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4 years ago

1 A. All master photographers are artists.

Mathematics

1 answer:

Ede4ka [16]4 years ago

7 0

Answer:

A is the appropriate option.

Step-by-step explanation:

The question given is a conditional statement.

With the condition that all master photographers are artist. This implies that any person who is a master photographer is automatically an artist.

A. Comparing the statement here, since Ansel Adam's is a master photographer, he is an artist.

B. Ansel Adams is an artist, but it is possible that not all artists are master photographer.

A is the correct option.

1. All master photographers are artists.

2. Ansel Adams is a master photographer.

Therefore, Ansel Adams is an artist.

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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

PLZ HELP I NEED HELP!!!!!;)

djverab [1.8K]

Step-by-step explanation:

can u give me braniest plss

8 0

3 years ago

Read 2 more answers

How long will it take to administer 1000 cc at a drop factor of 15 drop/ml and a drip rate of 50 drop/min

gayaneshka [121]

First, we are going to determine the number of drops that needs to be administered by dividing the total volume by the volume per drop. Since, 1 cc (cm³) is equal to 1 mL then, 1000 cc is equal to 1000 mL.

n = (1000 mL)(15 drop/1 mL) = 15000 drops

Then, divide the number of drops by the number of drops per minute.

N = (15000 drops)/ (50 drop/min) = 300 mins

Answer: 300 mins or 5 hours

5 0

3 years ago

Question in the picture

ki77a [65]

The answer is 3030.033.

4 0

4 years ago

-1/5 divided by -1 5/6 in fraction form.

Step2247 [10]

Answer:

6/5

Step-by-step explanation:

-1/5 / -1/6

-1/5 * -6

6/5

Hope this helps!

I would really appreciate if you mark me brainliest as I have been trying to get there for a veryyy long time now :)

5 0

3 years ago