ≥ 30, in this case, n₁= 116 and n₂=137

So the conditions are meet to estimate the difference per CI and test it through a hypothesis test.

I hope this helps!

8 0

3 years ago

A retired woman has $80,000 to invest but needs to make $3,000 a year from the interest to meet certain living expenses. One bon

Ilia_Sergeevich [38]

Answer:

Good Luck!

<em>Follow Me</em>

7 0

3 years ago

Joe believes the average battery life of a vehicle is equal to 9 years. A sample of 5 vehicles was taken and the following data

UkoKoshka [18]

Answer:

$t=\frac{7.26-9}{\frac{0.856}{\sqrt{5}}}=-4.54$

$p_v =2*P(t_{4}$

If we compare the p value and a significance level assumed $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we ccan say that the true mean is different from 9 years at 5% of significance.

Step-by-step explanation:

Data given and notation

Data : 6.2, 7.1, 8.5, 7.6,6.9

The sample mean and devation can be founded with the following formulas:

$\bar X =\frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X=7.26$ represent the average

$s=0.856$ represent the sample standard deviation

$n=5$ sample size

$\mu_o =76$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

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

State the null and alternative hypotheses.

We need to apply a two tailed test.

What are H0 and Ha for this study?

Null hypothesis: $\mu = 9$

Alternative hypothesis : $\mu \neq 9$

Compute the test statistic

The statistic for this case is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{7.26-9}{\frac{0.856}{\sqrt{5}}}=-4.54$

Give the appropriate conclusion for the test

First we need to find the degrees of freedom given by:

$df=n-1=5-1=4$

Since is a two tailed test the p value would be:

$p_v =2*P(t_{4}$

Conclusion

If we compare the p value and a significance level assumed $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we ccan say that the true mean is different from 9 years at 5% of significance.

7 0

3 years ago