What is the value of the 4 in the number 546210

In a school, only 13 out of 15 student participate in a competition.what is the probability of the student who do not makes it t

Papessa [141]

Answer:

$\frac{2}{15}$

Step-by-step explanation:

Number of students participating in the competition = 13.

Total number of students in the school = 15.

Thus ,

the number of students who did not make to this competition =

Total number of students - number of students participated

= 15-13 = 2.

Now ,

probability of the student who do not makes it to the competition

= fraction of students who did not make to this competition from the total number of students .

= $\frac{2}{15}$ .

4 0

3 years ago

The expression f(x) = 12(1.035)* models the monthly growth of membership in the new drama club at a school. According to the fun

olga nikolaevna [1]

Answer:

The monthly growth rate is 3.5%.

Step-by-step explanation:

The exponential growth function is given as follows:

$y=a(1+r)^{x}$

Here,

y = final value

a = initial value

r = growth rate

x = time taken

The provided expression for the monthly growth of membership in the new drama club at a school is:

$f(x) = 12\cdot(1.035)^{x}$

Comparing this function with the exponential growth function:

$a(1+r)^{x}=12(1.035)^{x}\\\\a(1+r)^{x}=12(1+0.035)^{x}$

Then value of r is 0.035 or 3.5%.

Thus, the monthly growth rate is 3.5%.

7 0

3 years ago

Sum of 2/3 + (-1/3) i cant find the answer

arsen [322]

Answer:

1/3

Step-by-step explanation:

2/3 plus -1/3 is the same as subtracting 1/3 from 2/3.

6 0

3 years ago

Read 2 more answers

Two catalysts may be used in a batch chemical process. Twelve batches were prepared using catalyst 1, resulting in an average yi

aalyn [17]

Answer:

a) $t=\frac{(91-85)-(0)}{2.490\sqrt{\frac{1}{12}}+\frac{1}{15}}=6.222$

$df=12+15-2=25$

$p_v =P(t_{25}>6.222) =8.26x10^{-7}$

So with the p value obtained and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

b) $(91-85) -2.79 *2.490 \sqrt{\frac{1}{12} +\frac{1}{15}} =3.309$

$(91-85) +2.79 *2.490 \sqrt{\frac{1}{12} +\frac{1}{15}} =8.691$

Step-by-step explanation:

Notation and hypothesis

When we have two independent samples from two normal distributions with equal variances we are assuming that

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

And the statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}}+\frac{1}{n_2}}$

Where t follows a t distribution with $n_1+n_2 -2$ degrees of freedom and the pooled variance $S^2_p$ is given by this formula:

$\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

This last one is an unbiased estimator of the common variance $\sigma^2$

Part a

The system of hypothesis on this case are:

Null hypothesis: $\mu_2 \leq \mu_1$

Alternative hypothesis: $\mu_2 > \mu_1$

Or equivalently:

Null hypothesis: $\mu_2 - \mu_1 \leq 0$

Alternative hypothesis: $\mu_2 -\mu_1 > 0$

Our notation on this case :

$n_1 =12$ represent the sample size for group 1

$n_2 =15$ represent the sample size for group 2

$\bar X_1 =85$ represent the sample mean for the group 1

$\bar X_2 =91$ represent the sample mean for the group 2

$s_1=3$ represent the sample standard deviation for group 1

$s_2=2$ represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

$\S^2_p =\frac{(12-1)(3)^2 +(15 -1)(2)^2}{12 +15 -2}=6.2$

And the deviation would be just the square root of the variance:

$S_p=2.490$

Calculate the statistic

And now we can calculate the statistic:

$t=\frac{(91-85)-(0)}{2.490\sqrt{\frac{1}{12}}+\frac{1}{15}}=6.222$

Now we can calculate the degrees of freedom given by:

$df=12+15-2=25$

Calculate the p value

And now we can calculate the p value using the altenative hypothesis:

$p_v =P(t_{25}>6.222) =8.26x10^{-7}$

Conclusion

So with the p value obtained and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

Part b

For this case the confidence interval is given by:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2} S_p \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}$

For the 99% of confidence we have $\alpha=1-0.99 = 0.01$ and $\alpha/2 =0.005$ and the critical value with 25 degrees of freedom on the t distribution is $t_{\alpha/2}= 2.79$