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3 years ago

How do I solve this equation

Mathematics

1 answer:

notka56 [123]3 years ago

6 0

Hello there!

$\frac{x}{y}=h-k$

Explanation:

↓↓↓↓↓↓↓↓

First you had to multiply by y from both sides of the equations.

$\frac{xy}{y}=hy-ky; y\neq0$

Simplify it should be the correct answer.

$x=hy-ky; y\neq0$

Answer⇒⇒⇒⇒⇒x=hy-ky; y≠0

Hope this helps!

Thank you for posting your question at here on Brainly.

Have a great day!

-Charlie

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the lateral surface area of right circular cylinder is 120 pi cm squared and the circumference is 12 cm find the height

konstantin123 [22]

Answer:

h=31.4 cm

Step-by-step explanation:

A=lateral surface area

c=circumference

h=height

A= c*h

solving for h we get:

h=A/c=120pi/12=10pi

6 0

3 years ago

Read 2 more answers

Let X denote the length of human pregnancies from conception to birth, where X has a normal distribution with mean of 264 days a

Kaylis [27]

Answer:

Step-by-step explanation:

Hello!

X: length of human pregnancies from conception to birth.

X~N(μ;σ²)

μ= 264 day

σ= 16 day

If the variable of interest has a normal distribution, it's the sample mean, that it is also a variable on its own, has a normal distribution with parameters:

X[bar] ~N(μ;σ²/n)

When calculating a probability of a value of "X" happening it corresponds to use the standard normal: Z= (X[bar]-μ)/σ

When calculating the probability of the sample mean taking a given value, the variance is divided by the sample size. The standard normal distribution to use is Z= (X[bar]-μ)/(σ/√n)

a. You need to calculate the probability that the sample mean will be less than 260 for a random sample of 15 women.

P(X[bar]<260)= P(Z<(260-264)/(16/√15))= P(Z<-0.97)= 0.16602

b. P(X[bar]>b)= 0.05

You need to find the value of X[bar] that has above it 5% of the distribution and 95% below.

P(X[bar]≤b)= 0.95

P(Z≤(b-μ)/(σ/√n))= 0.95

The value of Z that accumulates 0.95 of probability is Z= 1.648

Now we reverse the standardization to reach the value of pregnancy length:

1.648= (b-264)/(16/√15)

1.648*(16/√15)= b-264

b= [1.648*(16/√15)]+264

b= 270.81 days

c. Now the sample taken is of 7 women and you need to calculate the probability of the sample mean of the length of pregnancy lies between 1800 and 1900 days.

Symbolically:

P(1800≤X[bar]≤1900) = P(X[bar]≤1900) - P(X[bar]≤1800)

P(Z≤(1900-264)/(16/√7)) - P(Z≤(1800-264)/(16/√7))

P(Z≤270.53) - P(Z≤253.99)= 1 - 1 = 0

d. P(X[bar]>270)= 0.1151

P(Z>(270-264)/(16/√n))= 0.1151

P(Z≤(270-264)/(16/√n))= 1 - 0.1151

P(Z≤6/(16/√n))= 0.8849

With the information of the cumulated probability you can reach the value of Z and clear the sample size needed:

P(Z≤1.200)= 0.8849

$Z= \frac{X[bar]-Mu}{Sigma/\sqrt{n} }$

$Z*(Sigma/\sqrt{n} )= (X[bar]-Mu)$

$(Sigma/\sqrt{n} )= \frac{(X[bar]-Mu)}{Z}$

$Sigma= \frac{(X[bar]-Mu)}{Z}*\sqrt{n}$

$Sigma*(\frac{Z}{(X[bar]-Mu)})= \sqrt{n}$

$n = (Sigma*(\frac{Z}{(X[bar]-Mu)}))^2$

$n = (16*(\frac{1.2}{(270-264)}))^2$

n= 10.24 ≅ 11 pregnant women.

I hope it helps!

6 0

3 years ago

5 milligrams equals how many grams

katovenus [111]

5 milligrams are equal to .005 grams.

4 0

4 years ago

Read 2 more answers

At what point does the curve have maximum curvature? Y = 4ex (x, y) = what happens to the curvature as x → ∞? Κ(x) approaches as

MAXImum [283]

Answer-

At $x= \frac{1}{2304e^4-16e^2}$ the curve has maximum curvature.

Solution-

The formula for curvature =

$K(x)=\frac{{y}''}{(1+({y}')^2)^{\frac{3}{2}}}$

Here,

$y=4e^{x}$

Then,

${y}' = 4e^{x} \ and \ {y}''=4e^{x}$

Putting the values,

$K(x)=\frac{{4e^{x}}}{(1+(4e^{x})^2)^{\frac{3}{2}}} = \frac{{4e^{x}}}{(1+16e^{2x})^{\frac{3}{2}}}$

Now, in order to get the max curvature value, we have to calculate the first derivative of this function and then to get where its value is max, we have to equate it to 0.

${k}'(x) = \frac{(1+16e^{2x})^{\frac{3}{2} } (4e^{x})-(4e^{x})(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})}{(1+16e^{2x} )^{2}}$