Please help to solve this

Roger had a farm with a certain number of orange trees. he had to cut down 5 trees to control the insects. each of the remaining

Oksi-84 [34.3K]

41790/210 = 199
199 + 5 = 204.

3 0

2 years ago

Read 2 more answers

Can someone help me with this ASAP I am so close to the end of this assignment!!!

Sveta_85 [38]

Answer:

A

Step-by-step explanation:

$\sqrt{(-5 - (-1))^{2} + (7 - 2)^{2} } = \sqrt{41}$

8 0

2 years ago

Read 2 more answers

How do you put 27% ,miles per hour , 42% and 45 in a sentence

Serjik [45]

There are a number of possible combinations for this open-ended question, so here's an example question and answer, in which the two are non-related:
James' car is going at 45 miles per hour, and that is 42% of his maximum speed. How much slower does James have to go to be 27% of his maximum speed?
Amy's car used to be able to go at 45 miles per hour as 42% of her maximum speed, but due to the new upgrade, 45 miles per hour is now only 27% of her maximum speed.

7 0

2 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

There are 4 cards in the box with the numbers 2, 5, 6, and 8. Two cards are drawn one after the other. The second card drawn is

malfutka [58]

It is given that there are 4 cards with numbers 2, 5, 6, and 8.

1) It is required to list all the numbers available when two cards are drawn one after the other and the second card is placed at the right of the first card drawn.

To do this, find the sample space of the draw which are:

$25,26,28,52,56,58,62,65,68,82,85,86$

Hence, the required numbers.

2) Recall that probability is the ratio:

$Probability=\frac{\text{ number of favorable outcomes}}{\text{ total number of possible outcomes}}$

From the numbers listed, notice that the total number of possible outcomes is 12.

Since event A is "obtained number is even", notice that the number of even numbers in the listed numbers is 9 (26,28,52,56,58,62,68,82,86).

Hence, the number of favorable outcomes is 9.

Substituting into the probability ratio gives the required probability:

$P(A)=\frac{9}{12}=\frac{3}{4}$

Check the number of numbers that are a multiple of 5, notice that they are 3 (25,65,85).

Hence, the required probability of event B is:

$P(B)=\frac{3}{12}=\frac{1}{4}$

The number of multiple of 3 in the listed numbers is 0 since there is no number that is a multiple of 3.

Hence, the required probability of event C is:

$P(C)=\frac{0}{12}=0$

Answers:

1) {25,26,28,52,56,58,62,65,68,82,85,86}

2) P(A) = 3/4

P(B) = 1/4

P(C)= 0

7 0

7 months ago

Please help to solve this ​

Please help to solve this