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3 years ago

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-2x+3y=6 In slope intercept form

1 answer:

Helen [10]3 years ago

6 0

Answer:

-2x + 3y = 6

3y = 2x + 6

y = 2/3*x + 2

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HELP NO FILES PLEASE!!!

Gnom [1K]

Answer:

5.6 = AC

Step-by-step explanation:

We know that the length of BD is congruent to the length of AC, therefore we can say:

<em>BD ≅ AC</em>

4.8 * 1.6 = 3.2 * x

7.68 ≅ 3.2x

7.68 / 3.2 ≅ x

2.4 = x

Therefore length of AC would be:

<em>3.2 + x = AC</em>

3.2 + 2.4 = AC

<u>5.6 = AC</u>

Hope this helps!

5 0

3 years ago

In 2018, Mike Krzyewski and John Calipari topped the list of highest paid college basketball coaches (Sports Illustrated website

expeople1 [14]

From the data given, we estimate the population mean and population standard deviation. Then, we use this estimate to find a 95% confidence interval for the population variance and the population standard deviation.

Sample:

Salaries in millions of dollars: 2.2, 1.5, 0.5, 1.3, 2.4, 1.5, 2.7, 0.3, 2.0, 0.3

Question a:

The mean is the sum of all values divided by the number of values. So

$\overline{x} = \frac{2.2 + 1.5 + 0.5 + 1.3 + 2.4 + 1.5 + 2.7 + 0.3 + 2.0 + 0.3}{10} = 1.42$

The sample mean salary is of 1.42 million.

Question b:

The standard deviation is the square root of the difference squared between each value and the mean, divided by one less than the number of values.

So

$s = \sqrt{\frac{(2.2-1.42)^2 + (1.5-1.42)^2 + (0.5-1.42)^2 + (1.3-1.42)^2 + (2.4-1.42)^2 + (1.5-1.42)^2 + (2.7-1.42)^2 + ...}{9}} = 0.8772$

Thus, the estimate for the population standard deviation is of 0.8772 million.

Question c:

The sample size is $n = 10$

The significance level is $\alpha = 1 - 0.05 = 0.95$

The estimate, which is the sample standard deviation, is of $s = 0.8772$ .

Now, we have to find the critical values for the Pearson distribution. They are:

$\chi^2_{\frac{\alpha}{2},n-1} = \chi^2_{0.025,9} = 19.0228$

$\chi^2_{1-\frac{\alpha}{2},n-1} = \chi^2_{0.975,9} = 2.7004$

The confidence interval for the population variance is:

$\frac{(n-1)s^2}{\chi^2_{\frac{\alpha}{2},n-1}} < \sigma^2 < \frac{(n-1)s^2}{\chi^2_{1-\frac{\alpha}{2},n-1}}$

$\frac{9*0.8772^2}{19.0228} < \sigma^2 < \frac{9*0.8772^2}{2.7004}$

$0.3641 < \sigma^2 < 2.5646$

Thus, the 95% confidence interval for the population variance is (0.3641, 2.5646)

Question d:

Standard deviation is the square root of variance, so:

$\sqrt{0.3641} = 0.6034$

$\sqrt{2.5646} = 1.6014$

The 95% confidence interval for the population standard deviation is (0.6034, 1.6014).

For more on confidence intervals for population mean/standard deviation, you can check brainly.com/question/13807706

4 0

3 years ago

The school that Kayla goes to is selling tickets to a play. On the first day of ticket sales the school sold 6 senior citizen ti

Lera25 [3.4K]

Answer:

6S + 11C = $120

2S + 1 C = $24

So solve equation 2 for C

C = 24 - 2 S

Subsitute in to eq 1

6S + 11(24-2S) = 120

6S + 264 -22S = 120

264 -16S =120

264 -120 = 16S

S = (264-120)/16 = 144/16 = $9

c = 24 - 2S = 24 - 2x9 = 24-18 = $6

Senior Ticket is $9

Children Ticket is $6

Step-by-step explanation:

5 0

3 years ago

I need to know the answer

Bezzdna [24]

I think the answer is (3,5).

7 0

3 years ago

Read 2 more answers

32 divided by 9 =___with_____left over

rosijanka [135]

Answer:

3 with leftover of 5

Step-by-step explanation:

please give brainliest

8 0

3 years ago

Read 2 more answers