All categories

2 years ago

PLZ help with 6th math homework!

Mathematics

1 answer:

vitfil [10]2 years ago

3 0

I'm not doing your math homework, this is to help you learn and pass your test :) I will however show you a trick. ask your self this... "What can i factor out of 4x that i can also factor out of 8, which this would be 4. Now what you do if you divide everything by 4

4x/4 = x
8/4 = 2

so 4(x+2) is 4x+8 fully factored, NOW to check, use the distributive property, Multiply 4 back into x+2
4*x = 4x
4*2 = 8
so 4(x+2) = 4x+8

You might be interested in

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.

<u>Confidence interval </u>

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

2. Solve: 33 + 2(5+8)-(6-2)

mars1129 [50]

Answer:

Step-by-step explanation:

Add the numbers 5 + 8 and you get 13. Multiply 13 x 2 which is 26. Add 33 + 26 and you get 59. Then, subtract it by 4 and you get 55.

6 0

2 years ago

Una caja tiene 60 bombones eba se comio 1/5 bombones y ana 1/2 bombones cuantos bombones se comio eba y ana

DedPeter [7]

Responder:

42 bombones

Explicación paso a paso:

Dado que :

Número de bombones = 60

Cantidad consumida por Eba = 1/5 * 60 = 12

Cantidad comida por Ana = 1/2 * 60 = 30

Cantidad de chocolate que comieron Eba y Ana:

(12 + 30) = 42

De ahí que Ana y Eba se comieran 42 bombones

7 0

2 years ago

Which percent is equivalent to 1/20

solmaris [256]

Answer:

Step-by-step explanation:

1/20 = 0.05

0.05 = 5%

3 0

2 years ago

Read 2 more answers

108 students going on a field trip to pan for gold if they are going in bands that holds five students each how many Vans are ne

Alisiya [41]

If there are 108 students, and the vans can hold up to 5 students, then you would need 22 vans.
If you were to divide 108 by 5, then you would get 21.3 . That 3 is the remaining children, therefore, to fit them in the trip, you would need another van, hence, 22 vans. The van that is not full would have 3 children in it.

8 0

2 years ago