1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Kipish [7]
3 years ago
8

PLZZ HELP

Mathematics
2 answers:
Elan Coil [88]3 years ago
5 0
It is a function. This is because the x's don't repeat.

kifflom [539]3 years ago
4 0
This indeed is both a relation and a function because it is a set of ordered pairs and the x-axis does not repeat itself.
You might be interested in
I need help please it’s urgent I don’t get it
JulsSmile [24]

Answer: m∠RQS=104° and m∠TQS=76°

Step-by-step explanation:

4 0
3 years ago
A rectangle is inscribed in a semicircle of radius 10 as shown in the figure. Find a function that models the area A of the rect
monitta
So based on your question where there is a rectangle inscribed in a semicircle or a radius 10 as shown in your figure. The possible function that could represent the are of the rectangle is A=2x*sqrt(100-x^2) i hope you are satisfied with my answer and feel free to ask for more
5 0
3 years ago
Devin is going to deposit $3,000 in an account that earns 3.5% interest for 10 years. How much more INTEREST will he earn if the
Andrews [41]

Answer:

the answer will be rs.1050as more interest

7 0
3 years ago
For her exercise today, Keiko plans to both run and swim. Let r be the number of laps she runs and let s be the number of laps s
DerKrebs [107]

Answer:

5r + 3s ≥ 30

Step-by-step explanation:

We know Keiko wants to do exercise at least 30 minutes a day, and she wants ideally to mix up the activities, by combining running and swimming into her schedule.

Since each lap she runs takes 5 minutes and each lap she swims takes 3 minutes, these will be the left side of the inequality:

5r + 3s

She wants to exercise at least 30 minutes a day, so...

5r + 3s ≥ 30

she could do 3 laps of running and 5 laps of swimming for example.

6 0
3 years ago
Experian would like to test the hypothesis that the average credit score for an adult in Virginia is different from the average
aliya0001 [1]

Answer:

a. We fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

b. The 95% confidence interval for the true difference of means is -2.2468 and 36.2468. There is a probability of 95% that the true difference of means \mu_{1}-\mu_{2} is between -2.2468 and 36.2468. This confidence interval contains the number 0, this is consistent with the results we got in a. Because we fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

Step-by-step explanation:

Let \mu_{1}-\mu_{2} be the true difference between the average credit score for an adult in Virginia and the average credit score for an adult in North Carolina. We have the large sample sizes n_{1} = 40 and n_{2} = 35, the unbiased point estimate for \mu_{1}-\mu_{2} is \bar{x}_{1} - \bar{x}_{2}, i.e., 699-682 = 17.

The standard error is given by \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}, i.e.,

\sqrt{\frac{(44)^{2}}{40}+\frac{(41)^{2}}{35}} = 9.8198.

a. We want to test H_{0}: \mu_{1}-\mu_{2} = 0 vs H_{1}: \mu_{1}-\mu_{2} \neq 0 (two-tailed alternative). The rejection region is given by RR = {z | z < -1.96 or z > 1.96} where -1.96 and 1.96 are the 2.5th and 97.5th quantiles of the standard normal distribution respectively. The test statistic is Z = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}} and the observed value is z_{0} = \frac{17}{9.8198} = 1.7312. Because 1.7312 does not fall inside RR, we fail to reject the null hypothesis.

b. The endpoints for a 95% confidence interval for \mu_{1}-\mu_{2} is given by 17\pm (z_{0.05/2})9.8198, i.e., 17\pm (z_{0.025})9.8198 where z_{0.025} is the 2.5th quantile of the standard normal distribution, i.e., -1.96, so, we have 17-(1.96)(9.8198) and 17+(1.96)(9.8198), i.e., -2.2468 and 36.2468. There is a probability of 95% that the true difference of means \mu_{1}-\mu_{2} is between -2.2468 and 36.2468. This confidence interval contains the number 0, this is consistent with the results we got in a. Because we fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

3 0
3 years ago
Other questions:
  • Express your answer in simplest form. 5/9 + 1/6. what is the answer please
    11·1 answer
  • ᵂᴿᴵᵀᴱ ᴱᴬᶜᴴ ᶠᴿᴬᶜᵀᴵᴼᴺ ᴼᴿ ᴹᴵˣᴱᴰ ᴺᵁᴹᴮᴱᴿ ᴬˢ ᴬ ᴰᴱᶜᴵᴹᴬᴸ!!
    9·1 answer
  • 1 + 1? <br> .......……………………
    11·2 answers
  • The velocity of a turtle is recorded at 1 minute intervals (in meters per second). Use the right-endpoint approximation to estim
    11·1 answer
  • Divide £60 in a ratio of 2:1
    9·2 answers
  • Which is better, a really good desert, or a really yummy dinner? What is your favorite?
    15·2 answers
  • We have 500 mice and want to determine how effective a given cure for a virus. 10 mice are taken from the group and injected wit
    13·1 answer
  • Tres veces un número x, restado de 18 es menor que -90
    9·1 answer
  • I'm really confused about this problem. Can somebody help me?
    10·1 answer
  • Consider this equation <br><br> (thanks for the help!)
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!