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2 years ago

A roller-rink floor is shown below. Each end is semi-circle. What is its area? Use t = 3.14

Mathematics

1 answer:

liubo4ka [24]2 years ago

7 0

Answer:

6884

Step-by-step explanation:

8x8x5

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Svet_ta [14]

Harap awak faham hehehe kalau jawapan betul bagi brainliest yer hehehe

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2 years ago

Please help!! I'll give brainliest! <br>Identifying Triangle Congruence

ddd [48]

Answer:

1. SAS

2. SSS

3. ASA

4. AAS

5. RHS

7 0

2 years ago

Read 2 more answers

Help!!

lora16 [44]

False

(5,5)
y < x + 3
5 < 5 + 3
5 < 8 (correct)

(5,5)
y > -2x - 1
5 > -2(5) - 1
5 > -10 - 1
5 > -11 (correct)

so (5,5) is a solution..and there are many more

8 0

3 years ago

Two shooters, Rodney and Philip, practice at a shooting range. They fire rounds each at separate targets. The targets are mark

Sedaia [141]

Complete Question

Answer:

$SE = 0.66}$

$-3.29 < \mu_1 - \mu_2 < -0.70$

Step-by-step explanation:

From the question we are told that

The sample size is n = 60

The first sample mean is $\= x _1 = 8$

The second sample mean is $\= x _2 = 10$

The first variance is $v_1 = 0.25$

The first variance is $v_2 = 0.55$

Given that the confidence level is 95% then the level of significance is 5% = 0.05

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the first standard deviation is

$\sigma_1 = \sqrt{v_1}$

=> $\sigma_1 = \sqrt{0.25}$

=> $\sigma_1 = 0.5$

Generally the second standard deviation is

$\sigma_2 = \sqrt{v_2}$

=> $\sigma_2 = \sqrt{0.55}$

=> $\sigma_2 = 0.742$

Generally the first standard error is

$SE_1 = \frac{\sigma_1}{\sqrt{n} }$

$SE_1 = \frac{0.5}{\sqrt{60} }$

$SE_1 = 0.06$

Generally the second standard error is

$SE_2 = \frac{\sigma_2}{\sqrt{n} }$

$SE_2 = \frac{0.742}{\sqrt{60} }$

$SE_2 = 0.09$

Generally the standard error of the difference between their mean scores is mathematically represented as

$SE = \sqrt{SE_1^2 + SE_2^2 }$

=> $SE = \sqrt{0.06^2 +0.09^2 }$

=> $SE = 0.66}$

Generally 95% confidence interval is mathematically represented as

$(\= x_1 -\= x_2) -(Z_{\frac{\alpha }{2} } * SE) < \mu_1 - \mu_2 < (\= x_1 -\= x_2) +(Z_{\frac{\alpha }{2} } * SE)$

=> $(8 -10) -(1.96 * 0.66) < \mu_1 - \mu_2 < (8-10) +(Z_{\frac{\alpha }{2} } * 0.66)$

=> $-3.29 < \mu_1 - \mu_2 < -0.70$

5 0

3 years ago

Ira Lisetskai [31]

answer = ) 100.80

Step-by-step explanation:

I subtracted the people who payed on the 1st of the month which was 4 from the total 12 memebers. I got 8, then that 8 payed on the 2nd of the month so I multiplied 8 by 12.60

5 0

2 years ago