What is 4k-14+3k=21

F) If two angles are equal, they must be corresponding.

navik [9.2K]

Answer:

False

Step-by-step explanation:

This is an erroneous statement, so we presume it is supposed to be a true/false question. It is false.

__

Corresponding angles are only congruent in certain geometries. One such is where a transversal crosses parallel lines. There are a number of sets of corresponding angles in that geometry, and additional congruent angles that are not "corresponding."

Base angles in an isosceles triangle are equal, but not "corresponding." The presumption that equal angles are corresponding is FALSE.

8 0

3 years ago

A microwave manufacturing company has just switched to a new automated production system. Unfortunately, the new machinery has b

baherus [9]

Answer:

Null hypothesis: $\mu \leq 6$

Alternative hypothesis: $\mu > 6$

$t=\frac{6.5-6}{\frac{1.5}{\sqrt{36}}}=2$

$df=n-1=36-1=35$

$t_{crit}=1.690$ with the excel code:"=T.INV(0.95,35)"

$t_{crit}=1.306$ with the excel code:"=T.INV(0.90,35)"

$p_v =P(t_{(35)}>2)=0.0267$

If we compare the p value and the significance level given $\alpha=0.05,0.1$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the true mean it's significantly higher than 6 at 5% and 10% of significance.

Step-by-step explanation:

Data given and notation

$\bar X=6.5$ represent the mean time for the sample

$s=1.5$ represent the sample standard deviation for the sample

$n=36$ sample size

$\mu_o =6$ represent the value that we want to test

$\alpha=0.05,0.1$ 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 conduct a hypothesis in order to check if the mean is higher than 6 days, the system of hypothesis would be:

Null hypothesis: $\mu \leq 6$

Alternative hypothesis: $\mu > 6$

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic 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:

$t=\frac{6.5-6}{\frac{1.5}{\sqrt{36}}}=2$

Critical value and P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=36-1=35$

In order to calculate the critical value we need to find a quantile on the t distribution with 35 degrees of freedom that accumulates $\alpha$ on the right. Using the significance level of 0.05 we got:

$t_{crit}=1.690$ with the excel code:"=T.INV(0.95,35)"

And using the significance of 0.1 we got

$t_{crit}=1.306$ with the excel code:"=T.INV(0.90,35)"

Since is a one side right tailed test the p value would be:

$p_v =P(t_{(35)}>2)=0.0267$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05,0.1$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the true mean it's significantly higher than 6 at 5% and 10% of significance.

6 0

3 years ago

Endent Practice

Vikki [24]

Answer:

Part A: The unit that would be best to describes the situations is to the mile

Part B: 4 significant figures

Part C: 1,450 miles

Part D: He is not correct

Step-by-step explanation:

Part A: By unit conversion, we have;

1 mile = 1,760 yards

Given that the distance from Ottawa, Canada to Jackson Mississippi is 2,552,000 yards it would best simplified into miles

Part B: The number of significant digits in the reporters estimate are the non-zero digits before the zeros in the number figure which are 2, 5, 5, and 2 or 4 significant figures

Part C: 2,552,000 yards = 2,552,000 yards/(1,760 yards/mile) = 1,450 miles

Part D: Given that the distance from Jackson to Tucson = 7,159,680 feet, we have;

1 mile = 5,280 feet

Therefore;

7,159,680 feet = 7,159,680 feet/(5,280 feet/mile) = 1,356 miles

Therefore, the claim that the distance from Jackson to Tucson which is 1,356 miles is further than the distance from Ottawa, Canada to Jackson Mississippi which is 1,450 miles is not correct because;

The distance from Jackson to Tucson 1,356 miles $\ngtr$ 1,450 miles which is the distance from Jackson to Ottawa

6 0

3 years ago

What are the solutions of this quadratic equation ? x ^ 2 = 16x - 65 Substitute the values of a and b to complete the solutions

Karo-lina-s [1.5K]

Answer:

x = 8 + $\sqrt{58}$ or x = 8 − $\sqrt{58}$

Step-by-step explanation:

Hope This Helps

Please Mark Brainliest if correct

I do Not Have Time To Explain I'm SO Sorry

5 0

3 years ago

PLS Write a real world problem for the inequality 8-2x≤5.

Hatshy [7]

Step-by-step explanation:

8-2x< 5
-2x<5-8
x< -3/-2
x<1.5

hope it helps.

3 0

3 years ago

What is 4k-14+3k=21​

What is 4k-14+3k=21