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2 years ago

2. Round 459 to the nearest hundred. O 460 450 500 400

Mathematics

1 answer:

KonstantinChe [14]2 years ago

5 0

Answer:

The answer is 500

Step-by-step explanation:

When you round to the hundred, you look at the 10s place. If the 10s place is 5,6,7,8, or 9, you round the hundred up. In this case, it is 459 so you would round up to 500. Have a nice day!!!

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Which of the following are integer solutions to the inequality below?<br> −2≤×<3

Phoenix [80]

Answer:

-2, - 1, 0, 1, 2

Step-by-step explanation:

The greater than or equal to sign (≤) demonstrates that the unknown is equal to -2 and greater, but is less than 3, so the largest integer solution is 2, not 3

6 0

2 years ago

Which ordered pair is a solution of the equation 2x-3y=19

goldenfox [79]

D because 5 times 2 is 10 and - cuts off - so -3 times -3 would be 9 so 10+9 is 19 :)

7 0

2 years ago

Read 2 more answers

Suppose that the time (in hours) required to repair a machine is an exponentially distributed random variable with mean 1/0.41/0

lbvjy [14]

Answer:

a) P(t>3)=0.30

b) P(t>10|t>9)=0.67

Step-by-step explanation:

We have a repair time modeled as an exponentially random variable, with mean 1/0.4=2.5 hours.

The parameter λ of the exponential distribution is the inverse of the mean, so its λ=0.4 h^-1.

The probabity that a repair time exceeds k hours can be written as:

$P(t>k)=e^{-\lambda t }=e^{-0.4t}$

(a) the probability that a repair time exceeds 3 hours?

$P(t>3)=e^{-0.4*3}=e^{-1.2}= 0.30$

(b) the conditional probability that a repair takes at least 10 hours, given that it takes more than 9 hours?

The exponential distribution has a memoryless property, in which the probabilities of future events are not dependant of past events.

In this case, the conditional probability that a repair takes at least 10 hours, given that it takes more than 9 hours is equal to the probability that a repair takes at least (10-9)=1 hour.

$P(t>10|t>9)=P(t>1)$

$P(k>1)=e^{-0.4*1}=0.67$

6 0

3 years ago

-2y^2-19y+33 trail and error

Alexandra [31]

Answer: (-2y+3)(y+11)

Step-by-step explanation:

33 is 3×11 so expect something like

(y+3)(y+11) = y^2+14y+33

leading term has coefficient -2 so

one factor is like (-2y+__)

Try (-2y+3)(y+11) = -2y^2-19y+33

which happens to be the answer

7 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

2 years ago