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3 years ago

What is the image point of (7,0) after a translation right 5 units and down 5 units?

Mathematics

1 answer:

ivolga24 [154]3 years ago

7 0

Answer:

(12, - 5 )

Step-by-step explanation:

A translation of 5 units right means adding 5 to the x- coordinate

A translation of 5 units down means subtracting 5 from the y- coordinate

Thus

(7, 0 ) → (7 + 5, 0 - 5 ) → (12, - 5 )

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Please help pretty please.

Masteriza [31]

Answer:

reflection then translation

Step-by-step explanation:

you reflected of the y axis and then you translated 2 left and 5 up which brings you to 3

3 0

1 year ago

3 pounds of lawn seeds cover 1800 ft.². How many 4 pound bags are needed to cover 8400 ft.²

vlada-n [284]

1800/3 = 600
Every pound covers 600 square feet
8400/600 = 14
14/4 = 3.5 (round up) so it would become 4.


You would need 4 bags.

I hope this helps!

6 0

4 years ago

3 - x/2 = 6 (Solve each equation) Check your answer

Yanka [14]

Answer to question
3-x/2=6
Subtract 3 from both sides
-x/2=3
Multiply 3 by -2
Final Answer: x=-6

Work check
3-x/2=6
Substitute -6 for x since you solved for x in the last equation
3-(-6)/2
Convert 3-(-6/2) to 3+6/2
3+6/2
Divide 6 by 2
3+3
Add
Final Answer: 6 (This 6 is the same 6 we used to solve for x in the first equation).

6 0

4 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

$4,600 invested in a CD paying the investor 5% annual return for one year? what’s the answer

xxMikexx [17]

Answer:

230

Step-by-step explanation:

multiply 6,400 by 5% then multiply that by the 1 year and 230 will be the answer

Hope this helps :)

7 0

3 years ago