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9966 [12]
3 years ago
15

3/4 divided by 2 and 1/2??? Plssss thank youuu.

Mathematics
1 answer:
natita [175]3 years ago
7 0

Answer:

3/4 divided by 2 1/2 is 0.3

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Given that ∠Ф = 180°, determine the numerical value of this function.
Rudiy27

Answer:

The answer to your question is 3. 0

Step-by-step explanation:

For an angle of 180°:

- hypotenuse = 1

- Adjacent side = 1

- Opposite side = 0

Then, sin Ф = \frac{Opposite side}{Hypotenuse}

         sin 180° = \frac{0}{1}

         sin 180° = 0

8 0
3 years ago
I also need work for understanding on these questions (I will mark as brainliest)
stich3 [128]

Answer:

<em>24000 and 1620 for the first and fifth </em>

Step-by-step explanation:

  • <em>24 x 25 x 40 = 24000</em>
  • <em>9 x 6 x 30 = 1620</em>

<em></em>

<em></em>

<em></em>

  • <em>I only know one and five </em>
  • <em>i couldn't figure out the others</em>
  • <em>Hope this helps</em>
  • <em>The other three i didn't learn yet</em>
  • <em>Brainliest</em>
  • <em>Ask questions if wrong</em>

7 0
3 years ago
I need the answers for 8-11
Amiraneli [1.4K]
The answer is negative three ( -3 )
8 0
3 years ago
Read 2 more answers
My favorite songstress gives away 25% of the sales of her signature fragrance collection to her charity. Last month her sales we
guajiro [1.7K]

Answer:

300

Step-by-step explanation:

4=25%

12=1200

12/4=3

add the zeroes from 1200

300

to make sure...

300x4=1200

5 0
3 years ago
In order to ensure efficient usage of a server, it is necessary to estimate the mean number
juin [17]

Answer:

a. [36.19;39.21]

b. Reject the null hypothesis. The population mean of users that are connected at the same time is greater than 35.

Step-by-step explanation:

Hello!

Your study variable is,

X: "number of users of one server at a time"

The objective is to estimate the mean, for this, a sample of n=100 times was taken and the standard deviation S= 9.2 and the sample mean is X[bar]= 37.7 were calculated.

You need to study the population mean, for this you need your variable to have at least normal distribution. Since you don't have information about its distribution, but the sample is big enough (n≥30) you can apply the Central Limit Theorem and approximate the distribution of the sample mean X[bar] to normal:

X[bar]≈N(μ;σ²/n)

a. With this approximation, you can construct the 90% Confidence Interval using the approximate Z

[X[bar] ± Z_{1-\alpha /2} * S/√n]

Z_{1-\alpha /2} = Z_{0.95} = 1.64

[37.7± 1.64* 9.2/√100]

[36.19;39.21]

b. You need to test if the population mean is greater than 35 with a level of significance of 1%.

The hypothesis is:

H₀: μ ≤ 35

H₁: μ > 35

α: 0.01

This is a one-tailed test so you have only one critical level (right tail):

Z_{1\alpha } = Z_{0.99} = 2.33

This means that if the value of the calculated statistic is equal or greater than 2.33 you will reject the null Hypothesis.

If the value is less than 2.33 you will support the null hypothesis.

The statistic is:

Z=<u> X[bar] - μ </u>= <u> 37.7 - 35 </u> = 2.93

       S/√n           9.2/10

The value 2.93 > 2.33, so you reject the null hypothesis. This means that the population mean of users that are connected at the same time is greater than 35.

<u><em>Note: </em></u><em>To make the decision using the interval calculated on a), the hypothesis should have been two-tailed and the confidence and significance levels complementary.</em>

I hope it helps!

7 0
3 years ago
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