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andreyandreev [35.5K]
3 years ago
6

What is your favorite subject? ​

Mathematics
2 answers:
kakasveta [241]3 years ago
7 0

Answer: My fav subject is math because math is easy for me.

PIT_PIT [208]3 years ago
4 0

Answer:

Math and Biology

Step-by-step explanation:

I just find these subjects fascinating

You might be interested in
If sin Θ = 2 over 3 and tan Θ < 0, what is the value of cos Θ?
Scorpion4ik [409]
Sinθ = (2/3)    tanθ < 0  Implies tanθ = Negative.

For sin positive and tan negative, implies θ is in the second quadrant.

<span>Sinθ = (2/3) = opposite / Hypotenuse.

Adjacent side:

x</span>² + 2² = 3²

x² + 4 = 9<span>

</span>x<span>² = 9 - 4
</span>
x<span>² = 5
</span><span>
x = </span>√5 = Adjacent.
<span>
cos</span>θ  =  Adjacent / Hypotenuse =  √5 / 3
<span>
Since </span><span>θ is in the second quadrant, cosine is also negative in the second quadrant.
</span><span>

Therefore:

</span>cos<span>θ  =  - (√5/3)
</span>
The last option from the top :  <span>negative square root of 5 over 3</span>
6 0
3 years ago
Simplify 3/x²+14x+48 divided by 3/10x+60
stira [4]
C. 10/x+8
The easy way to do it would be to take it one step at a time.
8 0
3 years ago
Read 2 more answers
Since 1900, the magnitude of earthquakes that measure 0.1 or higher on the Richter Scale in CA are distributed normally with a m
Gwar [14]

Answer:

a) 3.59% probability that a randomly selected earthquake in CA has a magnitude greater than 7.1

b) 1.39% probability that a randomly selected earthquake in CA has a magnitude less than 5.1

c) 73.57% probability that ten randomly selected earthquakes in CA have mean magnitude greater than 6.1

d) 99.92% probability that ten randomly selected earthquakes in CA have mean magnitude between 5.7 and 7.22

e) 6.0735

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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.

Central Limit Theorem:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 6.2, \sigma = 0.5

a.) What is the probability that a randomly selected earthquake in CA has a magnitude greater than 7.1?

This is 1 subtracted by the pvalue of Z when X = 7.1. So

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

Z = \frac{7.1 - 6.2}{0.5}

Z = 1.8

Z = 1.8 has a pvalue of 0.9641

1 - 0.9641 = 0.0359

3.59% probability that a randomly selected earthquake in CA has a magnitude greater than 7.1

b.) What is the probability that a randomly selected earthquake in CA has a magnitude less than 5.1?

This is the pvalue of Z when X = 5.1. So

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

Z = \frac{5.1 - 6.2}{0.5}

Z = -2.2

Z = -2.2 has a pvalue of 0.0139

1.39% probability that a randomly selected earthquake in CA has a magnitude less than 5.1

c.) What is the probability that ten randomly selected earthquakes in CA have mean magnitude greater than 6.1?

Now n = 10, s = \frac{0.5}{\sqrt{10}} = 0.1581

This is 1 subtracted by the pvalue of  when X = 6.1. So

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

By the Central Limit Theorem

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

Z = \frac{6.1 - 6.2}{0.1581}

Z = -0.63

Z = -0.63 has a pvalue of 0.2643

1 - 0.2643 = 0.7357

73.57% probability that ten randomly selected earthquakes in CA have mean magnitude greater than 6.1

d.) What is the probability that a ten randomly selected earthquakes in CA have mean magnitude between 5.7 and 7.22

This is the pvalue of Z when X = 7.22 subtracted by the pvalue of Z when X = 5.7. So

X = 7.22

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

Z = \frac{7.22 - 6.2}{0.1581}

Z = 6.45

Z = 6.45 has a pvalue of 1

X = 5.7

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

Z = \frac{5.7 - 6.2}{0.1581}

Z = -3.16

Z = -3.16 has a pvalue of 0.0008

1 - 0.0008 = 0.9992

99.92% probability that ten randomly selected earthquakes in CA have mean magnitude between 5.7 and 7.22

e.) Determine the 40th percentile of the magnitude of earthquakes in CA.

This is X when Z has a pvalue of 0.4. So it is X when Z = -0.253.

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

-0.253 = \frac{X - 6.2}{0.5}

X - 6.2 = -0.253*0.5

X = 6.0735

6 0
3 years ago
Simplify root 32-6 divided by root 2 plus root 2​
Alex777 [14]

Answer:

5•362165924

Step-by-step explanation:

first make root of 32-6=5•099019514

then make root of 2+root2=1•84--

then divide upper by lower part answer comes

or

root32-6=root26

root 2+root 2=2root2

root26/root2root2

ans=3•0318---

7 0
3 years ago
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Is 100,60,50 a right triangle
zzz [600]

Answer:

I dont think so because a right triangle should have a 90 degree

7 0
4 years ago
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