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3 years ago

Is 36 a factor. yes or no

Mathematics

2 answers:

Leona [35]3 years ago

7 0

Yes because 6x6 is 36

kondaur [170]3 years ago

3 0

There is no such thing as 36 being a factor but it has factors and its a factor of 72

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Describe the difference between the calculation of population standard deviation and that of sample standard deviation.

taurus [48]

Answer:

View graph

Step-by-step explanation:

The population refers to the whole set under study, while the sample is a representative part of that population.

3 0

3 years ago

Need help please :( What is the second term of (s+v)^5?

Citrus2011 [14]

First compute the coefficient like this:
$C_5^1=\frac{5!}{1!(5-1)!}=\frac{5!}{4!}=\frac{5\times4!}{4!}$
Simplifying the fraction over 4! we get:
$C_5^1=5$
and the variables are $s^4v$ . So answer $5s^4v$ .

The correct answer is C then.

3 0

3 years ago

Select the quadratic equation that has roots x=8 and x=-5

stellarik [79]

Step-by-step explanation:

Since, x=8 and x=-5 are the roots.

Therefore, (x - 8) & (x - 5) will be factors.

Hence, required quadratic equation can be given as:

$(x - 8)(x - 5) =0 \\ \\ \therefore {x}^{2} + ( - 8 - 5) x+ ( - 8) \times ( - 5) = 0 \\ \\ \therefore {x}^{2} + ( - 13) x+ 40= 0 \\ \\ \huge \red{ \boxed{ \therefore {x}^{2} - 13x+ 40= 0 }} \\ is \: the \: required \: quadratic \: equation.$

5 0

3 years ago

8*11=(8*10)+(1*10) is the equation true

svet-max [94.6K]

Answer:

Step-by-step explanation:

8*10=80, 1*10=10

80+10=90

8*11=88

88 does not equal 90 so no

5 0

3 years ago

Read 2 more answers

he Nassau Bahamas Community College (NBC) claims that students who take statistics spend on average 19 hours a week working. The

Virty [35]

Using the t-distribution, it is found that since the test statistic is greater than the critical value for the right-tailed test, it is found that there is enough evidence to conclude that the students' claim that they work more than 19 hours is correct.

At the null hypothesis, it is tested if they spend on average 19 hours a week working, that is:

$H_0: \mu = 19$

At the alternative hypothesis, it is tested if they spend more than 19 hours a week working, that is:

$H_1: \mu > 19$

We have the standard deviation for the sample, hence, the t-distribution is used.

The test statistic is given by:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

The parameters are:

$\overline{x}$ is the sample mean.

$\mu$ is the value tested at the null hypothesis.

s is the standard deviation of the sample.

n is the sample size.

Searching the problem on the internet, it is found that the values of the parameters are:

$\overline{x} = 24.2, \mu = 19, s = 12.59, n = 25$

Hence, the value of the test statistic is:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{24.2 - 19}{\frac{12.59}{\sqrt{25}}}$

$t = 2.07$

The critical value for a right-tailed test, as we are testing if the mean is greater than a value, with a significance level of 0.05 and 25 - 1 = 24 df is of $t^{\ast} = 1.71$

Since the test statistic is greater than the critical value for the right-tailed test, it is found that there is enough evidence to conclude that the students' claim that they work more than 19 hours is correct.

To learn more about the t-distribution, you can take a look at brainly.com/question/13873630

3 0

3 years ago