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3 years ago

What is the slope intercept form of 16x+2y=12

1 answer:

8 0

So the slope intercept form is y=mx+b where m is slope and b is y intercept

so 16x+2y=12
subtract 16x from both sides
2y=12-16x
divide both sides by 2
y=-8x+6
slope is -8 and y intercept is 6

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The answer to 14 I don't understand how to classify it

Nikitich [7]

The classification would be a right triangle

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3 years ago

What is the third quartile, Q3, of the following distribution?

GREYUIT [131]

Answer:

The third quartile is:

$Q_3=29$

Step-by-step explanation:

First organize the data from lowest to highest

4, 5, 10, 12, 14, 16, 18, 20, 21, 21, 22, 22, 24, 26, 29, 29, 33, 34, 43, 44

Notice that we have a quantity of n = 20 data

Use the following formula to calculate the third quartile $Q_3$

For a set of n data organized in the form:

$x_1, x_2, x_3, ..., x_n$

The third quartile is $Q_3$ :

$Q_3=x_{\frac{3}{4}(n+1)}$

With n=20

$Q_3=x_{\frac{3}{4}(20+1)}$

$Q_3=x_{15.75}$

The third quartile is between $x_{15}=29$ and $x_{16}=29$

Then

$Q_3 =x_{15} + 0.75*(x_{16}- x_{15})$

$Q_3 =29 + 0.75*(29- 29)\\\\Q_3 =29$

5 0

3 years ago

Read 2 more answers

using the equation G = 10-1.25t. What is the x-intercept of the equation and what is its interpretation in the context of the pr

Masja [62]

Answer:

the x axis is 10, its interpretation is actually G, so on a graph instead of X and Y you would have G and T

8 0

3 years ago

Two machines are used for filling glass bottles with a soft-drink beverage. The filling process have known standard deviations s

stellarik [79]

Answer:

a. We reject the null hypothesis at the significance level of 0.05

b. The p-value is zero for practical applications

c. (-0.0225, -0.0375)

Step-by-step explanation:

Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.

Then we have $n_{1} = 25$ , $\bar{x}_{1} = 2.04$ , $\sigma_{1} = 0.010$ and $n_{2} = 20$ , $\bar{x}_{2} = 2.07$ , $\sigma_{2} = 0.015$ . The pooled estimate is given by

$\sigma_{p}^{2} = \frac{(n_{1}-1)\sigma_{1}^{2}+(n_{2}-1)\sigma_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(25-1)(0.010)^{2}+(20-1)(0.015)^{2}}{25+20-2} = 0.0001552$

a. We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} \neq 0$ (two-tailed alternative).

The test statistic is $T = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}}$ and the observed value is $t_{0} = \frac{2.04 - 2.07}{(0.01246)(0.3)} = -8.0257$ . T has a Student's t distribution with 20 + 25 - 2 = 43 df.

The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value $t_{0}$ falls inside RR, we reject the null hypothesis at the significance level of 0.05

b. The p-value for this test is given by $2P(T$ 0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.

c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)

$(\bar{x}_{1}-\bar{x}_{2})\pm t_{0.05/2}s_{p}\sqrt{\frac{1}{25}+\frac{1}{20}}$ , i.e.,

$-0.03\pm t_{0.025}0.012459\sqrt{\frac{1}{25}+\frac{1}{20}}$

where $t_{0.025}$ is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So

$-0.03\pm(2.0167)(0.012459)(0.3)$ , i.e.,

(-0.0225, -0.0375)

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2 years ago

What is the largest integer value less than 169?

MA_775_DIABLO [31]

168 :)

This is because an integer is a whole number, so the largest integer value that is less than 169 has to be 168

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2 years ago