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boyakko [2]
3 years ago
14

Using the formula Distance = rate . time, find the distance covered by an airplane flying at 300 mph for 3 hours.

Mathematics
1 answer:
Brut [27]3 years ago
6 0
900. If your formula is correct, you only have to multiply 300 by 3
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PLS HELP! (it will only let me do 31 points :() but you get brainliest.
MissTica

Answer:

107, there is two lines after that so 108 109 and the last line is 110 so your answer will be 107 degrees.

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Twice the difference of a number and seven
levacccp [35]

The difference of a number and seven is x - 7. So twice that, would be:

2(x - 7)

4 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(T0 (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
Can someone please help me i got stuck in this question
BartSMP [9]
You could set up a ratio given that it is a 5% solution (5% of the total, not 5% of water.

5/100 = x/473 You need to find out how much is vinegar. Cross multiply
5 * 473 = 100 x  Find the number on the left.
2365 = 100 x Divide by 100
2365 / 100 = x 
x = 23.65 mL of vinegar.
The water is found by subtraction
473 - 23.65 = 449.35 mL of water.

5 0
4 years ago
use green's theorem to evaluate the line integral along the given positively oriented curve. c 9y3 dx − 9x3 dy, c is the circle
Rina8888 [55]

The line integral along the given positively oriented curve is -216π. Using green's theorem, the required value is calculated.

<h3>What is green's theorem?</h3>

The theorem states that,

\int_CPdx+Qdy = \int\int_D(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y})dx dy

Where C is the curve.

<h3>Calculation:</h3>

The given line integral is

\int_C9y^3dx-9x^3dy

Where curve C is a circle x² + y² = 4;

Applying green's theorem,

P = 9y³; Q = -9x³

Then,

\frac{\partial P}{\partial y} = \frac{\partial 9y^3}{\partial y} = 27y^2

\frac{\partial Q}{\partial x} = \frac{\partial -9x^3}{\partial x} = 27x^2

\int_C9y^3dx-9x^3dy = \int\int_D(-27x^2 - 27y^2)dx dy

⇒ -27\int\int_D(x^2 + y^2)dx dy

Since it is given that the curve is a circle i.e., x² + y² = 2², then changing the limits as

0 ≤ r ≤ 2; and 0 ≤ θ ≤ 2π

Then the integral becomes

-27\int\limits^{2\pi}_0\int\limits^2_0r^2. r dr d\theta

⇒ -27\int\limits^{2\pi}_0\int\limits^2_0 r^3dr d\theta

⇒ -27\int\limits^{2\pi}_0 (r^4/4)|_0^2 d\theta

⇒ -27\int\limits^{2\pi}_0 (16/4) d\theta

⇒ -108\int\limits^{2\pi}_0 d\theta

⇒ -108[2\pi - 0]

⇒ -216π

Therefore, the required value is -216π.

Learn more about green's theorem here:

brainly.com/question/23265902

#SPJ4

3 0
2 years ago
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