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Sindrei [870]
2 years ago
11

This probability distribution shows the typical grade distribution for a Geometry course with 35 students. Find the probability

that a student earns a grade of A, B, or C. Enter a decimal rounded to the nearest hundredth.

Mathematics
2 answers:
iris [78.8K]2 years ago
5 0

Answer:

.86

Step-by-step explanation:

There are 5+10+15 = 30 students that earned and A, B or C

P( A, B or C) = number of students that earned and A ,B , C / total

                    =30/35

                    =.857142857

   Rounded to the nearest hundredth

                   =.86

iris [78.8K]2 years ago
5 0
  • Total marks=5+10+15+3+2=36
  • No of students earned A,B,C=5+10+15=30

\\ \rm\Rrightarrow P(A,B,C)

\\ \rm\Rrightarrow \dfrac{30}{36}

\\ \rm\Rrightarrow 0.85

\\ \rm\Rrightarrow 0.86

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In the xy-plane, the equations x + 2y = 10 and
konstantin123 [22]

By solving given equations, the value of c is 30.

Given two equations

x + 2y = 10 and

3x + 6y = c

These lines represents the same line for some constant c.

Value of c:

x + 2y = 10-------------(1)

3x + 6y = c-------------(2)

Dividing equation (2) by 3

\frac{3x}{3}+ \frac{6y}{3} =\frac{c}{3}

After solving the above equation, we get

x + 2y = c/3-----------(3)

Remember that a line is written as ax + by = c, in our case, both lines have a =1 and b = 2. Therefore, in orther that the two lines are equal, we need that, 10 = c/3

c = 10 × 3 = 30

c = 30

Therefore,

The value of c is 30.

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2 years ago
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3 years ago
In a genetics experiment on peas, one sample of offspring contained green peas and yellow peas. Based on those results, estimate
dem82 [27]

Complete Question

In a genetic experiment on peas, one sample of offspring contained 436 green peas and 171 yellow peas. Based on those results, estimate the probability of getting an offspring pea that is green. Is the result reasonably close to the value of 3/4 that was expected? The probability of getting a green pea is approximately: Is the probability reasonably close to 3/4?

Answer:

The  probability is  P(g) =0.72

Yes the result is reasonably close

Step-by-step explanation:

From the question we are told that

   The  number of  of  green peas is  g =  436

     The number of yellow peas is  y  = 171

   The sample size is  n  =  171 +  436 =  607

The probability of getting an offspring pea that is green is mathematically represented as

       P(g) = \frac{g}{n}

        P(g) = \frac{436}{607}

        P(g) =0.72

Comparing  P(g) =0.72  to   \frac{3}{4}  =  0.75 we see that the result is reasonably close

7 0
3 years ago
Find the surface area of x^2+y^2+z^2=9 that lies above the cone z= sqrt(x^@+y^2)
Mashcka [7]
The cone equation gives

z=\sqrt{x^2+y^2}\implies z^2=x^2+y^2

which means that the intersection of the cone and sphere occurs at

x^2+y^2+(x^2+y^2)=9\implies x^2+y^2=\dfrac92

i.e. along the vertical cylinder of radius \dfrac3{\sqrt2} when z=\dfrac3{\sqrt2}.

We can parameterize the spherical cap in spherical coordinates by

\mathbf r(\theta,\varphi)=\langle3\cos\theta\sin\varphi,3\sin\theta\sin\varphi,3\cos\varphi\right\rangle

where 0\le\theta\le2\pi and 0\le\varphi\le\dfrac\pi4, which follows from the fact that the radius of the sphere is 3 and the height at which the sphere and cone intersect is \dfrac3{\sqrt2}. So the angle between the vertical line through the origin and any line through the origin normal to the sphere along the cone's surface is

\varphi=\cos^{-1}\left(\dfrac{\frac3{\sqrt2}}3\right)=\cos^{-1}\left(\dfrac1{\sqrt2}\right)=\dfrac\pi4

Now the surface area of the cap is given by the surface integral,

\displaystyle\iint_{\text{cap}}\mathrm dS=\int_{\theta=0}^{\theta=2\pi}\int_{\varphi=0}^{\varphi=\pi/4}\|\mathbf r_u\times\mathbf r_v\|\,\mathrm dv\,\mathrm du
=\displaystyle\int_{u=0}^{u=2\pi}\int_{\varphi=0}^{\varphi=\pi/4}9\sin v\,\mathrm dv\,\mathrm du
=-18\pi\cos v\bigg|_{v=0}^{v=\pi/4}
=18\pi\left(1-\dfrac1{\sqrt2}\right)
=9(2-\sqrt2)\pi
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3 years ago
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