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3 years ago

HELP QUICKLY ANSWRR BRAINLIEST FOR QUICKEST AND CORRECT ANSWER

Mathematics

2 answers:

rosijanka [135]3 years ago

4 0

False because it would be 4(2)=4 which is not correct

tekilochka [14]3 years ago

3 0

Answer:

I believe your answer is False, because the ordered pair is not a solution to the equation.

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1. Ethan writes 1/6 of a page in 1/12 of a minute. how much time does it take him to write a full page?

Nata [24]

I believe that the answer is 1/2 of a minute.

8 0

4 years ago

Read 2 more answers

Help...................

melisa1 [442]

Answer:

10.63cm

Step-by-step explanation:

Pythagorean theorem

a^2+b^2=c^2

(square root of 32)^2+(9)^2=113^2

The square root of 113 is 10.63.

I hope this helps :)

5 0

3 years ago

All of the students of Music High School are in the band, the orchestra, or both. 80 percent of the students are in only one gro

il63 [147K]

Answer:

Option B.

Step-by-step explanation:

It is given that all of the students of Music High School are in the band, the orchestra, or both.

Students only one group = 80%

Students in both groups = 100%-80% = 20%

Students in band only = 50%

Students in band = Students in band only + Students in both groups

= 50% + 20% = 70%

Students in orchestra only = 100% - 70% = 30%

There are 119 students in the band.

Let total number of student be x.

$70\% of x=119$

$\dfrac{70}{100}x=119$

$0.7x=119$

$x=170$

Total number of students is 170.

We know that 30% of all students are in orchestra only.

$170\times \dfrac{30}{100}=51$

Therefore, the correct option is B.

5 0

3 years ago

What is the measure of

Zolol [24]

Answer:

pretty sure its 55° because the total should be 180* and I know that C is a right angle aka 90° and 90 + 35 = 125 and 180-125=55

3 0

3 years ago

Test the null hypothesis Upper H 0 : (mu 1 minus mu 2 )equals 0H0: μ1−μ2=0 versus the alternative hypothesis Upper H Subscript a

Law Incorporation [45]

Answer:

The test statistic t is t=2.9037.

The null hypothesis is rejected.

For a significance level of 0.05, there is enough evidence to support the alternative hypothesis.

Step-by-step explanation:

The question is incomplete:

The sample 1, of size n1=25 has a mean of 1.15 and a standard deviation of 0.31.

The sample 2, of size n2=25 has a mean of 0.95 and a standard deviation of 0.15.

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

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 α=0.05.

The difference between sample means is Md=0.2.

$M_d=M_1-M_2=1.15-0.95=0.2$

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

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2+\sigma_2^2}{n}}=\sqrt{\dfrac{0.31^2+0.15^2}{25}}\\\\\\s_{M_d}=\sqrt{\dfrac{0.119}{25}}=\sqrt{0.005}=0.069$

Then, we can calculate the t-statistic as:

$t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{0.2-0}{0.069}=\dfrac{0.2}{0.069}=2.9037$

The degrees of freedom for this test are:

$df=n_1+n_2-1=25+25-2=48$

This test is a two-tailed test, with 48 degrees of freedom and t=2.9037, so the P-value for this test is calculated as (using a t-table):

$P-value=2\cdot P(t>2.9037)=0.0056$

As the P-value (0.0056) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

8 0

3 years ago