How long did it take the 9000 N car to reach 100 km/hr?

Need help to solve this question

docker41 [41]

The answer is 73, simple math

6 0

3 years ago

PLS HELP ME ASAP!!!!!!!!!!!!!!

Anton [14]

The answer to that question is C

8 0

3 years ago

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Students in Miss Moseley's fourth grade class are learning multiplication, and they demonstrate mastery by passing assessments.

blagie [28]

Answer:
Each will have passed 17 tests and it will take 3 weeks.

Explanation: 
Let x be the number of weeks.
The number of tests Travis passes to begin with is 11.
We then add 2 tests per week, or 2x to that, giving us:
11+2x.

The number of tests Jenifer has passed to begin with is 2.
We then add 5 tests per week, or 5x to that, giving us:
2+5x.

Setting these equal, we have: 
11+2x=2+5x.

Subtract 2x from each side: 
11+2x-2x=2+5x-2x;
11=2+3x.

Subtract 2 from each side: 
11-2=2+3x-2;
9=3x.

Divide both sides by 3:
 $\frac{9}{3} = \frac{3x}{3}$ ;
3=x.

It will take 3 weeks.
In 3 weeks,
Travis will have passed:
11+2*3 = 11+6 = 17 tests.
Jenifer will have passed the same number, since she catches up with him at this point.

6 0

3 years ago

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Write the decimal 1,072.039 in words

Rina8888 [55]

One thousand seventy two point thirty nine thousandths.

3 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)

8 0

3 years ago