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3 years ago

Triangle ABC has side lengths: AB = 3.5 cm, BC = 2.4 cm, and AC = 4.2 cm ΔABC ≅ ΔHJK What is the length of side HJ?

1 answer:

4 1

For this you'll want to draw out Triangle ABC and label it as the problem gives you.

Draw a similar triangle to the first one you drew and label it the exact same way as "ABC" and then match the measurement "AB" to "HJ".

Hope this helps.

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A cone-shaped container has a height of 9 in. and diameter of 2 in. It is filled with a liquid that is worth $2 per cubic inch.

ELEN [110]

Maybe it is $9.00
2 x 9 = 18
$2 per inch
18 divided by 2
$9.00?

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3 years ago

WILL GIVE BRAINEST <br>Leo had $91, which is 7 times as much......

vivado [14]

C) because the 7 and x right next to each other represents multiplication. so it's:
7x=91

7X(Allison's money)=91

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Question on angles, photo attached.

MArishka [77]

I think the answer is D.

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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)

8 0

3 years ago

How do you end up with 0=0

Eva8 [605]

Ok so first plug in y for in the equation 2x-2y=12
Therefore 2x - 2(x -6)=12
Then distribute the two to the x and the -6
2x-2x+12=12
The 2x's cancel leaving you with
12=12
From there you could subtract the 12 from both sides to get 0=0, I assume

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3 years ago

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