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lyudmila [28]
3 years ago
10

PLEASE HELP ME!! =) NEED DONE NOW.

Mathematics
2 answers:
sergij07 [2.7K]3 years ago
5 0
The answer is a translation of 2 units to the left and 7 units up

Cloud [144]3 years ago
5 0

Answer:

D. a translation of 2 units to the left and 7 units up

just trust me. i got 100!

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I will show the options, for example how I wish it to be is like, "1 Acute" something like that, I will show you the answers.
Olin [163]

Answer:

1. Acute

2. Obtuse

3. Isosceles

7 0
3 years ago
Read 2 more answers
What is the quotient in the polynomial form?
Oxana [17]

First step of a synthetic divison is that we need to carry down the leading coefficient. Here the leading coefficient is 2. So, carry down 2 at the bottom.

Next step is to multiply the divisor -3 with this carry down number 2. So, we have got 3*(-2)= -6 which will place atthe bottom of the next coefficient 4.

Next step is to add this column.

Now repeat the same method again till the last colum.

At the end we have got 0 after the addition. Which means the remainder is 0.

So, the quotient is 2x^2-2x+2.

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
A) find the measure of arc AB<br> b)find angle measure APC
Aleks04 [339]

Answer:

Step-by-step explanation:

5 0
2 years ago
The figure below is divided into 100 small squares of equal size.
Bess [88]
60/100 / 60 are shaded out of the hundred. Simplified would be 3/5. The decimal would be 0.6.
5 0
2 years ago
Read 2 more answers
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