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solong [7]
3 years ago
15

Waht is 8/5 times 7/4 pleeaaaaseeee help LAST QUESTION PLZ

Mathematics
1 answer:
saveliy_v [14]3 years ago
8 0

Answer:

below

Step-by-step explanation:

improper fraction: 14/5

as a decimal: 2.8

proper fraction: 2 4/5

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Sonbull [250]

The perimeter is 12x+10

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What's 10 percent off of 10.35
MissTica

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I need help on this plz ​
Shalnov [3]

Answer:

x = 3

Step-by-step explanation:

You can use cross multiply.

8x = 6 x 4

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7 0
3 years ago
Read 2 more answers
Consider the probabilities of people taking pregnancy tests. Assume that the true probability of pregnancy for all people who ta
Valentin [98]

Using conditional probability, it is found that there is a 0.8462 = 84.62% probability that a woman who gets a positive test result is truly pregnant.

<h3>What is Conditional Probability?</h3>

Conditional probability is the probability of one event happening, considering a previous event. The formula is:

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

  • P(B|A) is the probability of event B happening, given that A happened.
  • P(A \cap B) is the probability of both A and B happening.
  • P(A) is the probability of A happening.

In this problem, the events are:

  • Event A: Positive test result.
  • Event B: Pregnant.

The probability of a positive test result is composed by:

  • 99% of 10%(truly pregnant).
  • 2% of 90%(not pregnant).

Hence:

P(A) = 0.99(0.1) + 0.02(0.9) = 0.117

The probability of both a positive test result and pregnancy is:

P(A \cap B) = 0.99(0.1)

Hence, the conditional probability is:

P(B|A) = \frac{0.99(0.1)}{0.117} = 0.8462

0.8462 = 84.62% probability that a woman who gets a positive test result is truly pregnant.

You can learn more about conditional probability at brainly.com/question/14398287

7 0
2 years ago
Find the area of the surface. The part of the sphere x2 + y2 + z2 = a2 that lies within the cylinder x2 + y2 = ax and above the
sineoko [7]

Answer:

The area of the sphere in the cylinder and which locate above the xy plane is \mathbf{ a^2 ( \pi -2)}

Step-by-step explanation:

The surface area of the sphere is:

\int \int \limits _ D \sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )   } \ dA

and the cylinder x^2 + y^2 =ax can be written as:

r^2 = arcos \theta

r = a cos \theta

where;

D = domain of integration which spans between \{(r, \theta)| - \dfrac{\pi}{2} \leq \theta  \leq \dfrac{\pi}{2}, 0 \leq r \leq acos \theta\}

and;

the part of the sphere:

x^2 + y^2 + z^2 = a^2

making z the subject of the formula, then :

z = \sqrt{a^2 - (x^2 +y^2)}

Thus,

\dfrac{\partial z}{\partial x} = \dfrac{-2x}{2 \sqrt{a^2 - (x^2+y^2)}}

\dfrac{\partial z}{\partial x} = \dfrac{-x}{ \sqrt{a^2 - (x^2+y^2)}}

Similarly;

\dfrac{\partial z}{\partial y} = \dfrac{-2y}{2 \sqrt{a^2 - (x^2+y^2)}}

\dfrac{\partial z}{\partial y} = \dfrac{-y}{ \sqrt{a^2 - (x^2+y^2)}}

So;

\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )}  = \sqrt{\begin {pmatrix} \dfrac{-x}{\sqrt{a^2 -(x^2+y^2)}} \end {pmatrix}^2 + \begin {pmatrix} \dfrac{-y}{\sqrt{a^2 - (x^2+y^2)}}   \end {pmatrix}^2+1}\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )}  = \sqrt{\dfrac{x^2+y^2}{a^2 -(x^2+y^2)}+1}

\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )}  = \sqrt{\dfrac{x^2+y^2+a^2 -(x^2+y^2)}{a^2 -(x^2+y^2)}}

\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )}  = \sqrt{\dfrac{a^2}{a^2 -(x^2+y^2)}}

\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )}  = {\dfrac{a}{\sqrt{a^2 -(x^2+y^2)}}

From cylindrical coordinates; we have:

\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )}  = {\dfrac{a}{\sqrt{a^2 -r^2}}

dA = rdrdθ

By applying the symmetry in the x-axis, the area of the surface will be:

A = \int \int _D \sqrt{ (\dfrac{\partial z}{\partial x})^2+ (\dfrac{\partial z}{\partial y})^2+1} \ dA

A = \int^{\dfrac{\pi}{2}}_{-\dfrac{\pi}{2}} \int ^{a cos \theta}_{0} \dfrac{a}{\sqrt{a^2 -r^2 }} \ rdrd \theta

A = 2\int^{\dfrac{\pi}{2}}_{0} \begin {bmatrix} -a \sqrt{a^2 -r^2} \end {bmatrix}^{a cos \theta}_0 \ d \theta

A = 2\int^{\dfrac{\pi}{2}}_{0} \begin {bmatrix} -a \sqrt{a^2 - a^2cos^2 \theta} + a \sqrt{a^2 -0}} \end {bmatrix} d \thetaA = 2\int^{\dfrac{\pi}{2}}_{0} \begin {bmatrix} -a \ sin \theta +a^2 } \end {bmatrix} d \theta

A = 2a^2 [ cos \theta + \theta ]^{\dfrac{\pi}{2} }_{0}

A = 2a^2 [ cos \dfrac{\pi}{2}+ \dfrac{\pi}{2} - cos (0)- (0)]

A = 2a^2 [0 + \dfrac{\pi}{2}-1+0]

A = a^2 \pi - 2a^2

\mathbf{A = a^2 ( \pi -2)}

Therefore, the area of the sphere in the cylinder and which locate above the xy plane is \mathbf{ a^2 ( \pi -2)}

6 0
3 years ago
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