All categories

3 years ago

May I please receive help?

Mathematics

1 answer:

Natalija [7]3 years ago

6 0

Answer:

<h2>Line 1 and line 2: ●Parallel ○Perpendicular ○Nether</h2><h2>Line 1 and line 3: ○Parallel ○Perpendicular ●Nether</h2><h2>Line 2 and line 3: ○Parallel ●Perpendicular ○Nether</h2>

<h2>The Black circle is the answer</h2>

hope you add brainliest and 5 stras

<h2>Step-by-step explanation:</h2><h2>Goodbluck</h2>

dont forget add brainliest

and 5 stars

<h2>THANK YOU!!!</h2>

You might be interested in

If y = kx, and x = 3 and y = 15, calculate k.

kenny6666 [7]

Answer:

k = 5

Step-by-step explanation:

Since the equation is y = kx and x equals 3 while y equals 15, in order to solve this equation we are going to have to substitute the variables with their answers:

x = 3

y = 15

15 = 3k

Now we divide 3 from both 3x and 15 to segregate the variable x:

15/3 = 3k/3

5 = k

Now to check our work we are going to substitute 5 for the variable k, 3 for the variable x and 15 for the variable y.

y = kx

15= (5)(3)

15 = 15

Tada, I hope this helps you!

6 0

3 years ago

Read 2 more answers

Perform the indicated operation. (-6)(4)

Vadim26 [7]

Answer:

-24

Step-by-step explanation:

parentheses always mean multiplication.

its just -6 x 4

which is -24

(an odd number of negative signs means you get a negative solution)

6 0

3 years ago

Read 2 more answers

If a watch has a 0.12 defect rate. what is the probability that the watch has fewer than 3 defects in a run of 75 chimes?

Maslowich

Answer:

$(\frac{88}{100})^{75} + \binom{75}{1} (\frac{88}{100})^{74}(\frac{12}{100}) + \binom{75}{2} (\frac{88}{100})^{73}(\frac{12}{100})^{2}$

Step-by-step explanation:

If a watch has fewer than three defects, then either 1.) It has no defects, 2.) it has exactly 1 defect, or 3.) it has exactly 2 defects.

1.) The probability that the watch has no defects is $(\frac{88}{100})^{75}$ , because for every chime there is a probability of $\frac{88}{100}$ that there is no defect

2.) The probability that the watch has exactly 1 defect is $(\frac{88}{100})^{74}(\frac{12}{100})$ times the number of ways you can choose 1 of 75 of the chimes to be defective, which is $\binom{75}{1}$ , so the probability that the watch has exactly 1 defect is $\binom{75}{1} (\frac{88}{100})^{74}(\frac{12}{100})$ .

3.) For the same reason as 2.), the probability that the watch has exactly two defects is $\binom{75}{2} (\frac{88}{100})^{73}(\frac{12}{100})^{2}$

Since 1.), 2.), and 3.) are mutually exclusive events, the probability of their union is simply the probability of each of them added together, which is $(\frac{88}{100})^{75} + \binom{75}{1} (\frac{88}{100})^{74}(\frac{12}{100}) + \binom{75}{2} (\frac{88}{100})^{73}(\frac{12}{100})^{2}$

8 0

3 years ago

Arandom sample of n1 =12 students majoring in accounting in a college of business has a mean grade-point average of 2.70 (where

antiseptic1488 [7]

Answer:

We accept the null hypothesis that the mean gpa's are equal, with the 5 percent level of significance.

Step-by-step explanation:

We have these following hypothesis:

Null

Equal means

$\mu_{1} = \mu_{2}$

Alternative

Different means

$\mu_{1} \neq \mu_{2}$

Our test statistic is:

$\frac{\overline{Y_{1}} - \overline{Y_{2}}}{\sqrt{\frac{s_{1}^{2}}{N_{1}} + \frac{s_{2}^{2}}{N_{2}}}}$

In which $\overline{Y_{1}}, \overline{Y_{2}}$ are the sample means, $N_{1}, N_{2}$ are the sample sizes and $s_{1}, s_{2}$ are the standard deviations of the sample.

In this problem, we have that:

$\overline{Y_{1}} = 2.7, s_{1} = 0.4, N_{1} = 12, \overline{Y_{2}} = 2.9, s_{2} = 0.3, N_{2} = 10$

$T = \frac{2.7 - 2.9}{\sqrt{\frac{0.4}^{2}{12} + \frac{0.3}^{2}{10}}} = -1.3383$

What to do with the null hypothesis?

We will reject the null hypothesis, that is, that the means are equal, with a significante level of $\alpha$ if

$|T| > t_{1-\frac{\alpha}{2},v}$

In which v is the number of degrees of freedom, given by

$v = \frac{(\frac{s_{1}^{2}}{N_{1}} + \frac{s_{2}^{2}}{N_{2}})^{2}}{\frac{\frac{s_{1}^{2}}{N_{1}}}{N_{1}-1} + \frac{\frac{s_{2}^{2}}{N_{2}}}{N_{2} - 1}}$