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nevsk [136]
3 years ago
15

What is the domain of this relation? (14, 2) (5, 17) (-1, 8) (9, 3)?

Mathematics
1 answer:
koban [17]3 years ago
5 0
Domain is the collection of the x values: {14, 5, -1, 9}
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emmainna [20.7K]

Answer:

x = 12.8

Step-by-step explanation:

i use this calculator to find triangle measurements as long as you know like three values it works

https://www.calculator.net/triangle-calculator.html?vc=90&vx=&vy=10&va=38&vz=&vb=&angleunits=d&x=79&y=19

ik sending a link is hecka sus but you can just look up triangle calculator and it'll pop up

7 0
3 years ago
Need answer as soon as possible
sergejj [24]

Answer:

I think it would be 65 degrees

Step-by-step explanation:

If the side of Line q and line k is 115 degrees and its the same for line m and line k, then all you have to do is take 180 and subtract 115 from it to get the other side of the angle.

7 0
3 years ago
Please help solve please
irina1246 [14]
You just solved it.... do u need the work shown?
8 0
3 years ago
A company wants to find out if the average response time to a request differs across its two servers. Say µ1 is the true mean/ex
lorasvet [3.4K]

Answer:

a) The null and alternative hypothesis are:

H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2\neq 0

Test statistic t=0.88

The P-value is obtained from a t-table, taking into acount the degrees of freedom (419) and the type of test (two-tailed).

b)  A P-value close to 1 means that a sample result have a high probability to be obtained due to chance, given that the null hypothesis is true. It means that there is little evidence in favor of the alternative hypothesis.

c) The 95% confidence interval for the difference in the two servers population expectations is (-0.372, 0.972).

d) The consequences of the confidence interval containing 0 means that the hypothesis that there is no difference between the response time (d=0) is not a unprobable value for the true difference.

This relate to the previous conclusion as there is not enough evidence to support that there is significant difference between the response time, as the hypothesis that there is no difference is not an unusual value for the true difference.

Step-by-step explanation:

This is a hypothesis test for the difference between populations means.

The claim is that there is significant difference in the time response for the two servers.

Then, the null and alternative hypothesis are:

H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2\neq 0

The significance level is 0.05.

The sample 1, of size n1=196 has a mean of 12.5 and a standard deviation of 3.

The sample 2, of size n2=225 has a mean of 12.2 and a standard deviation of 4.

The difference between sample means is Md=0.3.

M_d=M_1-M_2=12.5-12.2=0.3

The estimated standard error of the difference between means is computed using the formula:

s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{3^2}{196}+\dfrac{4^2}{225}}\\\\\\s_{M_d}=\sqrt{0.046+0.071}=\sqrt{0.117}=0.342

Then, we can calculate the t-statistic as:

t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{0.3-0}{0.342}=\dfrac{0.3}{0.342}=0.88

The degrees of freedom for this test are:

df=n_1+n_2-1=196+225-2=419

This test is a two-tailed test, with 419 degrees of freedom and t=0.88, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=2\cdot P(t>0.88)=0.381

As the P-value (0.381) is greater than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that there is significant difference in the time response for the two servers.

<u>Confidence interval </u>

We have to calculate a 95% confidence interval for the difference between means.

The sample 1, of size n1=196 has a mean of 12.5 and a standard deviation of 3.

The sample 2, of size n2=225 has a mean of 12.2 and a standard deviation of 4.

The difference between sample means is Md=0.3.

The estimated standard error of the difference is s_Md=0.342.

The critical t-value for a 95% confidence interval and 419 degrees of freedom is t=1.966.

The margin of error (MOE) can be calculated as:

MOE=t\cdot s_{M_d}=1.966 \cdot 0.342=0.672

Then, the lower and upper bounds of the confidence interval are:

LL=M_d-t \cdot s_{M_d} = 0.3-0.672=-0.372\\\\UL=M_d+t \cdot s_{M_d} = 0.3+0.672=0.972

The 95% confidence interval for the difference in the two servers population expectations is (-0.372, 0.972).

7 0
3 years ago
PLEASE ANSWER ASAP NEED HELP REALLY BAD CALLING OUT TO ALL MATH PROS WHO CAN HELP ME PLEASE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
sergeinik [125]

Answer:

<abd is 90

<abc is 44

this isnt that hard im in middle school and doing this

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
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