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astraxan [27]
3 years ago
10

Tan 2t=0 on an interval [0,2pi) solve and simplify

Mathematics
1 answer:
sasho [114]3 years ago
3 0
There an infinite number of solutions that can be generated by giving different values
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PLEASE HELP WILL GIVE BRANLIEST pythagorean theorem
nikklg [1K]

Answer:

It is easy...

replace the value...

like 1st. ..

3^{2} + 4^{2}  =   {5}^{2} \\ or \: 9 + 16 = 25 \\or \:  25 = 25 \\ proved

5 0
3 years ago
At yesterdays ball game, 2,842 tickets were sold. The price of each ticket was $7.50. Which amount is closest to the total amoun
Margarita [4]

Answer:

C. $20,000

Step-by-step explanation:

2,824*$7.50=$21,315

6 0
3 years ago
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Assume that weights of adult females are normally distributed with a mean of 79 kg and a standard deviation of 22 kg. What perce
LenKa [72]

Answer:

14.28% of individual adult females have weights between 75 kg and 83 ​kg.

92.82% of the sample means are between 75 kg and 83 ​kg.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

Assume that weights of adult females are normally distributed with a mean of 79 kg and a standard deviation of 22 kg. This means that \mu = 79, \sigma = 22.

What percentage of individual adult females have weights between 75 kg and 83 ​kg?

This percentage is the pvalue of Z when X = 83 subtracted by the pvalue of Z when X = 75. So:

X = 83

Z = \frac{X - \mu}{\sigma}

Z = \frac{83 - 79}{22}

Z = 0.18

Z = 0.18 has a pvalue of 0.5714.

X = 75

Z = \frac{X - \mu}{\sigma}

Z = \frac{75- 79}{22}

Z = -0.18

Z = -0.18 has a pvalue of 0.4286.

This means that 0.5714-0.4286 = 0.1428 = 14.28% of individual adult females have weights between 75 kg and 83 ​kg.

If samples of 100 adult females are randomly selected and the mean weight is computed for each​ sample, what percentage of the sample means are between 75 kg and 83 ​kg?

Now we use the Central Limit THeorem, when n = 100. So s = \frac{22}{\sqrt{100}} = 2.2.

X = 83

Z = \frac{X - \mu}{s}

Z = \frac{83 - 79}{2.2}

Z = 1.8

Z = 1.8 has a pvalue of 0.9641.

X = 75

Z = \frac{X - \mu}{s}

Z = \frac{75-79}{2.2}

Z = -1.8

Z = -1.8 has a pvalue of 0.0359.

This means that 0.9641-0.0359 = 0.9282 = 92.82% of the sample means are between 75 kg and 83 ​kg.

8 0
3 years ago
<img src="https://tex.z-dn.net/?f=2x%20%3D%2044" id="TexFormula1" title="2x = 44" alt="2x = 44" align="absmiddle" class="latex-f
FrozenT [24]

Answer:

<h2>4</h2>

Step-by-step explanation:

2x = 44

=  > x =  \frac{44}{2}

=  > x = 22

Now if the 2 at the end is an power then it will turn to

22 =   {2}^{2}

= 2 \times 2

= 4

So the answer will be 4.

5 0
3 years ago
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A pie has a diameter of 9 inches. What is the circumference of the pie
IgorC [24]

Answer:

pretty sure it is 56.55  

Step-by-step explanation:

4 0
3 years ago
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