Simplify the expressions:

A Web music store offers two versions of a popular song. The size of the standard version is 2.1 megabytes (MB). The size of the

liq [111]

If the number of downloads of the standard version is x, and the high quality is x, 2.1*x+4.1*y=2761 (not 1010 due to that this is multiplied by 2.1 and 4.1, therefore representing the total amount of megabytes) In addition, there are 1010 total downloads, and it's either 2.1 MB or 4.1 MB, so x+y=1010.

We have
2.1x+4.1y=2761
x+y=1010

Multiplying the second equation by -2.1 and adding it to the first equation, we get 2y=2761-1010*2.1=640 and by dividing both sides by 2 we get y=320 downloads of the high quality version

5 0

3 years ago

COME ON PLSSSSSS HELP I'VE POSTED THIS BEFORE :(

mr Goodwill [35]

Here are your points (-9,-8), (-6,-2), (-2,-6)

5 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.

Confidence interval

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

The following is a geometric sequence. - 1/2 , 1/4, -1/8 , 1/16 true false

seraphim [82]

Answer:

True

Step-by-step explanation:

It has a common ratio of -1/2

Each term after the first is obtained by multiplying the previous one by -1/2:-

-1/2 * 1/2 = 1/4 ; 1/4 * -1/2 = -1/8 and so on.

6 0

3 years ago

Simplify cosx ( tanx+cotx)

Soloha48 [4]

The first step to solving this is to use tan(t) = $\frac{sin(t)}{cos(t)}$ to transform this expression.
cos(x) × $( \frac{sin(x)}{cos(x)} + cot(x) )$
Using cot(t) = $\frac{cos(x)}{sin(x)}$ ,, transform the expression again.
cos(x) × $( \frac{sin(x)}{cos(x)} + \frac{cos(x)}{sin(x)} )$
Next you need to write all numerators above the least common denominator (cos(x)sin(x)).
cos(x) × $\frac{sin(x)^{2} + cos(x)^{2} }{cos(x)sin(x)}$
Using sin(t)² + cos(t)² = 1,, simplify the expression.
cos(x) × $\frac{1}{cos(x)sin(x)}$
Reduce the expression with cos(x).
$\frac{1}{sin(x)}$
Lastly,, use $\frac{1}{sin(t)}$ = csc(t) to transform the expression and find your final answer.
csc(x)
This means that the final answer to this expression is csc(x).
Let me know if you have any further questions.
:)

4 0

3 years ago