Can someone plz help me with this. 100 points and brainliest to the correct answer.

What is the answer 0,12km² a m² show work

seraphim [82]

Answer:

120 000 m^2

Step-by-step explanation:

I ASSUME you're being asked to convert from km^2 to m^2.

For this you need to identify the conversion rate, which is 1 000 000 m^2 / 1 km ^2. Multiply by the thing you're converting from and cancel the units.

(1 000 000 m^2 / 1 km ^2) x 0,12 km^2 = 120 000 m^2 (the km^2 cancels out, leaving you only with your target unit of m^2)

8 0

2 years ago

Read 2 more answers

A student wants to use evidence from "The Treasure of Lemon Brown" to support the idea that Greg becomes more

Nastasia [14]

Answer:

URGENT

Debug the following:any one free out

sin theta equals to cos theta then the value of sin 2 theta

8 0

2 years ago

Read 2 more answers

Solve for X.<br><br>I've tried and it's confusing I'm not really that great at geometry

Nutka1998 [239]

Set as a proportion
16 = 12
------------------
5x+2 24

cross multiply
12(5x+2) = 16 * 24
60x +24 = 384
60x +24-24 = 384-24
60x =360
60x/60=360/60
x =6

To figure AB plug 6 into 5x+2
5x +2
5(6) + 2
30+2
32
AB = 32

4 0

3 years ago

Seven clowns hold 4 balloons each at the fair. Draw and label a tape diagram to show the total number of balloons the clowns hol

Inessa05 [86]

Answer:

7 times 4 which is 28 at each fair.

8 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.

<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