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3 years ago

11

I go to school on the 4th of June Monday and i summer term is on 29th June i have 3 days a week at school Monday Tuesday Wednesd

ay
how meany of my schooldays are left

1 answer:

Virty [35]3 years ago

3 0

11 school days are left.

In the attached picture of a calendar, I marked the days the person has left. There are 11.

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Question 20: Please help what are sin x and cos x?

sweet [91]

Answer:

A is the correct answer.

Step-by-step explanation:

$\tan x = \frac{6}{8} = \frac{opposite \: side \: to \: \angle \: x}{adjacent \: side \: to \: \angle \: x} \\ \\ hypotenuse \\ = \sqrt{ {(opp. \: side)}^{2} + {(adj. \: side)}^{2} } \\ = \sqrt{ {6}^{2} + {8}^{2} } \\ = \sqrt{36 + 64} \\ = \sqrt{100} \\ = 10 \\ \\ \because \sin (x) = \frac{opposite \: side \: to \: \angle \: x}{hypotenuse} \\ \\ \huge{ \red{ \boxed{\therefore \: \sin (x) = \frac{6}{10}}}} \\ \\ \cos (x) = \frac{adjacent \: side \: to \: \angle \: x}{hypotenuse} \\ \\ \huge{ \purple{ \boxed{\therefore \: \cos (x) = \frac{8}{10} }}} \\$

Thus, option A is the correct answer.

8 0

3 years ago

The sum of twice a number and three equals five. Find the number

Anna71 [15]

2n + 3 = 5
2n = 2
n = 1
the sum of 2 times a number and three means you add 2n + 3 and since it is equal to 5 you then solve for n.

6 0

3 years ago

Read 2 more answers

What is the p-value? -- Researcher Jessie is studying how the fear of going to the dentist affects an adult's actual number of v

Vesnalui [34]

Answer:

Two, one for the 14 responses (number of visits) by the adults who fear going to the dentist and one for the 31 responses (number of visits) by the adults who do not fear going to the dentist.

Step-by-step explanation:

Hello!

1)

You want to test if the average visits to the dentist of people who fear to visit it are greater than the average visits of people that don't fear it.

In this case, the statistic to use is a pooled Student t-test. The reason I've to choose this test is that one of your sample sizes is small (n₁= 14) and the t-test is more accurate for small samples. Even if the second sample is greater than 30, if both variables are normally distributed, the pooled t-test is the one to use.

H₀: μ₁ = μ₂

H₁: μ₁ > μ₂

α: 0.10

t=<u> (X₁[bar]-X₂[bar]) - (μ₁ - μ₂)</u> ~ t $_{n₁+n₂-2}$

Sₐ√(1/n₁+1/n₂)

Where

X₁[bar] and X₂[bar] are the sample means of both groups

Sₐ is the pooled standard deviation

This is a one-tailed test, you will reject the null hypothesis to big numbers of t. Remember: The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis), and in this case, is also one-tailed.

P(t $_{n₁+n₂-2}$ ≥ t $_{H0}$ ) = 1 - P(t $_{n₁+n₂-2}$ < t $_{H0}$ )

Where t $_{H0}$ is the value of the calculated statistic.

Since you didn't copy the data of both samples, I cannot calculate it.

2)

Well there was one sample taken and separated in two following the criteria "fears the dentist" and "doesn't fear the dentist" making two different samples, so this is a test for two independent samples. To check if both variables are normally distributed you need to make two QQplots.

I hope it helps!

3 0

3 years ago

Alex took 31 exams for 5 years studying at the MaPhAs University. Each year, he took more exams than the previous year. During h

katen-ka-za [31]

Answer:

The fourth year he took 8 exams

Step-by-step explanation:

We must do it by trial and error:

We must start from the first year:

A = 1 exam

If the first year he took 1 exam, the last year he did 3, but this answer is not possible because the rule that the more the years pass, the more exams he does, is not fulfilled.

A = 2 exams

In this case, if it is true that each year increases, but it does not meet the total rule, since:

Year 1 = 2

Year 2 = 3

Year 3 = 4

Year 4 = 5

Year 5 = 6

2 + 3 + 4 + 5 + 6 = 20

A = 3 exams

Therefore the fifth year would be 9, therefore it would be:

Year 1 = 3

Year 2 = 4

Year 3 = 5

Year 4 = 6

Year 5 = 9

3 + 4 + 5 + 6 + 9 = 27

This means that we are missing 4 units to reach 31, therefore we must add 4 units between years 2, 3 and 4, but always bearing in mind that each year should increase with respect to the previous one.

If we add 2 exams to year 4, 1 exams to year 3 and 1 exam to year 2, we would have the following results:

Year 1 = 3

Year 2 = 5

Year 3 = 6

Year 4 = 8

Year 5 = 9

3 + 5 + 6 + 8 + 9 = 31

Which means that in the fourth year he took 8 exams

8 0

3 years ago

Some one please me a-e

Akimi4 [234]

Answer:

a) yes

b)yes

c)no

d)no

e) cannot see it properly

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers