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Delicious77 [7]

3 years ago

10

Which of the following equations represents a line that is perpendicular to the line represented by x + 2y = 6 and passes throug

h the point (3 , –4)?

1 answer:

Andru [333]3 years ago

6 0

Answer:

y=2x-10

Step-by-step explanation:

First change to equation into slope intercept form

2y=-x+6

y=-1/2x+3

slope=2 (beacause it is perpendicular)

y--4=2(x-3)

y+4=2x-6

y=2x-10

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1 a.) 20cm^2 to mm^2

mixas84 [53]

Answer:

a) 200 mm^2

b) 500 mm^2

c) 3.75e + 8

d) 6390 cm^2

e) 650 cm^2

f) 20 m/s

Step-by-step explanation:

For a) and b): 1 cm = 10 mm, multiply 20 (and 50) by 10

For c): the formula is to multiply 1e + 8

For d): 1 Liter = 1000 Cubic Centimeters (hopefully that's what you're referring to)

For e): 10 mm = 1 cm, divide by 10

For f): divide the value by 3.6

6 0

3 years ago

There are 5 people who will share a 24 pound bag of apples.how many pounds of apples should each person get .Between which two w

Darya [45]

Each person should get 4 4/5. That will be between the numbers 4 and 5.

4 0

3 years ago

Read 2 more answers

A cow can give 5 gallons of milk if it eats 2 pounds of grass. Express x in terms of y, where y is the number of gallons of milk

spayn [35]

Answer:

x = 2.5y

Step-by-step explanation:

If a cow can give 5 gallons of milk after eating 2 pounds of grass.

Let's consider 'x' is the number of pounds of grass it eats and 'y' is the number of gallons of milk the cow gives.

Hence,

5y = 2x

We will simplify it to get what x equal to y by dividing 2 on both side we find,

x = 2.5y

x is equal to 2.5 y

By eating 1 pound of grass cow can give 2.5 gallons of milk.

7 0

3 years ago

A random sample of 25 ACME employees showed the average number of vacation days taken during the year is 18.3 days with a standa

Norma-Jean [14]

Answer:

a) Null hypothesis: $\mu \leq 15$

Alternative hypothesis: $\mu > 15$

b) $df=n-1=25-1=24$

For this case the p value is given $p_v = 0.0392$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 15 at 5% of signficance.

c) Type I error, also known as a “false positive” is the error of rejecting a null hypothesis when it is actually true. Can be interpreted as the error of no reject an alternative hypothesis when the results can be attributed not to the reality.

So for this case a type I of error would be reject the hypothesis that the true mean is less or equal than 15 and is actually true.

Step-by-step explanation:

Data given and notation

$\bar X=18.3$ represent the sample mean

$s=3.72$ represent the sample standard deviation

$n=25$ sample size

$\mu_o =15$ represent the value that we want to test

$\alpha=0.05$ 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)

Part a: State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean for vacation days is higher than 15, the system of hypothesis would be:

Null hypothesis: $\mu \leq 15$

Alternative hypothesis: $\mu > 15$

If we analyze the size for the sample is < 30 and 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".

Part b: P-value and conclusion

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

$df=n-1=25-1=24$

For this case the p value is given $p_v = 0.0392$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 15 at 5% of signficance.

Part c

Type I error, also known as a “false positive” is the error of rejecting a null hypothesis when it is actually true. Can be interpreted as the error of no reject an alternative hypothesis when the results can be attributed not to the reality.

So for this case a type I of error would be reject the hypothesis that the true mean is less or equal than 15 and is actually true.

3 0

3 years ago

Please answer this<br><br> (I don’t know what else to put, it need 20 characters)

Vinvika [58]

Answer:

15yd^2

Step-by-step explanation:

Triangle area formula is A = bh/2

A = 6*5/2

A = 30/2

A = 15

Best of Luck!

6 0

3 years ago

Read 2 more answers