Please someone help!5m^2

There is 1/3 of a carrot in every container of salad.

snow_tiger [21]

Answer:

1/6

Step-by-step explanation:

should be the answer if I'm not mistaken

6 0

2 years ago

Yogi is 6 years older than Michelle. The sum of their ages is 26. Write a system of linear equations to represent this informati

postnew [5]

Answer:

10 and 16, x+(x+6)=26

Step-by-step explanation:

Michelle has an age we don't know, so we put her age as x.

Yogi is 6 years older than her, so her age is x+6

Michelle=x

Yogi=x+6

we know both their ages equal 26. so we set it up as

x+(x+6)=26

combining like terms we get

2x+6=26

subtract 6 from both sides

2x=20

divide both sides by 2

x=10

now that we have the value for x, we plug it into their original ages

Michelle is 10, because her age is just x.

Yogi is 16, because her age is x+6

8 0

2 years ago

A rectangular pool contains 920 cubic meters of water. the pool is .4 meter deep and 100 meters wide. How long is the pool?

LekaFEV [45]

Hi there,

Sol:
Length of the swimming pool = 260 m
Breadth of the swimming pool = 140 m

Let the height of the water in the swimming pool after pouring water be h meters.

Volume of the water poured = 54600 m3

l x b x h = 54600
260 x 140 x h = 54600
h = (54600) / (260 x 140)
h = 1.5 m = 150 cm

Hope it helps

5 0

2 years ago

Alex purchased

Aleks [24]

Answer:

1/11 of a gallon

Step-by-step explanation:

He used 2/11 of 1/2 gallon

2/11 * 1/2 = 1/11 of a gallon

5 0

2 years ago

Read 2 more answers

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

Please someone help!5m^2 - 26m -24 =0