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3 years ago

Janelle draws line segments AB and BC in the same plane such that AB = 8 cm and BC = 6 cm. Then Janelle draws

Mathematics

2 answers:

MrMuchimi3 years ago

7 0

Answer:

Step-by-step explanation:

if ABC forms a triangle, AB=8 and BC=6

a possible value of AC=10cm

if ABC is a line segment n C is between A n B

AC + BC = AB

AC = 8 - 6 = 2cm

Dovator [93]3 years ago

4 0

Answer:

Step-by-step explanation:

• AB, BC, and AC form a triangle. Enter a possible value of AC....

So it asks for only a possible value of AC as there are many possible values.

Given AB = 8 cm and BC = 6 cm, they are in the ratio of 3:4.

Line segments of 3, 4 and 5 length will form a right-angled triange.

A possible value of AC = 5*2 = 10cm

• Points A, B, and C lie on the same line, and C lies between A and B.

So AC+CB = AB

AC+6 = 8

AC = 2cm

Enter this value of AC in the second

response box.

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What is the order of the set A={-1,20,3,17,5}

laiz [17]

Hey!

Hope this helps...

~~~~~~~~~~~~~~~~~~~~~~~~~~~

I'm assuming your talking about least to greatest, OR greatest to least. If so, that's what I'll do...

Least to Greatest: -1, 3, 5, 17, 20

Greatest to Least: 20, 17, 5, 3, -1

~~~~~~~~~~~~

Unfortunately I did not completely understand the question, for future reference please give AS MUCH detail as possible so that your fellow brainly members can help you ASAP...

And thanks for using Brainly!

3 0

3 years ago

I need help writing the first 5 terms of the sequence!

Angelina_Jolie [31]

The first five terms of the sequence are 1, 4, 7, 10, 13.

Solution:

Given data:

$a_{1}=1$

$a_{n}=a_{n-1}+3$

General term of the arithmetic sequence.

$a_{n}=a_{n-1}+d$ , where d is the common difference.

d = 3

$a_{n}=a_{n-1}+3$

Put n = 2 in $a_{n}=a_{n-1}+3$ , we get

$a_{2}=a_1+3$

$a_{2}=1+3$

$a_2=4$

Put n = 3 in $a_{n}=a_{n-1}+3$ , we get

$a_{3}=a_2+3$

$a_{3}=4+3$

$a_3=7$

Put n = 4 in $a_{n}=a_{n-1}+3$ , we get

$a_{4}=a_3+3$

$a_{4}=7+3$

$a_4=10$

Put n = 5 in $a_{n}=a_{n-1}+3$ , we get

$a_{5}=a_4+3$

$a_{5}=10+3$

$a_5=13$

The first five terms of the sequence are 1, 4, 7, 10, 13.

3 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

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

-2/3 divided by (-1/2)

denis23 [38]

Answer:

4/3

Step-by-step explanation:

multiply by the reciprocal

$-\frac{2}{3}$ × $-\frac{2}{1}$ = $\frac{4}{3}$