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4 years ago

Select the correct answer from each drop-down menu.

Mathematics

1 answer:

REY [17]4 years ago

6 0

Answer:

Try to re write this question so I can help. I can not help without the answer choices

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Pls help me with these math problems

user100 [1]

2. W<5
4. D<6 (or equal to)

3 0

4 years ago

Read 2 more answers

Please anyone help me on this question

vazorg [7]

Answer:

a) 60

b) 0.24

c) 150

Step-by-step explanation:

a) P(green) = ⅕

Expected no. = 300 × ⅕

= 60

b) total = 300

P(purple) = 72/300

= 6/25 = 0.24

c) P(blue) = 45/300

= 3/20 = 0.15

Expected no.: 1000 × 0.15

= 150

5 0

3 years ago

2/7 • 9 1/10 Please help

nordsb [41]

$\frac{2}{7} * 9 \frac{1}{10} = \frac{2}{7} * \frac{91}{10} = \frac{2~*~91}{7~*~10} = \frac{182}{70} = 2 \frac{42}{70} = 2 \frac{3}{5}$

3 0

3 years ago

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R-18=27 Solve this equation

Flauer [41]

Step-by-step explanation:

r - 18 = 27

r = 27 + 18

r = 45

...

6 0

3 years ago

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A group of researchers are interested in the possible effects of distracting stimuli during eating, such as an increase or decre

Dmitry [639]

Using the t-distribution, it is found that since the p-value of the test is of 0.0302, which is less than the standard significance level of 0.05, the data provides convincing evidence that the average food intake is different for the patients in the treatment group.

At the null hypothesis, it is tested if the consumption is not different, that is, if the subtraction of the means is 0, hence:

$H_0: \mu_1 - \mu_2 = 0$

At the alternative hypothesis, it is tested if the consumption is different, that is, if the subtraction of the means is not 0, hence:

$H_1: \mu_1 - \mu_2 \neq 0$

Two groups of 22 patients, hence, the standard errors are:

$s_1 = \frac{45.1}{\sqrt{22}} = 9.6154$

$s_2 = \frac{26.4}{\sqrt{22}} = 5.6285$

The distribution of the differences is has:

$\overline{x} = \mu_1 - \mu_2 = 52.1 - 27.1 = 25$

$s = \sqrt{s_1^2 + s_2^2} = \sqrt{9.6154^2 + 5.6285^2} = 11.14$

The test statistic is given by:

$t = \frac{\overline{x} - \mu}{s}$

In which $\mu = 0$ is the value tested at the null hypothesis.

Hence:

$t = \frac{25 - 0}{11.14}$

$t = 2.2438$

The p-value of the test is found using a two-tailed test, as we are testing if the mean is different of a value, with t = 2.2438 and 22 + 22 - 2 = 42 df.

Using a t-distribution calculator, this p-value is of 0.0302.

Since the p-value of the test is of 0.0302, which is less than the standard significance level of 0.05, the data provides convincing evidence that the average food intake is different for the patients in the treatment group.

A similar problem is given at brainly.com/question/25600813

4 0

3 years ago