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Reil [10]
2 years ago
6

I need help finding how to get ABCD to QRST

Mathematics
1 answer:
Margarita [4]2 years ago
5 0

mgerijdi iskl tsl ifjb idvjk ufjk kd on ifd

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If the hemisphers has a radius of 3 feet, how many cubic feet of water can the water tank hold?
Eduardwww [97]

Answer:

volume \ of \ hemisphere = \frac{2}{3} \pi r^3 = \frac{2}{3}\pi \cdot (3)^3  = \frac{2}{3} \pi 27 = 2\pi \cdot9 = 18\pi cubicfeet

8 0
3 years ago
The time that it takes a randomly selected job applicant to perform a certain task has a distribution that can be approximated b
zepelin [54]

Answer:

If the task is performed in less than or equal to 130.8 seconds, then, the individuals qualify for advanced training.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 150 sec

Standard Deviation, σ = 15 sec

We are given that the distribution of time taken is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

We have to find the value of x such that the probability is 0.10

P(X < x)  

P( X < x) = P( z < \displaystyle\frac{x - 150}{15})=0.10  

Calculation the value from standard normal z table, we have,  

\displaystyle\frac{x - 150}{15} = -1.282\\x = 130.8  

Thus, if the task is performed in less than or equal to 130.8 seconds, then, the individuals qualify for advanced training.

8 0
3 years ago
Calculus, question 5 to 5a​
Llana [10]

5. Let x = \sin(\theta). Note that we want this variable change to be reversible, so we tacitly assume 0 ≤ θ ≤ π/2. Then

\cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - x^2}

and dx = \cos(\theta) \, d\theta. So the integral transforms to

\displaystyle \int \frac{x^3}{\sqrt{1-x^2}} \, dx = \int \frac{\sin^3(\theta)}{\cos(\theta)} \cos(\theta) \, d\theta = \int \sin^3(\theta) \, d\theta

Reduce the power by writing

\sin^3(\theta) = \sin(\theta) \sin^2(\theta) = \sin(\theta) (1 - \cos^2(\theta))

Now let y = \cos(\theta), so that dy = -\sin(\theta) \, d\theta. Then

\displaystyle \int \sin(\theta) (1-\cos^2(\theta)) \, d\theta = - \int (1-y^2) \, dy = -y + \frac13 y^3 + C

Replace the variable to get the antiderivative back in terms of x and we have

\displaystyle \int \frac{x^3}{\sqrt{1-x^2}} \, dx = -\cos(\theta) + \frac13 \cos^3(\theta) + C

\displaystyle \int \frac{x^3}{\sqrt{1-x^2}} \, dx = -\sqrt{1-x^2} + \frac13 \left(\sqrt{1-x^2}\right)^3 + C

\displaystyle \int \frac{x^3}{\sqrt{1-x^2}} \, dx = -\frac13 \sqrt{1-x^2} \left(3 - \left(\sqrt{1-x^2}\right)^2\right) + C

\displaystyle \int \frac{x^3}{\sqrt{1-x^2}} \, dx = \boxed{-\frac13 \sqrt{1-x^2} (2+x^2) + C}

6. Let x = 3\tan(\theta) and dx=3\sec^2(\theta)\,d\theta. It follows that

\cos(\theta) = \dfrac1{\sec(\theta)} = \dfrac1{\sqrt{1+\tan^2(\theta)}} = \dfrac3{\sqrt{9+x^2}}

since, like in the previous integral, under this reversible variable change we assume -π/2 < θ < π/2. Over this interval, sec(θ) is positive.

Now,

\displaystyle \int \frac{x^3}{\sqrt{9+x^2}} \, dx = \int \frac{27\tan^3(\theta)}{\sqrt{9+9\tan^2(\theta)}} 3\sec^2(\theta) \, d\theta = 27 \int \frac{\tan^3(\theta) \sec^2(\theta)}{\sqrt{1+\tan^2(\theta)}} \, d\theta

The denominator reduces to

\sqrt{1+\tan^2(\theta)} = \sqrt{\sec^2(\theta)} = |\sec(\theta)| = \sec(\theta)

and so

\displaystyle 27 \int \tan^3(\theta) \sec(\theta) \, d\theta = 27 \int \frac{\sin^3(\theta)}{\cos^4(\theta)} \, d\theta

Rewrite sin³(θ) just like before,

\displaystyle 27 \int \frac{\sin(\theta) (1-\cos^2(\theta))}{\cos^4(\theta)} \, d\theta

and substitute y=\cos(\theta) again to get

\displaystyle -27 \int \frac{1-y^2}{y^4} \, dy = 27 \int \left(\frac1{y^2} - \frac1{y^4}\right) \, dy = 27 \left(\frac1{3y^3} - \frac1y\right) + C

Put everything back in terms of x :

\displaystyle \int \frac{x^3}{\sqrt{9+x^2}} \, dx = 9 \left(\frac1{\cos^3(\theta)} - \frac3{\cos(\theta)}\right) + C

\displaystyle \int \frac{x^3}{\sqrt{9+x^2}} \, dx = 9 \left(\frac{\left(\sqrt{9+x^2}\right)^3}{27} - \sqrt{9+x^2}\right) + C

\displaystyle \int \frac{x^3}{\sqrt{9+x^2}} \, dx = \boxed{\frac13 \sqrt{9+x^2} (x^2 - 18) + C}

2(b). For some constants a, b, c, and d, we have

\dfrac1{x^2+x^4} = \dfrac1{x^2(1+x^2)} = \boxed{\dfrac ax + \dfrac b{x^2} + \dfrac{cx+d}{x^2+1}}

3(a). For some constants a, b, and c,

\dfrac{x^2+4}{x^3-3x^2+2x} = \dfrac{x^2+4}{x(x-1)(x-2)} = \boxed{\dfrac ax + \dfrac b{x-1} + \dfrac c{x-2}}

5(a). For some constants a-f,

\dfrac{x^5+1}{(x^2-x)(x^4+2x^2+1)} = \dfrac{x^5+1}{x(x-1)(x+1)(x^2+1)^2} \\\\ = \dfrac{x^4 - x^3 + x^2 - x + 1}{x(x-1)(x^2+1)^2} = \boxed{\dfrac ax + \dfrac b{x-1} + \dfrac{cx+d}{x^2+1} + \dfrac{ex+f}{(x^2+1)^2}}

where we use the sum-of-5th-powers identity,

a^5 + b^5 = (a+b) (a^4-a^3b+a^2b^2-ab^3+b^4)

4 0
2 years ago
An index that is a standardized measure used in observing infants over time is approximately normal with a mean of 90 and a stan
weqwewe [10]

Answer:

The proportion of children that have an index of at least 110 is 0.0478.

Step-by-step explanation:

The given distribution has a mean of 90 and a standard deviation of 12.

Therefore mean, \mu = 90 and standard deviation, \sigma = 12.

It is given to find the proportion of children having an index of at least 110.

We can take the variable to be analysed to be x = 110.

Therefore we have to find p(x < 110), which is left tailed.

Using the formula for z which is p( Z < \frac{x - \mu}{\sigma}) we get p(Z < \frac{110 - 90}{12} = 1.67).

So we have to find p(Z ≥ 1.67) = 1 - p(Z < 1.67)

Using the Z - table we can calculate p(Z < 1.67)  = 0.9522.

Therefore p(Z ≥ 1.67) = 1 - 0.9522 = 0.0478

Therefore the proportion of children that have an index of at least 110 is 0.0478

8 0
3 years ago
Jonah owned a pizzeria he made 14 12in pizza's for Sallies party. sally and her friends ate 132 in of the pizzas, how much many
shtirl [24]
They ate 11 because 132/12=11
6 0
2 years ago
Read 2 more answers
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